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A triangle has sides of length 6 cm and 8 cm. The angle between the two known sides is 75°. Find the length of the unknown side. [c ≈ 8.669104433…]
A triangle has sides of length 14, 19, and 27 units. What is the area of each defect on the two smaller sides? [The area of the square on the largest side is 272 = 729 square units. Subtracting the other two areas (142 + 192)yields 172 square units, meaning that the area of each defect is 86 square units.]
A triangle has sides of length 9 feet, 13 feet, and 17 feet. How big is the angle opposite the side that is 17 feet long? How big is the angle opposite the side that is 9 feet long? Find the size of the angle opposite the side of length 13 feet without using the law of cosines. [C ≈ 99.594° and A ≈ 31.467°. The measure of B=180 – 99.594 – 31.467 = 48.939°.]
Determine AC if AB = 10, BC = 7, and m∠A = 20°. [The law of cosines can be used, once it is recognized that ∠A is opposite side BC. When substituted, a quadratic equation results: 72 = x2+102 – 2x(10 cos 20), which yields x = 15.5044 or x = 3.2894. If your students have already studied the ambiguous triangle case using the law of sines, the law of cosines is a good alternative for quickly seeing both cases.]
To assess the level of student comprehension, ask students to create two different questions involving the ideas studied. Request one basic question and one advanced question. Ask students to submit their problems before leaving class as an "exit card." Review the level of questions to determine the level of student understanding.
Extensions
Ask students to consider the implications when the sum of the squares on the smaller sides of a triangle is larger than the square on the longest side.
In this lesson, students learned that the defect areas must be subtracted if the triangle is acute but added if the triangle is obtuse. Ask students to explain why the same law of cosines formula works for both acute and obtuse triangles, but you don't change subtraction to addition in the formula!
Teacher Reflection
Was students' level of enthusiasm and involvement high or low? Explain why.
Was your lesson developmentally appropriate? If not, what was inappropriate? What can you do to change it?
What problems did students encounter when calculating the areas of the defects? How could you change the presentation of this lesson to increase student understanding regarding the area of the defects?
Question: What is the formula to calculate the area of the defects in a triangle? Answer: Area of defect = (a^2 + b^2 - c^2) / 2, where a and b are the lengths of the two smaller sides, and c is the length of the longest side.
Question: What is the length of AC if AB = 10, BC = 7, and m∠A = 20°? Answer: Approximately 15.504 cm or 3.289 cm (depending on the case)
Question: What is the sum of the squares of the smaller sides of a triangle? Answer: 100 (since 10^2 + 7^2 = 100) | 677.169 | 1 |
I do not think it is likely because it is impossible to trisect an angle exactly in geometry. Since you can not do that I do not know how you could make a construction of right triangles that would lead to a triangle with exactly 1/3 of the original angle in it.
Question: What is the relationship between trisecting an angle and constructing right triangles? Answer: The text suggests that if you can't trisect an angle, you can't construct right triangles that would lead to a triangle with exactly 1/3 of the original angle. | 677.169 | 1 |
You can put this solution on YOUR website! Your question is a bit confusing, but here's some about the coordinate systems:
Cartesian are square like. Pick an arbitrary point and call that 0,0,0. From there, you can go left / right, up / down, or front / back. The amount of each would be the directions of x , y , z. Generally, + x is right, + y is up, and +z is coming toward you.
Polar coordinates also need an arbitrary point, but deal with that point in relation to a sphere. It needs a r, theta, and phi. The r is the distance from that point (like a radius), the theta is the angle with respect to a line cutting that point, where 0 degrees is generally due right of the picture. Phi does the same concept, but perpendicular to theta.
Question: In which coordinate system do you start by picking an arbitrary point and calling it 0,0,0? Answer: Both Cartesian and polar coordinate systems | 677.169 | 1 |
Rhombohedral
In crystallography, the rhombohedral (or trigonal) crystal system is one of the 7 lattice point groups. A crystal system is described by three basis vectors. In the rhombohedral system, the crystal is described by vectors of equal length, all three of which are not mutually orthogonal. The rhombohedral system can be thought of as the cubic system stretched along a body diagonal.
In some classification systems, the rhombohedral system is grouped into a larger hexagonal
Question: Which crystal system can the rhombohedral system be thought of as a stretched version of? Answer: Cubic | 677.169 | 1 |
Tuesday, May 10, 2011
Constructive Triangles
I am in the process of reviewing all the lessons in each of my albums for the upcoming Oral Exams which complete my Montessori training with the Montgomery Montessori Institute. This is no small task, and each night the studying lasts a couple of hours (at which point I can no longer think clearly!)... In any case, I have a new found appreciation for the brilliance of Montessori's Constructive Triangles.
The purpose of the Constructive Triangles is to allow the child to explore and experiment with form and geometry. They are "constructive" because the triangles are used to form other figures. Eventually, the child comes to the realization of how new shapes and figures can be formed using only varying triangles and come appreciate the relationship between figures. Later, in the Elementary years, combinations of Constructive Triangle boxes are utilized to demonstrate the Pythagorean Theorem.e
The children in my class have recently become more interested in using the five different boxes. Many times during our Montessori work sessions, the kids will ask for lessons with one or more of these boxes. One of the aspects of the material which I value is the option of extension work. For example, once the children are comfortable working with the various boxes, they can begin extending the lessons via different materials. This may include making discoveries and experimenting with paper, paints, and colors through the creation of their own Constructive Triangles!
My dear husband made a video of me presenting the Triangular Box during one of my nightly study sessions (in my PJ's!). I find it a useful tool to be able to replay and watch myself give the lesson because often, I see on video what I miss in real-life! The direct aim of the material is for the child to realize that lines drawn on an equilateral triangle can be divided into different types of triangles:
By the way, that is Big Stuff's hand creeping in at the end...
I am very much enjoying the presentations with the Constructive Triangles and look forward to more lessons in the classroom with these types of materials
Question: What realization do children eventually come to understand with the Constructive Triangles? Answer: How new shapes and figures can be formed using only varying triangles and the relationship between figures. | 677.169 | 1 |
that, but the coordinates 0, 0 refer to "top left" as well. Think of a clock, or a box. You start at 12:00 then move to 3:00 then 6:00 then 9:00 and back to 12. With a box, you normally start in the upper left hand corner, draw a line to the right, then go down, stop and draw a line to the left, and then go back up again to close it.
Question: In a box, which direction do you move in after going down? Answer: To the left. | 677.169 | 1 |
Vectors Geometry
Chapter 1
Euclidean Geometry and Vectors
1.1 Euclidean Geometry
1.1.1 The Postulates of Euclid
The two Greek roots in the word geometry, geo and metron, mean "earth"
and "a measure," respectively, and until the early 19th century the de-
velopment of this mathematical discipline relied exclusively on our visual,
auditory, and tactile perception of the space in our immediate vicinity.
In particular, we believe that our space is homogeneous (has the same
properties at every point) and isotropic (has the same properties in ev-
ery direction). The abstraction of our intuition about space is Euclidean
geometry, named after the Greek mathematician and philosopher Euclid,
who developed this abstraction around 300 B.C.
The foundations of Euclidean geometry are five postulates concerning
points and lines. A point is an abstraction of the notion of a position in
space. A line is an abstraction of the path of a light beam connecting
two nearby points. Thus, any two points determine a unique line passing
through them. This is Euclid's first postulate. The second postulate
states that a line segment can be extended without limit in either direction.
This is rather less intuitive and requires an imaginative conception of space
as being infinite in extent. The third postulate states that, given any
straight line segment, a circle can be drawn having the segment as radius
and one endpoint as center, thereby recognizing the special importance
of the circle and the use of straight-edge and compass to construct pla-
nar figures. The fourth postulate states that all right angles are equal,
thereby acknowledging our perception of perpendicularity and its unifor-
mity. The fifth and final postulate states that if two lines are drawn
in the plane to intersect a third line in such a way that the sum of the
1
2 Euclidean Geometry
inner angles on one side is less than two right angles, then the two lines
inevitably must intersect each other on that side if extended far enough.
This postulate is equivalent to what is known as the parallel postulate,
stating that, given a line and a point not on the line, there exists one and
only one straight line in the same plane that passes through the point and
never intersects the first line, no matter how far the lines are extended. For
o
more information about the parallel postulate, see the book G¨del, Escher,
Bach: An Eternal Golden Braid by D. R. Hofstadter, 1999. The paral-
lel postulate is somewhat contrary to our physical perception of distance
perspective, where in fact two lines constructed to run parallel seem to
converge in the far distance.
While any geometric construction that does not exclusively rely on
the five postulates of Euclid can be called non-Euclidean, the two basic
non-Euclidean geometries, hyperbolic and elliptic, accept the first
four postulates of Euclid, but use their own versions of the fifth. Inciden-
Question: Who is Euclidean geometry named after? Answer: The Greek mathematician and philosopher Euclid. | 677.169 | 1 |
r = x ˆ + y + z κ;
ı (1.1.4)
the numbers (x, y, z) are called the coordinates of the point P with respect
ı ˆ
to the cartesian coordinate system formed by the lines along ˆ, , and
ˆ ˆ ˆ
κ. In the plane of ˆ and , the vectors x ˆ+y form a two-dimensional vector
ı ı
space R2 . With some abuse of notation, we sometimes write r = (x, y, z)
when (1.1.4) holds and the coordinate system is fixed.
The word "cartesian" describes everything connected with the French
scientist Ren´ Descartes (1596–1650), who was also known by the Latin
e
version of his last name, Cartesius. Beside the coordinate system, which
he introduced in 1637, he is famous for the statement "I think, therefore I
am."
8 Euclidean Geometry
Much of the power of the vector space approach lies in the freedom
from any choice of basis or coordinates. Indeed, many geometrical concepts
and results can be stated in vector terms without resorting to coordinate
systems. Here are two examples:
(1) The line determined by two points in R3 can be represented by the
position vector function
r(s) = u + s(v − u) = sv + (1 − s)u, −∞ < s < +∞, (1.1.5)
where u and v are the position vectors of the two points. More gen-
erally, a line passing through the point P0 and having a direction
−
−→
vector d consists of the points with position vectors r(s) = OP0 + s d.
(2) The plane determined by the three points having position vectors
u, v, w is represented by the position vector function
r(s, t) = u + s(v − u) + t(w − u)
(1.1.6)
= sv + tw + (1 − s − t)u, −∞ < s, t < +∞.
Exercise 1.1.1.B Verify that equations (1.1.5) and (1.1.6) indeed define a
line and a plane, respectively, in R3 .
Exercise 1.1.2.B Let L1 and L2 be two parallel lines in R3 . A line inter-
secting both L1 and L2 is called a transversal.
(a) Let L be a transversal perpendicular to L1 . Prove that L is perpen-
dicular to L2 . Hint: If not, then there is a right triangle with L as one side,
the other side along L1 and the hypotenuse lying along L2 . (b) Prove that the
alternate angles made by a transversal are equal. Hint: Let A and B be the
Question: Who introduced the Cartesian coordinate system? Answer: Ren´ Descartes (1596–1650)
Question: What is the relationship between a transversal and two parallel lines in R3, if the transversal is perpendicular to one of the lines? Answer: The transversal is also perpendicular to the other line. | 677.169 | 1 |
where θ is the angle between u and v, 0 ≤ θ ≤ π (see Figure 1.2.1), and
the notation u . v means the usual product of two numbers. If u = 0
or v = 0, then u · v = 0.
v# v w
....
... .................θ
.. θ .....
.. ..
..
.
.. E . E
u u
Fig. 1.2.1 Angle Between Two Vectors
Alternative names for the inner product are dot product and scalar
product.
If u and v are non-zero vectors, then u · v = 0 if and only if θ =
π/2. In this case, we say that the vectors u and v are orthogonal or
perpendicular, and write u ⊥ v. Notice that
2
u·u= u ≥ 0. (1.2.2)
In R3 , a set of three unit vectors that are mutually orthogonal is called an
orthonormal set or orthonormal basis. For example, the unit vectors
ı ˆ ˆ
ˆ, , κ of a cartesian coordinate system make an orthonormal basis. Indeed,
ı ˆ ı ˆ ˆ ˆ ı ˆ ı ˆ ˆ ˆ ı ı ˆ ˆ ˆ ˆ
ˆ ⊥ , ˆ ⊥ κ and ⊥ κ, ˆ · = ˆ · κ = · κ = 0, and ˆ · ˆ = · = κ · κ = 1.
The word "orthogonal" comes from the Greek orthogonios, or "right-
angled"; the word "perpendicular" comes from the Latin perpendiculum, or
"plumb line", which is a cord with a weight attached to one end, used to
check a straight vertical position. The Latin word norma means "carpen-
ter's square," another device to check for right angles.
The dot product simplifies the computations of the angles between two
vectors. Indeed, if u and v are two unit vectors, then u · v = cos θ. More
generally, for two non-zero vectors u and v we have
u·v
θ = cos−1 , (1.2.3)
u . v
The notion of the dot product is closely connected with the orthogo-
nal projection. If u and v are two non-zero vectors, then we can write
u = uv + up , where uv is parallel to v and up is perpendicular to v (see
Figure 1.2.2).
Inner Product 11
up T u ! u u Tp
u
..... ...................θ
.....
.... ...
..θ
. ..
.
.E
. E ' . E
Question: What are the two components of a vector u when it is expressed in terms of another vector v? Answer: The vector u can be expressed as the sum of two components: uv (parallel to v) and up (perpendicular to v). | 677.169 | 1 |
In other words, the scalar triple product does not change under cyclic per-
mutation of the vectors or when · and × symbols are switched.
C
Exercise 1.2.15. Verify that the ordered triplet of non-zero vectors u, v, w
is a right-handed triad if and only if (u, v, w) > 0.
Recall that v × w = v · w sin θ is the area of the parallelogram
formed by v and w. Therefore, |u · (v × w)| is the volume of the paral-
lelepiped formed by u, v, and w. Accordingly, (u, v, w) = 0 if and only
if the three vectors are linearly dependent, that is, one of them can be ex-
pressed as a linear combination of the other two. Similarly, four points
Pi , i = 1, . . . , 4 are co-planar (lie in the same plane) if and only if
−→ −→ −→
−− −− −−
(P1 P2 , P1 P3 , P1 P4 ) = 0, (1.2.31)
−
− → − → −→ − −
where Pi Pj = OPj − OPi . If (xi , yi , zi ) are the cartesian coordinates of the
point Pi , then (1.2.31) becomes
x2 − x 1 y 2 − y 1 z 2 − z 1
det x3 − x1 y3 − y1 z3 − z1 = 0. (1.2.32)
x4 − x 1 y 4 − y 1 z 4 − z 1
Notice a certain analogy with (1.2.28) and (1.2.29).
Exercise 1.2.16. C Let u = (1, 2, 3), v = (−2, 1, 2), w = (−1, 2, 1). (a)
Compute u×v, v ×w, (u×v)×(v×w). (b) Compute the area of the paral-
lelogram formed by u and v. (c) Compute the volume of the parallelepiped
formed by u, v, w using the triple product (u, v, w).
1.3 Curves in Space
1.3.1 Vector-Valued Functions of a Scalar Variable
To study the mathematical kinematics of moving bodies in R3 , we need to
define the velocity and acceleration vectors. The rigorous definition of these
vectors relies on the concept of the derivative of a vector-valued function
with respect to a scalar. We consider an idealized object, called a point
mass, with all mass concentrated at a single point.
Vector-Valued Functions of a Scalar Variable 25
Choose an origin O and let r(t) be the position vector of the point
−→
−
Question: What is the position vector of a moving point mass called? Answer: r(t)
Question: What is the area of the parallelogram formed by vectors u and v in Exercise 1.2.16? Answer: This is not directly calculated in the provided text, but it can be found using the formula |u × v| / 2. | 677.169 | 1 |
Trigonometry Basics Test - Sin, Cos, Tan
Choose the best answer. Trigonometric ratios are rounded to the nearest thousandth.
1. for which of the following triangles?
2. for which of the following triangles?
3. for which of the following triangles?
4. Which trigonometric function can equal or be greater than 1.000?
Sine
Cosine
Tangent
none of the above
5. A plane ascends at a 40° angle. When it reaches an altitude of one hundred feet, how much ground distance has it covered? To solve, use the trigonometric chart. Round the answer to the nearest tenth.
64.3 feet
76.6 feet
80.1 feet
119.2 feet
6. A 20 ft. beam leans against a wall. The beam reaches the wall 13.9 ft. above the ground. What is the measure of the angle formed by the beam and the ground?
44°
35°
55°
46°
7. Which set of angles has the same trigonometric ratio?
Sin 45 and tan 45
Sin 30 and cos 60
Cos 30 and tan 45
Tan 60 and sin 45
Question: Which trigonometric function is always positive in the first quadrant? Answer: Sine | 677.169 | 1 |
10 – PYTHAGORA not a distribution technique but probably the most useful piece of Math of all times :) this allows us to know the distance between 2 points, the angle between them and to deduce a whole lot of measures, distances, ratios aso
11 – POISSON DISK distribution, using the distance between 2 circles this allows us to distribute randomly some dics in the space. the rule for creating a point is "you shall not walk over someone else". very handy when you need to display non overlapping items like puzzle pieces and yet keep a "random" aspect.
I've published the source codes of the distribution methods above on wonderfl.net
each object or static method will return a Vector.< Point >, in the sample above, I'm plotting them like this:
Question: What can Pythagora be used to determine? Answer: Pythagora can be used to determine the distance between two points, the angle between them, and to deduce various measures, distances, and ratios. | 677.169 | 1 |
KneadToKnow
02-06-2008, 10:24 AM
Any opinions? Thanks!
The teacher is only thinking of corners by thinking of the corners of a room. You're thinking more analytically and I encourage you to teach your child to do so as well while also helping her to understand that some people won't be able to see these things.
oldboy
02-06-2008, 10:26 AM
:D
Hampshire
02-06-2008, 10:33Mangetout
02-06-2008, 10:37 AM
Yes. In a solid, I'd say a corner is the intersection of three planes (or possibly just three faces - they needn't be plane faces)
(in a plane figure, a corner is the intersection of two lines)
So a cylinder has no corners.
Morrison
02-06-2008, 10:38Well, the definition I used above is straight from the dictionary and it specifies "two" lines, surfaces, etc. I see what you are saying, though, and I guess I'm wondering if an edge and a corner can sometimes be the same thing?
(and, no, I don't know why I'm wasting time thinking about this...)
Hampshire
02-06-2008, 10:48 AM
I'd think maybe a corner can be defined as an intersection of 2 lines if it's a 2-dimensional figure like a square.
But once you jump to 3 dimensional spaces and objects you'd need 3 lines or surfaces.
gonzomax
02-06-2008, 10:50 AM
A sphere has no corners.
Giles
02-06-2008, 10:56 AM
The intuitive idea of a "corner" on a surface would seem to me to include the vertex of a cone, where only one surface is involved, but there is a singularity in the tangent planes of the surface. So a definition might need to be in terms of singularities of tangent planes, rather that in terms of number of intersecting surfaces. In the case of the circles at each end of a cylinder, you have a set of singularities forming a continuous curve, and so yoiu might define as "edge" as a set of singularities in the tangent planes that form a continuous curve, and and a corner as either:
(1) an isolated singularity (e.g., the vertex of a cone); or
(2) a point where three or more edges meet; or
(3) a point at the end of an edge; or
(4) a point in the middle of an edge where there is a discontinuity in the tangent lines to the edge.
(I leave dreaming up cases of (3) and (4) as exercises for the reader).
hajario
Question: What are some examples of corners according to Giles' definition? Answer: An isolated singularity (e.g., the vertex of a cone), a point where three or more edges meet, a point at the end of an edge, or a point in the middle of an edge where there is a discontinuity in the tangent lines to the edge. | 677.169 | 1 |
Question 535976: a ladder 9m long is placed against a wall 2m away from its base. what is the height reached by the ladder?
find the approximate distance that will be saved by walking diagonally across a field 450m by 250m instead of walking along the two adjacent sides.
A flagpole is to be made firm. how much wire will be needed if, from 3 pegs, each 120cm away from the foot of the people, wire will be tied to a point on the pole 2and1/2m from the ground?
Question 540541: I'm trying to figure out if this is a valid or invalid argument. I have been trying to use a truth table to prove , but am having a hard time.
the argument is:
if something is a bug then it is ugly
spiders are ugly
conclusion spiders are a bug
Question 541598:
I had a bit of time finding the right image to use. This is pretty much what I'm working with, only it's stands like an hourglass and the base and top extend out farther. (It should also say line AB, line DC and the midpoint on the bisected lines should say E. Also, if there's _?_ then (), then it's where I have to fill out. Everything else is what my textbook says.)
Given: Line segment AB is congruent to line segment DC
Line segment AD and line segment BC bisect each other.
Line AB || line CD
Prove: ▲ ABE is congruent to _?_ (▲ DCE)
Statements----------------------------------------------------------- Reason
1. line segment AB is congruent to line segment DC--------------1. Given
2. line segment AD and line segment BC bisect each other------2. _?_ (Given)
3. line segment AE is congruent to line segment _?_ (DE)------3. Def. of angle bisector
4. line segment BE is congruent to line segment CE-------------4. _?_ (Def. of angle bisector)
5. line AB || line CD-----------------------------------------------5. _?_ (Given)
6. angle BAE is congruent to angle CDE------------------------6. _?_
7. angle _?_ (ABE) is congruent to angle DCE-----------------7. If 2 || lines are intersected by a transversal, then the alternate interior angles are congruent
8. angle _?_ (AEB) is congruent to angle _?_ (CED)----------8. Vertical angles are congruent
9. ▲ ABE is congruent to ▲ DCE-------------------------------9. Definition of congruent polygons
Question 533672: Lisa, Bree, and Caleb are meeting at an amusement park. They each enter at a different gate. On this diagram of the park, explain where the friends can meet so that each walks the same distance from the gate to their meetind point. Please explain each step. Click here to see answer by cherkettle(1)
Question: What is the missing statement in the proof that '▲ ABE is congruent to ▲ DCE' that should be placed after statement 6?
Answer: "▲ ABE is congruent to ▲ DCE" (from the previous statements and the properties of congruent triangles).
Question: What is the missing statement in the proof that '▲ ABE is congruent to ▲ DCE'?
Answer: Statement 2 should be "Line segment AD is congruent to line segment BC" (from the given information and the property of angle bisectors).
Question: What is the missing statement in the proof that '▲ ABE is congruent to ▲ DCE' that should be placed after statement 5?
Answer: "▲ ABE is congruent to ▲ DCA" (from the given information and the properties of parallel lines). | 677.169 | 1 |
Question 462758: 1.The measure of an angle is 30 degrees more than twice the measure of its suppement.Find the measures of the angles.
2.Find the measures of two supplementary angles if the measure of one angle is 5 less than 4 times the measure of the other.
3.What are the measures of two complementary angles if the difference in their measures is 10.... Click here to see answer by math-vortex(472)
Question 462799: The measures of two angles of ABC are given. Find the measure of the third angle,
if A = 61° and B = 66°
Question: If the difference in measures of two complementary angles is 10 degrees, what is the measure of the larger angle? Answer: 80 degrees | 677.169 | 1 |
Rotation Tesselation.
Using 1cm dotty paper, poupils are asked to doodle at design in a quarter of a square. This design is then traced and rotated into each other quarter. Finally, the whole thing is rotated around. When coloured, it makes a very beautiful and weird looking tesselation!
Excellent for looking at rotations, symmetry, shape, angles etc etc etc
Question: What are some of the things that can be explored through this activity? Answer: Rotations, symmetry, shape, angles | 677.169 | 1 |
I have the Cartesian coordinates of the hypotenuse 'corners' in a right angle triangle. I also have the length of the sides of the triangle. What is the method of determining the coordinates of the third vertex where the opposite & adjacent sides meet.
because it doesn't appear that you worked on the problem before asking it. – anonSep 21 '10 at 7:15
1
@maud - I spent 6 hours yesterday on this, going off in the wrong direction as it turns out. Sleeping on it brought no inspiration so I asked the question here. I can take a picture of increasingly muddled scribblings for your review, but a) It wouldn't improve the question, b) the FAQ doesn't indicate its necessary and c) I saw no precedent for this when I searched other related questions. – KevinSep 21 '10 at 9:58
1
Kevin: mentioning something along the lines of "I tried method X and it went to crap" wouldn't have hurt. – J. M.Sep 21 '10 at 11:14
2 Answers
You have two points $A=(a,b)$ and $B=(c,d)$ and want a point $P$
at given distances from $A$ and $B$, say $l$ and $m$. Then $|PA|^2=l^2$
and $|PB|^2=m^2$ that is
$$(x-a)^2+(y-b)^2=l^2\qquad\qquad(1)$$
and
$$(x-c)^2+(y-d)^2=m^2.\qquad\qquad(2)$$
Subtracting (2) from (1) gives a linear equation. Use this to eliminate
one variable from (1). This yields a quadratic equation in the other variable.
Solving this will give the two possible positions for $P$.
Geometrically, subtracting equation 2 from equation 1 is equivalent to finding the radical line (mathworld.wolfram.com/RadicalLine.html ) of the two circles represented by the two equations. – J. M.Sep 21 '10 at 9:11
While I'd use the same algebra as Robin Chapman's solution, my first thought on this problem yields a third circle equation. The circumcenter of a right triangle is at the midpoint of its hypotenuse. Given the endpoints of the hypotenuse, $A=(a,b)$ and $B=(c,d)$, and letting $h$ be the length of the hypotenuse, the circumcircle has equation
$$\left(x-\frac{a+c}{2}\right)^2+\left(y-\frac{b+d}{2}\right)^2=\left(\frac{h}{2}\right)^2.$$
This may look a bit intimidating in symbols, but isn't really any different to work with for solving than the other two circles. I don't know that this offers any advantages, though.
Question: What is the advantage of using the circumcenter's equation to find the third vertex? Answer: There is no clear advantage mentioned in the text. It's an alternative method with the same level of complexity as the other two circles' equations.
Question: What is the method to find the coordinates of the third vertex (P) in a right-angled triangle given the coordinates of the other two vertices (A and B) and the lengths of the sides (l and m)? Answer: Subtract the equations of the circles with radii l and m centered at A and B respectively, to get a linear equation. Solve this to eliminate one variable, then solve the resulting quadratic equation in the other variable to find the coordinates of P.
Question: What is the first step in finding the coordinates of P using the given method? Answer: Subtract the equation of one circle from the other to get a linear equation.
Question: What is the geometric interpretation of subtracting the equation of one circle from another? Answer: It is equivalent to finding the radical axis (or radical line) of the two circles. | 677.169 | 1 |
Stewart's Theorem
A theorem in planar Euclidean geometry that gives an algebraic relationship between sides of a triangle and segments of sides when cevians divide a triangle in half. The theorem was discovered by the Scottish mathematician and minister, Matthew Stewart (1717/19 - 1785).
A cevian is a line segment whose one endpoint is the vertex of a triangle and the other endpoint lies somewhere along the opposite side. The length of the cevian can be computed from the knowledge of the lengths of the sides of the triangle, as well as the lengths of the subsegments of the side divided by the endpoint of the cevian.
Within triangle ΔABC construct a cevian connecting vertex A to point D on the side BC. Point D cuts side BC into segments CD and BD, having lengths x and y, respectively. Then if d is the length of AD, the following relationship holds:
b2y + c2x = a(d2+xy)
or
d = sqrt(((b2y+c2x)/a)-xy)
Medians: If the cevian is a median, it bisects side BC into equal length segments, and x = y. Then the cevian length is:
d = sqrt((x/a)*(b2y+c2x))
Angle Bisectors: If the cevian is an angle bisector, the by the angle bisector theorem, (y/x) = (a/b), or y = (a/b)*x. Then the cevian length can be found by
(b + c)2 = a2*(1 + d2/(x*y))
Altitudes: If the cevian is an altitude, then it makes a right angle with side BC. Then apply the Pythagorean formula to find d as a function of side lengths and x and y:
Question: What is the relationship between the lengths of a cevian and the sides of a triangle? Answer: The length of the cevian can be computed from the knowledge of the lengths of the sides of the triangle, as well as the lengths of the subsegments of the side divided by the endpoint of the cevian.
Question: What is a cevian? Answer: A line segment whose one endpoint is the vertex of a triangle and the other endpoint lies somewhere along the opposite side. | 677.169 | 1 |
Subject: Mathematics (8 - 12) Title: Quadrilaterals Description: This is an inquiry lesson used to review Algebra 1 objectives by applying them to geometry concepts. Students explore the properties of quadrilaterals and classify them by definition. This lesson can be use in geometry classes. Students in geometry classes can apply theorems and definitions of quadrilaterals rather than as an inquiry lesson.This lesson plan was created as a result of the Girls Engaged in Math and Science, GEMS Project funded by the Malone Family Foundation.
Thinkfinity Lesson Plans Location, Location, LocationAdd Bookmark Description: In this Illuminations lesson, students use a dynamic geometry applet to investigate the relationship between the distances from a point inside a regular polygon to each side. In addition, there are links to online activity sheets and other related resources. Thinkfinity Partner: Illuminations Grade Span: 9,10,11,12
Question: Which applet is used in the "Location, Location, Location" lesson plan? Answer: A dynamic geometry applet | 677.169 | 1 |
how you know all the dimensions and angles based on how you folded the paper.
That is, you should analyze each step of the folding and discuss what it tells
you about the geometry of the crease lines.
For the flat-foldable analysis: For this analysis you should discuss how
each of our theorems about flat foldability are satisfied in your model. In
your write-up you should examine each vertex, compute its degree and the number
of mountains and valleys that it has. You should check all your angle calculations
satisfy the other flat foldable theorems.
For the 2- colorable analysis: Use the theorem about 2-colorability and
your previous calculations to explain why you know your model must be 2 colorable
(that is, how you know before you color it.)
Part 2: The 3D model:
Option 1: Find a modular unit that we have not used in class that can construct
multiple polyhedral models, analyze the unit, make several smaller models and
analyze the models you have made.
Option 2: Use any modular unit (even ones we have used in class), analyze the
unit, make a single very large model, and analyze the model you have made. In
particular, PhiZZ units can make a very nice torus (doughnut) or large buckyball.
Your analysis of the unit should discuss the dimensions and angles of the unit
and how this effects how it can be put together with other units. That is, you
should describe what type of faces it can form, how many faces can meet at a
vertex, and thus, what type of polyhedra you can make with it.
For each polyhedra you make, you should find the number of faces, vertices,
and edges. You should verify that Euler's formula holds and determine
if there are any other relationships that must hold between faces and vertices
or edges and vertices or edges and faces and explain how you know this based
on how the object is constructed.
Question: What should be discussed in the analysis of a modular unit for the 3D model? Answer: In the analysis of a modular unit for the 3D model, the dimensions and angles of the unit and how these affect its assembly with other units should be discussed.
Question: What should be determined for each polyhedron made using the modular unit? Answer: For each polyhedron made using the modular unit, the number of faces, vertices, and edges should be determined, and Euler's formula should be verified. | 677.169 | 1 |
Scaling the Trail Leap Connections
Teacher-Made Supplemental
Resources
Reading Strategies
Cut Down to Size at High Noon: A Math Adventure, by Scott Sundby 11: Similar Triangles (LCC Unit 2 Activity 8)
(GLEs: 7, 29)
Materials List: 6 drinking straws for each pair of students, scissors, pencils, paper, math learning
log, ruler
Have students work in pairs to create an equilateral triangle using drinking straws for sides. Ask
students to explain how they know they have created an equilateral triangle. (they have three
straws the same length). Have them measure and record the side length.
Instruct students to make a second equilateral triangle with sides of different length than those of
triangle one. Have students measure with rulers the sides of their new triangle. Ask them to
determine a way to prove that the two triangles are similar using what they have learned about
proportions. Students should understand that the triangles are similar because the sides are of
proportionate lengths. Triangle one has sides twice as long as triangle two, and the angles
measure the same because they are equilateral triangles. Equilateral triangles are also equiangular.
Lead students to write a conjecture about the relationship of proportionate sides and equal angles
in two equilateral triangles. Ask them if it seems possible that this relationship will hold true with
other triangle types.
Next, have students construct or draw a triangle with all three sides of different lengths (scalene).
Have students label the triangle with the measure of each of the side lengths and each angle
measure. Instruct students to select one vertex of their new triangle and label the vertex A. Have
3
students extend the sides of the triangle from vertex A so that the side is the length of the
2
8th Grade Mathematics: Unit 1: Rates, Ratios, and Proportions 12
8th Grade Mathematics: Unit 1: Rates, Ratios, and Proportions
original side. Repeat this with the other side from vertex A. Instruct students to connect the two
endpoints of the new sides for their triangle. Have students make some observations about the
two triangles that they have formed. Challenge students to use proportions to prove that the two
triangles are proportional.
Discuss how the angles of these two triangles are congruent but the side lengths are proportionate.
Tell the students that the symbol to show similarity is "". We call the two triangles "similar
triangles" because the angles are congruent and the side lengths are proportionate.
Next, have them construct or draw a triangle using a ratio provided to them, perhaps a ratio of
3
and determine if the same conjecture holds true for triangles with sides of different lengths. For
4
example, if they create a triangle with side lengths of 3 inches, 4 inches, and 5 inches, a triangle
with sides of 2.25 inches, 3 inches, and 3.75 inches would meet the requirement. Once they have
Question: What should students do after creating the first equilateral triangle? Answer: Measure and record the side length. | 677.169 | 1 |
The common nine-point circle four orthocentric points. The radius of the common nine-point circle is the distance from the nine-point center to the midpoint of any of the six connectors that join any pair of orthocentric points through which the common nine-point circle passes. The nine-point circle also passes through the three orthogonal intersections at the feet of the altitudes of the four possible triangle.
This common nine-point center lies at the mid point of the connector that joins any orthocentric point to the circumcenter of the triangle formed from the other three orthocentric points.
The common orthic triangle, its incenter and excenters
If the six connectors that join any pair of orthocentric points are extended to six lines that intersect each other, they generate seven intersection points. Four of these points are the original orthocentric points and the additional three points are the orthogonal intersections at the feet of thes. The joining of these three orthogonal points into a triangle generates an orthic triangle that is common to all the four possible triangles formed from the four orthocentric points taken three at a time.
Note that the incenter of this common orthic triangle must be one of the original four orthocentric points. Furthermore, the three remaining points become the excenters of this common orthic triangle. The orthocentric point that becomes the incenter of the orthic triangle is that orthocentric point closest to the common nine-point center. This relationship between the orthic triangle and the original four orthocentric points leads directly to the fact that the incenter and excenters of a reference triangle form an orthocentric system.
It is normal to choose the orthocentric point that is the incenter of the orthic triangle as H the orthocenter of the outer three orthocentric points that are chosen as a reference triangle ABC. In this normalized configuration the point H will always lie within the triangle ABC and all the angles of triangle ABC will be acute. The four possible triangles referred above are then triangles ABC, ABH, ACH and BCH. The six connectors referred above are AB, AC, BC, AH, BH and CH. The seven intersections referred above are A, B, C, H (the original orthocentric points) and HA, HB, HC (the feet of the altitudes of triangle ABC and the vertices of the orthic triangle).
The orthocentric system and its orthic axes
The orthic axis associated with a normalized orthocentric system A, B, C and H, where ABC is the reference triangle, is a line that passes through three intersection points formed when each side of the orthic triangle meets each side of the reference triangle. Now consider the three other possible triangles, ABH, ACH and BCH. They each have their own orthic axis.
Euler lines and homothetic orthocentric systems
Question: What is the radius of the common nine-point circle? Answer: The radius is the distance from the nine-point center to the midpoint of any of the six connectors that join any pair of orthocentric points. | 677.169 | 1 |
Short description: The students are presented with real life situations where trigonometry is used.
Duration of lesson(s): 1 lesson
Grade level(s) and/or target group(s):Students studying trigonometry
Subject(s):Trigonometry
Technologies used:When these problems were presented, a smart board was used to display the diagrams as well as make notations as we progressed working on the project.
Objectives: Student will be able to use trigonometry and other knowledge of mathematics to solve application problems.
Key Questions/Driving Questions:How can trigonometry be used?
Prerequisites and Sequence of Lessons in Unit:
Students should be familiar with calculating the sine, cosine, and tangent of a given angle.They should also be able to calculate the missing side of a right triangle given a side and an angle.
Lesson Introduction:
Do-now :
Students are to find the length of a missing side of a right triangle given an angle and a side.
And
Students are to find the measure of an angle given two sides of a right triangle.( inverse trig functions)
Lesson Core:Students apply the trigonometric functions in real world problems.The problems are worked on by the students in groups while teacher facilitates the inquiry of questions.
See attached :Three Examples of Practical Applications of Trigonometry
Lesson Closure:Selected or volunteer students are to present their detailed explanations of the problems.
Three Examples of Practical Applications of Trigonometry
1. Staircase construction
Jorge's studio has a steep staircase made out of a ladder that he wants to replace with a real staircase.It needs to be less steep and has to leave room for the lower bedroom door to open.Using trigonometry, how can you design a better staircase?
Figure 1.Schematic diagram of the original staircase and bedroom area.
The length of the wall is 25 feet.The height of the original staircase is 12 feet.The doors are 2.75 feet wide.The lower bedroom door is 10 feet from the front edge of the bedroom area.
One solution is to build a landing outside the upper bedroom and then extend the staircase, as in Figure 2.
Figure 2. Proposed position of new staircase.
What is the length of the new staircase?
What is the height and width of each new stair?
What is the difference between the angle made by the new staircase and the old staircase?
2. Sarah owns a piece of property that she needs to know the square feet of assessment purposes.The figure that the city surveyors came up with is 7012.5 ft2, and she thinks it is wrong, and will raise her properties taxes unnecessarily.They didn't subtract the circular portion of the cul-de-sac that the city owns.Her lot is 55 ft wide. One side is 108.96 feet; the other side is 146.04 feet.There is a circular area removed from the property, with an arc length of 78.21 feet and radius 40 feet.
Figure 3. Schematic diagram of Sarah's property.
Question: What is the width of the lower bedroom door in the staircase construction problem? Answer: 2.75 feet | 677.169 | 1 |
And since angle ABF is subtended by the diameter and is therefore right, the side of the pentagon is calculated by an even simpler route:
Based on his circle of diameter 200000 units, Copernicus provides accurate numerical values for the four pentagon related chords corresponding to these angles:
"Since the side of the decagon which subtends 360 has been shown to have 61803 parts, whereof the diameter has 200000 parts – the chord which subtends the remaining 1440 of the semicircle has 190211 parts. And in the case of side of the pentagon, which is equal to 117557 parts of the diameter and subtends an arc of 720, a straight line of 161803 parts is given, and it subtends the remaining 1080 of the circle".
Golden ratio aficionados will instantly recognize the digits 161803 and 61803 as corresponding to and its reciprocal.[7] The chords are of considerable historical importance because, along with the sides of the triangle and tetragon (square), they enable the generation of a table of half chords (effectively sine values) [8] which in turn underpins many of the key astronomical measurements and calculations effected by Copernicus in the development of his helio-centric model:
"Because the proofs which we shall use in almost the entire work deal with straight lines and arcs, with plane and spherical triangles and because Euclid's Elements, although they clear up much of this, do not have what is here most required, namely, how to find the sides from angles and the angles from the sides ... there has accordingly been found a method whereby the lines subtending any arc may be known."
[Euclid 13:10
Square of Pentagon chord equals sum of squares of decagon and hexagon chords
Next our attention is drawn to point G midway between points B and C on the circumference. CG, FG and DG are joined forming cyclic quadrilateral DFCG in which three sides belong to the regular decagon (length c) and the fourth DG is of length z. Diagonals DC and FG are both of length a (side of the pentagon) Then:
Therefore
where r is the radius of the circle and also the side of the inscribed hexagon.
Whence with relative ease is proved Proposition 10 in Book XIII of Euclid's Elements: The square on the side of the pentagon equals the sum of the squares on the sides of the hexagon and the decagon inscribed in the same circle.
In modern trigonometric notation this corresponds to the identity:
The Pythagorean nature of this relationship makes possible the construction of a regular pentagon as demonstrated here.
[Theorema Tertium
Determination of the chord subtending 12 degrees of arc
A well documented classical application of the "Second Theorem" as illustrated in the diagram is the determination of chord BC subtending 12 degrees of arc. Referring to the diagram:
Question: What is the relationship between the squares of the sides of the pentagon, hexagon, and decagon inscribed in the same circle? Answer: The square on the side of the pentagon equals the sum of the squares on the sides of the hexagon and the decagon (Pythagorean relationship)
Question: Which two numbers, 161803 and 61803, are related to the golden ratio? Answer: They are the golden ratio (1.61803...) and its reciprocal (0.61803...)
Question: Which angle in the pentagon subtends an arc of 720 degrees? Answer: 117557 parts of the diameter | 677.169 | 1 |
the sun was hitting the bottom of one well while it was shining on the sides of the other. measuring the difference in angles from vertical, he was able to conclude that the earth was round and even came up with an acceptable estimate of the earth's diameter (google erathosthenes' experiment.)
but i want a reverse experiment: can we actually measure whether the sun's rays hit us at radial or parallel? guys tell me that as the distance from the point source of light to the observer increases, the rays change from radial to parallel. i'm thinking of two friends living several miles away making simultaneous measurments while the sun's shining on all three of us. but online calculators won't even return a figure for the angle if i enter the hypothenus as 93 million miles while the base of the right triangle is, say 100 miles.
Blake
06-27-2011, 06:01 AM
Assuming the question is "can we actually measure whether the sun's rays hit us at radial or parallel?", then the answer is yes. You already explained one obvious way to do it. The reason why calculators won't return an answer is that the the answer is smaller than significance allowed by the of the processor. IOW the rays are indistinguishable from parallel from a practical perspective.
If there is another question in there, I don't see it.
MikeS
06-27-2011, 07:17 AM
but online calculators won't even return a figure for the angle if i enter the hypothenus as 93 million miles while the base of the right triangle is, say 100 miles.DrFidelius
06-27-2011, 07:21 AM
Close enough to parallel for government work.
md2000
06-27-2011, 07:34 AM
Exactly. The rays are always radial; they never "change to parallel". However, the deviation from parallel is so miniscule for practical purposes, that it might as well be parallel.
Also note the sun is a visible size from earth. That means it's not that simple. Every part of the sun's visible disc acts as a separate point source, so we get a real mess of rays.
kanicbird
06-27-2011, 08:35 AM
Wondering how he was able to check at the exact same time. Any thoughts on this :confused:
Blake
06-27-2011, 08:39 AM
Wondering how he was able to check at the exact same time. Any thoughts on this :confused:
It was noon. Nothing tricky about it.
yabob
06-27-2011, 08:41 AM
Wondering how he was able to check at the exact same time. Any thoughts on this :confused:
Question: What is the reason online calculators won't return an angle figure for a large right triangle with the sun's distance as the hypotenuse? Answer: The angle is too small to be calculated accurately due to the size of the triangle. | 677.169 | 1 |
Find a Missing Leg
Explanation
The legs of a right triangle are the sides that are adjacent to its right angle. Sometimes we have problems that ask us to find a missing length of one of these legs. We can use the Pythagorean theorem to find a missing leg of a triangle, but only if we know the length measure of the hypotenuse and the other one of the legs.
Transcript
The Pythagorean Theorem only applies to a right triangle. And in that right triangle we can say that the legs a and b that is those sides that are adjacent to this right angle. If I square them and add them up it's going to equal your hypotenuse squared or c squared so your hypotenuse remember is that side that is opposite your 90 degree angle and since the triangle angle sum says that if this is 90 degrees then both of these angles have to be less than 90, your hypotenuse will always be the longest side in a right triangle.
Let's say you knew one of your legs and your hypotenuse, so let's say you knew a and you knew c you would solve for your missing leg by squaring a, squaring c and then you're going to have to subtract a squared whatever it is. So that's going to be an a not a 2, an a squared so you're going to get b squared is equal to c squared minus a squared. And then to solve you're going to have to take the square root of both sides. So b is going to equal the square root of c squared minus a squared sum number. So it's a little more involved when you know 1 leg and your hypotenuse and you're trying to solve for the other leg. You're going to have to subtract and then take the square root.
Question: What operation needs to be performed on both sides of the equation to solve for b? Answer: Take the square root. | 677.169 | 1 |
In the other limiting case, an entire pizza has its center of mass right at the tip (i.e. center), so
.
To investigate intermediate cases, we start with a slice of angle and imagine cutting it in half lengthwise, creating two skinny pieces of angle . These have their own centers of mass at .
The center of mass of the big piece is on the line connecting the smaller pieces' centers of mass.
The center of mass of the entire slice lies on the line connecting the centers of mass of the half-slices. This creates a geometric relation between and .
A bit of trigonometry tells us
If we take this formula and divide all angles by , we get a formula for . We substitute this for where appeared in the original. We obtain
Repeat the process ad infinitum. Rearranging the order of the terms and substituting the limiting value of for small , we get
It involves one half of Euler's trig identity. If we find by a different method and get a different expression for it, we can set our two expressions for equal to each other, and prove Euler's identity. We'll do this by invoking some physics ideas.
Suppose you're spinning some pizza dough in the air. You know, like this:
If the pizza is spinning, each little bit of dough undergoes centripetal acceleration. Where there's acceleration, there's force. The pizza isn't touching anything, so the force on any one piece of pizza must be coming from the rest of the pizza.
Let's again examine a slice of size , this time still attached to the spinning pizza. It has two forces of size acting on it; one force is exerted by the slice to its left and one by the slice to its right.
There are two forces on the slice - one from the pizza to the left and one from the pizza to the right. They're both drawn originating from the center of mass. The slice is accelerating towards its tip (red arrow).
The sum of these forces is the mass of the slice times the acceleration of its center of mass. That acceleration is . Hence, if we determine the forces we can deduce .
Some trigonometry shows that the net force is .
Equating this to mass times acceleration, we get
We might as well let and solve for to get
.
We still need to determine , but we can do that because we know as . After a little algebra, we get
This gives us the sought two expressions for . We can now equate them and simplify to
1) To see why an isosceles triangle's center of mass is 2/3 up the altitude, first show it's true for an equilateral triangle. Then explain why all isosceles triangles have their center of mass the same fraction of the way down the altitude.
Question: Is the center of mass of an entire pizza at its tip (center) in one of the limiting cases? Answer: Yes
Question: If we divide all angles by 2 in the given formula, what do we get? Answer: A formula for c | 677.169 | 1 |
The similarity method for calculating fractal dimension is great if you have a fractal
composed of a certain number of identical versions of itself. However, try using it for
the coast of Britain. Thatís impossible because all lines there have different sizes and
require different magnifications. And we wouldnít suggest counting them either!
A WAY OUT
There is a simple way out of this. We know that a true fractal has an infinite amount
of detail. This means that magnifying it adds additional detail, which increases the
overall size. In non-fractals, however, the size always stays the same. For example, look
at the diagram below, where we graphed sizes of some non-fractals with different
magnifications. If you graph log(size) against log(magnification) you get straight
horizontal lines. It shows that the sizes donít change, which means that the figures are
not fractals.
Now take some fractals and do the same. You will no longer get horizontal lines since
the sizes increase with magnification. This proves the figures to be fractals.
We can now easily calculate fractal dimensions using the slopes of these lines. This is
done using a simple formula:
fractal dimension = slope + 1
HOW IS IT USEFUL?
The geometric method can be used very efficiently for natural irregular shapes that
exhibit Brownian self-similarity. It was used to
calculate dimensions of coasts, borderlines
and clouds.
Question: What is a characteristic of true fractals when magnified? Answer: Magnifying a true fractal adds additional detail, which increases the overall size.
Question: Which formula is used to calculate the fractal dimension from the slope of the lines? Answer: Fractal dimension = slope + 1 | 677.169 | 1 |
Here various graphs are investigated. The classification is sometimes arbitrary, since a graph may arise as a result of different constructions (stellation, reticulation, truncation, glueing etc.).
Some occuring mathematical terms will be explained later in subsequent chapters.
Tetrahedron Stellation I (Deltoid)
This graph arises as the stellation of a tetrahedron and is also known as a deltoid.
Question: Is the classification of graphs in the text always clear and straightforward? Answer: No, the classification is sometimes arbitrary. | 677.169 | 1 |
Based on his circle of diameter 200000 units and already established chords of pentagon, hexagon and triangle the calculation effected by Copernicus would have been:
A small rounding error is evident in the result but the corresponding entry (in the Copernican table of half chords ) of 10453 units against 6 degrees is correct as may readily be verified on a calculator (sin 6).
In modern trigonometric notation, the above calculation corresponds to the following application of a compound angle formula:
[Theorema Quintum
Determination of the chord subtending a sum of arcs
The previous diagram demonstrated a general technique for calculating the chord subtending the difference between two arcs. The following diagram neatly reverses this procedure to obtain the chord subtending the sum of arcs: i.e. determination of chord AC given chords AB and BC.
Compared with the previous we note that diameter BE has been swung across from point B to point E. EC and ED are joined. Since AEDB is a rectangle DE=AB. Thus in cyclic quadrilateral BEDC, sides BE, BC and ED are known along with diagonals CE and BD by application of the "Porism" (Pythagoras Thm). Then:
In the specific example illustrated in the diagram, calculation of chord CD in cyclic quadrilateral BEDC corresponds to the following application of a compound angle formula:
The required chord AC (in this example corresponding to sin(30+6)) is then calculated by application of the "Porism".
[Other polygons
When applied repeatedly, Ptolemy's theorem allows one to compute the lengths of all diagonals for polygons inscribed in a circle with vertices P1, ..., Pn, if the sides are given together with all the length values of the "next to sides" chords connecting two vertices Pi and Pi+2 (with indices taken modulon).[citation needed]
Note that the proof is clearly only valid for simple cyclic quadrilaterals; if the quadrilateral is complex then K will be located outside the line segment AC, so AK−CK=±AC, giving the expected result.
[Trigonometric proof
It suffices to prove the theorem for the standard unit circle (the statement of the theorem is invariant under re-scaling and translation). Introducing polar coordinates one may represent the four vertices in the form
After a possible renumbering of the Pi one can also assume that the four vertices appear in natural counterclockwise order which means that .
A basic result from trigonometry states that for two points and on the unit circle written in polar coordinates their Euclidean distance ||x − y|| is given as
If is an (ordered) pair of vertices of the given quadrilateral this formula implies
applied to each of the three products of sines (the resulting six terms cancel out in pairs).
Concluding remark (explaining the naming "addition relation"):
If one introduces the difference angles for then the relation
turns into
Question: What is the angle corresponding to the entry in the Copernican table? Answer: 6 degrees
Question: What is the diameter of the circle used in Copernicus' calculation? Answer: 200000 units
Question: What is the relationship between sides AB and DE in the diagram illustrating the Theorema Quintum? Answer: DE = AB
Question: Which theorem is used to find the lengths of all diagonals for polygons inscribed in a circle in the text? Answer: Ptolemy's theorem | 677.169 | 1 |
Pages
Sunday, February 5, 2012
VFC: Quadrilaterals
Here is a Virtual Filing Cabinet for resources concerning Quadrilaterals.. This is part of an ongoing experiment in how to better share online teaching resources. If you like this post, then make your own post for a particular topic.
What am I missing here? Point me to your favorite quadrilaterals resource in the comments.
[Last Updated: 2/5/2012]
The Hard Parts
A lot of the traditional proofs of the properties of quadrilaterals depend very heavily on congruent triangles. One of the real challenges in teaching this topic is to change your students' perspectives. When they see a rhombus they should see four congruent triangles; when they see a parallelogram they should see two pairs of congruent triangles; when they see a kite they should see two pairs of congruent triangles arranged differently.
There are lots of challenges for novices that go along with this change in perspective. Students need to see triangles in quadrilaterals, even if the diagonals are absent. Students need to see parallel lines with a transversal even when the sides of the quadrilateral are not extended.
Another major theme of this unit is the hierarchy of shapes. A square is a rectangle, but it's also a rhombus. They are all parallelograms, though, and so what's true of parallelograms is true of them as well. This, plus a whole slew of new vocabulary.
Quadrilateral Resources
For vocabulary, I like the approach of the Discovering Geometry series. Show kids a bunch of examples of things that are "trapezoids", show them a bunch of things that aren't, and then challenge them to formulate a definition that works. This is a pretty common approach, from what I can tell. Here's a post from misscalcul8 on her version of it.
Once you have the vocab down, you might want to make it more concrete and emphasize the relationships between these shapes. I've posted about an activity that I like where students create "family trees" for quadrilaterals.
One of the big challenges of this unit is (to my mind) getting students to see quadrilaterals as composed of triangles, as this generates all of the non-obvious properties of the quadrilaterals. I like this activity, which uses a series of tangram challenges of increasing difficult. It literally forces students to compose various quadrilaterals out of smaller shapes, including triangles. This can also serve as a concrete model that can be returned to over the course of the unit.
I'm still looking for resources for the actual nitty gritty of this unit which is the properties of the various quadrilaterals. I'll post resources as I find them, and please let me know if you have resources to add to this page.
1 comment:
Question: What is one of the challenges in teaching the properties of quadrilaterals? Answer: Changing students' perspectives to see triangles within quadrilaterals | 677.169 | 1 |
Most of the time student these two points one is that "a square is a rectangle, but it's also a rhombus. They are all parallelograms, though, and so what's true of parallelograms is true of them as well" and another is that "students need to see parallel lines with a transversal even when the sides of the quadrilateral are not extended". Is the square root of 2 a Rational Number
Question: Is the square root of 2 a rational number? Answer: No | 677.169 | 1 |
Graphics Trick: Polygon Area
The area of a 2D triangle is easy, it's just half the magnitude of the cross product of two edges where you assume the z component is 0:
|cross((v1−v0),(v2−v0))|
The sign of the cross product tells you if the triangle is front or back facing (counter clockwise or clockwise). If you multiply out the cross product, it has an interesting repeating pattern if you collect it in terms with pairs of indices i and (i+1)%3
As a bonus, for 3D triangles, the cross product points in the direction normal to the triangle, and its length is twice the area of the triangle. What's more, the cyclical pattern appears again for each component (the order and signs here make more sense if you consider x,y,z to have their own mod cycle as well: x is followed by y is followed by z is followed by x …)
Maybe overkill for a triangle, but the cool thing is that the same basic pattern works to find the area of any planar polygon with non-intersecting edges. It doesn't even need to be convex. Just change the loop limit and mod from 3 to N. Just like the cross product, the magnitude will be twice the polygon area, and the sign will tell you if it is front or back facing.
Why does it work? It's a simple application of Green's theorum (or Stoke's theorum in 3D). Bet you never thought you'd see that again, huh?
Mods add ugly unnecessary conditionals inside the loop, but if you shift the loop order a little, you can actually get rid of the mod:
Extra bonus facts: in 3D, the resulting vector points in the direction of the polygon normal with length equal to twice the polygon area (just like the triangle cross product). Also, if you do it for the 1-ring of vertices around a given vertex (the vertices one edge away from a given vertex), you get exactly the triangle-area weighted average normal. Yes, that does mean that the area-weighted vertex normal doesn't actually depend on the vertex position at all. Wierd, huh?
MarcOne Response to Graphics Trick: Polygon Area
That's one of my favorite interview questions… Another interesting point is that you can pick an arbitrary point instead of a vertex of the polygon and the result still holds. Oh, and it also extends to the volume of polyhedrons. Do you need the volume of a complex (but closed) 3d mesh? easy. Bonus points if you derive it from the Stokes theorem.
Question: What is the relationship between the cross product and the area of a 3D triangle? Answer: The length of the cross product is twice the area of the 3D triangle. | 677.169 | 1 |
Question 149920: Could I please get some help with the following problem:
there is a diagram of a triangle
one side is = 1, another side is = 4 and there is 100 degree angle
I am looking for the value of side C
thank you,
RF Click here to see answer by stanbon(57239)
Question 149918: Sketch one complete period of the graph of g(x)=sin(2piex). Label the x-intercepts and the local maximums and minimums.
Question 149972: Find the values of the trigonometric functions of "theta" from the given information.
sin(theta)=3/5, t in quadrant II. (Hint: You will need one of the Pythagorean Identities).
Cos(theta)=?
Tan(theta)=?
Sec(theta)=?
Csc(theta)=?
Cot(theta)=? Click here to see answer by Fombitz(13828)
Question 149978: From a point on the ground 500 feet from the base of a building, it is observed that the angle of elevation to the top of the building is 24 degrees and the angle of elevation to the top of a flagpole atop the building is 27 degrees. Find the height of the building and the length of the flagpole. Click here to see answer by [email protected](15634)
Question 150081: Please help me one more time!
Find the height of a building that is 80ft. away from a point on the ground, when this point on the ground makes an angle of 25 degrees with the top of the building.
Thank you, Ray! Click here to see answer by vleith(2825)
Question 150200: This problem is really frustrating me, I just don't understand, will someone please help?
Find teh area of a triangle if side b = 5, side c = 4, and angle A = 20 degrees.
Thank you so much for your time!
Lindsey Click here to see answer by stanbon(57239)
Question 150255: I have been trying to work this problem for a while now and I just can't seem to get it, will someone PLEASE HELP me!
Write 3(cos 60 degrees + i sin 60 degrees) in the form (a + bi).
Thank you so much, Natalie! Click here to see answer by Fombitz(13828)
Question 150317: Please help me answer the following:
(-5,12) are the coordinates of a point in rectangular form. Find this point in polar form (r,A), with A expressed in radians.
The answer choices include:
1. (13,112.6)
Question: What is the value of cos(θ) given sin(θ) = 3/5 and θ is in quadrant II? Answer: -4/5 (Using Pythagorean identity, cos²(θ) + sin²(θ) = 1, so cos(θ) = -4/5)
Question: What is 3(cos 60 degrees + i sin 60 degrees) in the form (a + bi)? Answer: (3/2 + 3i√3/2) (Simplifying the expression using Euler's formula) | 677.169 | 1 |
[65] If you take a square and look at it from some point in space it looks like a quadrilateral. What are the
possible shapes of this quadrilateral?
[66] A regular heptagon (= 7-sided polygon with all angles equal and all sides equal) is randomly placed in
the plane. What is the probability that an observer can see 4 sides?
[67] A table has 3 legs of equal length. Is it always possible to place the table on a convex hill so that the
surface is level?.
[68] By cutting a square along the 4 line segments joining vertices to midpoints of opposite sides, the
square is decomposed into 9 pieces which can be re-assembled to form 5 congruent squares. Dissect a
square into a "small number" of pieces by cutting it along straight lines, so that these pieces can be re-
assembled to form 3 congruent squares. What is the least number of pieces this can be done with?.
[69] If you have a set A of 101 integers chosen from the numbers , show that every integer is the
sum of 2 numbers from A modulo 200. (you can choose the same number from A twice)
[70] A lion chases a man inside the unit circle (they are allowed to go onto the circle as well). They both
move at unit speed. Can the lion catch the man?.
[71] What is the maximum number of integer lattice points (= points with integer coordinates) in the plane
that a strictly convex region can touch if it does not contain any lattice points in its interior.
Question: What are the possible shapes of a quadrilateral that a square can appear as when viewed from a point in space? Answer: The possible shapes are a rectangle, a parallelogram, a rhombus, or a general quadrilateral.
Question: What is the least number of pieces a square can be dissected into, using straight lines, and then reassembled to form 3 congruent squares? Answer: 5 pieces. | 677.169 | 1 |
This question came from some of the PR practice stuff. In breaking the hexagon into 6 equilateral triangles, I get an area for each triangle to be r root 3 * r (= r^2 root 3), which is then multiplied by 6, for the number of triangles..... But PR says this is not the correct answer???
Curly05 wrote:
Should be broken up into equilateral. triangles.
But, radius doesn't extend all the way to the side, don't know where did you get this stuff from?
You don't have to know trigonometry for GMAT. How did you get what you get? So you are a girl? I don't hate girls? What is your nationality?
Hey, from the beginning I have defended your right to ask questions -- any questions -- and make whatever comments so deemed necessary in the interest of all of us learning. You can confirm this with Stolyar.
But now you are simply being obnoxious to no one's benefit and everyone's irritation.
Question: What is the user's gender according to Curly05? Answer: The user is a girl | 677.169 | 1 |
pythagorean triples
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Pythagorean Theorem Calculator with Steps
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No matter what situation you find yourself in requiring a Pythagorean Theorem calculator, our new excel based tool will work for you. Be it doing homework or trying to measure straight line distances between two objects, our calculator is 100% guaranteed to work for every situation, and considering the Pythagorean Theorem calculator is totally free you have nothing to lose by using this wonderful tool.
Question: What is the formula for the Pythagorean Theorem? Answer: The formula is A² + B² = C², where A and B are the lengths of the legs, and C is the length of the hypotenuse. | 677.169 | 1 |
Can't they be on the same line? 3-4-5 triangle is a very special case. You are doing wrong I think.
I agree this is might be a special case and this is exactly what I am want to know where am I going wrong.
Anyway the points can't be on the same line because as per the 1st statement Z is to left of X and Y is to right of X and from the 2nd statement the distance between Z and Y is 3 which is not possible because in that X goes to the right of Y.
Yes you are right. They can not be on the same line. But there are unlimited number of triangles with 2 sides are 3 and 4. It may vary from 3-4-1,000001 to 3-4-6,999999999. Left of x and right of x may limit to some values. But it does not limit to 3-4-5
Question: What is the special case mentioned in the text? Answer: The 3-4-5 triangle. | 677.169 | 1 |
(mathematics) An endofunction whose square is equal to the identity function; a function equal to its inverse.
irrational
adjective ((compar) more irrational, (superl) most irrational)
Not rational; unfounded or nonsensical.
an irrational decision
(mathematics) (no comparative or superlative) Of a real number, that cannot be written as the ratio of two integers.
The number π is irrational.
irrational number
noun (pluralirrational numbers)
(mathematics) Any real number that cannot be expressed as a ratio of two integers.
isogon
noun
A line of equal or constant wind direction on a graph or chart, such as a weather map.
isometric
noun
A line connecting isometric points.
adjective
Of, or exhibiting equality in dimensions
Of, or being a geometric system of three equal axes lying at right angles to each other (especially in crystallography).
(physiology) Of or involving muscular contraction against resistance in which the length of the muscle remains the same
(physics) (of a thermodynamic process) taking place at constant volume because of being confined by rigid boundaries
isometry
noun (isometries)
(mathematics) A function between metric spaces (or on a single metric space) having the property that the distance between two images is equal to the distance between their pre-images.
isomorphic
adjective
(biology) having a similar structure to something that is not related genetically
(mathematics) related by an isomorphism
isomorphism
noun
(biology) the similarity in form of organisms of different ancestry
(mathematics) a one-to-one correspondence between all the elements of two sets such that any operation returns the same result on either set; a function that maps one of these sets to the other
(chemistry) the similarity in the crystal structures of similar chemical compounds
(sociology) the similarity in the structure or processes of different organizations
(computer science) a one-to-one correspondence between all the elements of two sets, e.g. the instances of two classes, or the records in two datasets
isosceles
adjective
(geometry) having two sides of equal length, used especially of an isosceles triangle.
Question: Which of the following is NOT a definition of isosceles? A) Having two sides of equal length B) Having three equal sides C) Having all sides of different lengths D) Having two angles of equal measure Answer: C) Having all sides of different lengths
Question: Is π (pi) an irrational number? Answer: Yes, π is an irrational number. | 677.169 | 1 |
Angle alpha is the angle of longitude of the observer (how far north, or south, of the equator they are). For this example, I will be using a longitude of 50 degrees north (northern France/southern England).
The angle at B is 90 degrees (the angle between the radius from A and the horizon) plus the angle of elevation, beta, which we are looking to solve.
Lenth a is the straight-line distance from the observer to the satellite
Length b is the distance from the centre of the earth to the satellite, re (radius of earth) plus rs (altitude of satellite, measured from earth's surface)
Length c is the radius of the earth
We only know two of the lengths (b and c) and the included angle, alpha, so we must start solving the triangle by using the cosine rule:
In order to find angle B, and hence beta, we will first need to find length a.Substituting the known lengths and angle into the cosine rule, we get:
Now that we know all three sides, we can use the sine rule to determine an angle by knowing one other angle and the two opposite sides. I will calculate angle C and then subtract A + C from 180 degrees to find C.
a / sin A = c / sin C
a = 38,376,585 m (as calculated above)
A = angle of longitude of the observer
c = distance from Earth centre to geostationary satellite, which was calculated previously as 42,164,000 m.
C = angle on diagram; angle at satellite between centre of Earth and observer on the ground.
There is an alternative route to finding angle beta, and that's by dividing the triangle ABC into two right-angled triangles by dropping a perpendicular from B onto the line AC, see below. Angle Z = 90 degrees.
Firstly, calculate the distance BZ, which is common to both triangles ABZ and BCZ. This can be done by simple trigonometry since angle Z is 90 degrees, and angle A is known (or determined by the observer):
sin A = BZ/ re and where A = 50 degrees, BZ = 4,885,831 m.
Next, calculate AZ in the same triangle ABZ:
cos A = AZ/ re and where A = 50 degrees, AZ = 4099699 m.
As we now know AZ, we can calculate CZ, and hence identify two of the sides of triangle BCZ.
We want to know the angle between the observer's horizon and the satellite; since the angle AB and the horizon is 90 degrees ('the angle between a radius and a tangent is 90 degrees') this is simply 122.685 degrees - 90 degrees = 32.685 degrees... which agrees with the result from the first method.
Wednesday, 30 January 2013
Question: What is the radius of the Earth (length 'c')? Answer: 42,164,000 m
Question: What is the angle 'A' in the triangle ABC? Answer: 50 degrees
Question: Which two methods are described to find angle beta? Answer: Using the cosine rule and dividing the triangle into two right-angled triangles
Question: What is the angle of longitude of the observer in this example? Answer: 50 degrees north | 677.169 | 1 |
The point where the axes meet is the common origin of the two number lines and is simply called the origin. It is often labeled O and if so then the axes are called Ox and Oy. A plane with x and y-axes defined is often referred to as the Cartesian plane or xy plane. The value of x is called the x-coordinate or abscissa and the value of y is called the y-coordinate or ordinate.
The choices of letters come from the original convention, which is to use the latter part of the alphabet to indicate unknown values. The first part of the alphabet was used to designate known values.
In the Cartesian plane, reference is sometimes made to a unit circle or a unit hyperbola.
[edit] Cartesian coordinates in three dimensions
Choosing a Cartesian coordinate system for a three-dimensional space means choosing an ordered triplet of lines (axes), any two of them being perpendicular; a single unit of length for all three axes; and an orientation for each axis. As in the two-dimensional case, each axis becomes a number line. The coordinates of a point p are obtained by drawing a line through p perpendicular to each coordinate axis, and reading the points where these lines meet the axes as three numbers of these number lines.
Alternatively, the coordinates of a point p can also be taken as the (signed) distances from p to the three planes defined by the three axes. If the axes are named x, y, and z, then the x coordinate is the distance from the plane defined by the y and z axes. The distance is to be taken with the + or − sign, depending on which of the two half-spaces separated by that plane contains p. The y and z coordinates can be obtained in the same way from the (x,z) and (x,y) planes, respectively.
[edit] Generalizations
One can generalize the concept of Cartesian coordinates to allow axes that are not perpendicular to each other, and/or different units along each axis. In that case, each coordinate is obtained by projecting the point onto one axis along a direction that is parallel to the other axis (or, in general, to the hyperplane defined by all the other axes). In those oblique coordinate systems the computations of distances and angles is more complicated than in standard Cartesian systems, and many standard formulas (such as the Pythagorean formula for the distance) do not hold.
[edit] Notations and conventions
The Cartesian coordinates of a point are usually written in parentheses and separated by commas, as in (10,5) or (3,5,7). The origin is often labelled with the capital letter O. In analytic geometry, unknown or generic coordinates are often denoted by the letters x and y on the plane, and x, y, and z in three-dimensional space. w is often used for four-dimensional space, but the rarity of such usage precludes concrete convention here. This custom comes from an old convention of algebra, to use letters near the end of the alphabet for unknown values (such as were the coordinates of points in many geometric problems), and letters near the beginning for given quantities.
Question: What is the point where the x and y axes meet called? Answer: The origin | 677.169 | 1 |
A glide reflection is the composition of a reflection across a line followed by a translation in the direction of that line. It can be seen that the order of these operations does not matter (the translation can come first, followed by the reflection).
[edit] General matrix form of the transformations
These Euclidean transformations of the plane can all be described in a uniform way by using matrices. The result of applying a Euclidean transformation to a point is given by the formula
where A is a 2×2 orthogonal matrix and b = (b1, b2) is an arbitrary ordered pair of numbers;[5] that is,
where
[Note the use of row vectors for point coordinates and that the matrix is written on the right.] To be orthogonal, the matrix A must have orthogonal rows with same Euclidean length of one, that is,
and
This is equivalent to saying that A times its transpose must be the identity matrix. If these conditions do not hold, the formula describes a more general affine transformation of the plane provided that the determinant of A is not zero.
The formula defines a translation if and only if A is the identity matrix. The transformation is a rotation around some point if and only if A is a rotation matrix, meaning that
A reflection or glide reflection is obtained when,
Assuming that translation is not used transformations can be combined by simply multiplying the associated transformation matrices.
[edit] Affine Transformation
Another way to represent coordinate transformations in Cartesian coordinates is through affine transformations. In affine transformations an extra dimension is added and all points are given a value of 1 for this extra dimension. The advantage of doing this is that then all of the euclidean transformations become linear transformations and can be represented using matrix multiplication. The affine transformation is given by:
[Note the A matrix from above was transposed. The matrix is on the left and column vectors for point coordinates are used.] Using affine transformations multiple different euclidean transformations including translation can be combined by simply multiplying the corresponding matrices.
[edit] Scaling
An example of an affine transformation which is not a Euclidean motion is given by scaling. To make a figure larger or smaller is equivalent to multiplying the Cartesian coordinates of every point by the same positive number m. If (x,y) are the coordinates of a point on the original figure, the corresponding point on the scaled figure has coordinates
If m is greater than 1, the figure becomes larger; if m is between 0 and 1, it becomes smaller.
[edit] Shearing
A shearing transformation will push the top of a square sideways to form a parallelogram. Horizontal shearing is defined by:
Shearing can also be applied vertically:
[edit] Orientation and handedness
Main article: Orientation (mathematics) [edit] In two dimensions
Fixing or choosing the x-axis determines the y-axis up to direction. Namely, the y-axis is necessarily the perpendicular to the x-axis through the point marked 0 on the x-axis. But there is a choice of which of the two half lines on the perpendicular to designate as positive and which as negative. Each of these two choices determines a different orientation (also called handedness) of the Cartesian plane.
Question: What is the formula for applying a Euclidean transformation to a point? Answer: The formula is x' = A * x + b, where A is a 2x2 orthogonal matrix and b is an ordered pair of numbers.
Question: Does the order of operations in a glide reflection matter? Answer: No, the order of operations does not matter.
Question: What type of transformation is described by the formula if A is the identity matrix? Answer: A translation.
Question: What is an affine transformation? Answer: An affine transformation is a way to represent coordinate transformations in Cartesian coordinates by adding an extra dimension and using matrix multiplication. | 677.169 | 1 |
The usual way of orienting the axes, with the positive x-axis pointing right and the positive y-axis pointing up (and the x-axis being the "first" and the y-axis the "second" axis) is considered the positive or standard orientation, also called the right-handed orientation.
A commonly used mnemonic for defining the positive orientation is the right hand rule. Placing a somewhat closed right hand on the plane with the thumb pointing up, the fingers point from the x-axis to the y-axis, in a positively oriented coordinate system.
The other way of orienting the axes is following the left hand rule, placing the left hand on the plane with the thumb pointing up.
When pointing the thumb away from the origin along an axis, the curvature of the fingers indicates a positive rotation along that axis.
Regardless of the rule used to orient the axes, rotating the coordinate system will preserve the orientation. Switching any two axes will reverse the orientation.
[edit] In three dimensions
Once the x- and y-axes are specified, they determine the line along which the z-axis should lie, but there are two possible directions on this line. The two possible coordinate systems which result are called 'right-handed' and 'left-handed'. The standard orientation, where the xy-plane is horizontal and the z-axis points up (and the x- and the y-axis form a positively oriented two-dimensional coordinate system in the xy-plane if observed from above the xy-plane) is called right-handed or positive.
The name derives from the right-hand rule. If the index finger of the right hand is pointed forward, the middle finger bent inward at a right angle to it, and the thumb placed at a right angle to both, the three fingers indicate the relative directions of the x-, y-, and z-axes in a right-handed system. The thumb indicates the x-axis, the index finger the y-axis and the middle finger the z-axis. Conversely, if the same is done with the left hand, a left-handed system results.
Figure 7 depicts a left and a right-handed coordinate system. Because a three-dimensional object is represented on the two-dimensional screen, distortion and ambiguity result. The axis pointing downward (and to the right) is also meant to point towards the observer, whereas the "middle" axis is meant to point away from the observer. The red circle is parallel to the horizontal xy-plane and indicates rotation from the x-axis to the y-axis (in both cases). Hence the red arrow passes in front of the z-axis.
Figure 8 is another attempt at depicting a right-handed coordinate system. Again, there is an ambiguity caused by projecting the three-dimensional coordinate system into the plane. Many observers see Figure 8 as "flipping in and out" between a convex cube and a concave "corner". This corresponds to the two possible orientations of the coordinate system. Seeing the figure as convex gives a left-handed coordinate system. Thus the "correct" way to view Figure 8 is to imagine the x-axis as pointing towards the observer and thus seeing a concave corner.
[edit] Representing a vector in the standard basis
Question: Which hand is used to determine the direction of the axes in a right-handed coordinate system? Answer: The right hand. | 677.169 | 1 |
Related Products
Catenary defines the shape of a line strung between two points. It is not the same math as a parabolic curve. I am looking for a program that will do catenary calculations to determine the change in sag of a line if the line is longer or shorter and takes into account the different elevations of the end points. Can you help me find such a program??
Question: What does the user ask at the end of the text? Answer: Can you help me find such a program? | 677.169 | 1 |
AB = BC = AC => ABC is an equilateral triangle=> angle BAC is 60 degree
that means angle DAC = 60 degree
As per the rule - angle made by the arc on center is twice the angle made by the same arc on the circumference. Here the angle DAC made by the arc DC on the center is 60 degree. So the DBC = 30.
As angle ABC = 60 degree, the angle ABD = x = 30 degree
from given info and figure it is clear that AB=BC=AC=AD=DC. Here basically BD connects two opposite vertices of 2 equi. triangles with common base AC. Hence, it should be dividing the the angles B and D into half. i.e. 60/2 = 30
Question: Is ABC an equilateral triangle? Answer: Yes | 677.169 | 1 |
In this lively introduction to shapes and polygons, a bored triangle is turned into a quadrilateral after a visit to the shapeshifter. Delighted with his new career opportunities--as a TV screen and a picture frame--he decides the more angles the better, until an accident teaches him a lesson. Includes special teaching section. Full color.
Reviews
PreS-Gr 1'An offbeat introduction to geometry. When a triangle tires of having only three sides, he asks the shapeshifter to change him first into a quadrilateral, then a pentagon, a hexagon, and so forth until he realizes he is happiest as a triangle: he can hold up a roof, be a slice of a pie and, best of all, slip into place when people put their hands on their hips. ``That way I always hear the latest news...which I can tell my friends.'' The text is clever and shows more than the usual places to find shapes'part of a computer screen, a section of a soccer ball, a floor tile. The acrylic and colored-pencil illustrations are colorful, abstract, and filled with smiling shapes done in shades of turquoise, pink, and yellow. A two-page spread of suggestions for adults to reinforce the math lessons featured is included at the end of the book.'Christine A. Moesch, Buffalo and Erie County Public Library, NY
The author of The I Hate Mathematics Book celebrates geometric shapes in this informative but visually cluttered addition to the Marilyn Burns Brainy Day series. Her main character, a triangle with gleaming black eyes and a perky grin, leads a full life-it can take the shape of a slice of pie or rest in an elbow's angle ``when people put their hands on hips.'' Yet the triangle aspires to greater complexity, so it asks a ``shapeshifter'' to turn it into a quadrilateral (the shape of a TV or a book's page), then into a pentagon (a house's facade) and so forth. Burns fails to show that the triangle is ``greedy''; it's just adventurous. But her story successfully introduces basic polygons, and her afterword to adults suggests ways of teaching children some of the finer points about geometry (e.g., the concept of a plane or rhomboids). For his picture book debut, Silveria chooses tart shades of yellow, orange, lavender and green. His airbrushed colored-pencil compositions have suitably angular details; speckled paint and multicolored doodles soften the effect but create a sense of disorder. If the art as a whole is somewhat jumbled, readers still come away from this volume noticing and naming the shapes of the objects around them. Ages 6-9. (Mar.)
I read this story to my class after teaching shapes but it would have been a perfect launch prior to starting to teach children about shapes. The story was great because it had pictures that linked to real life contexts as well as how adding 1 angle and 1 side can change a shape. The story is good because some parts are repetitive which the children liked.
You can earn a 5% commission by selling The Greedy Triangle
Question: What are some real-life examples of triangles mentioned in the story? Answer: A slice of pie, a roof, and when people put their hands on their hips
Question: What does the triangle initially ask the shapeshifter to do? Answer: Change it into a quadrilateral | 677.169 | 1 |
What are an Oval and an Egg Curve? There is no clear definion. Mostly you define:
......
An oval is a closed plane line, which is like an ellipse or like the
shape of the egg of a hen.
An egg curve only is the border line of a hen egg.
The hen egg is smaller at one end and has only one symmetry axis.
The oval and the egg shaped curve are convex curves, differentiate twice
and has a positive curvature.
......
You distinguish between the oval, the ovoid and the oval shape in the
same way as between the circle, the figure of the circle and the sphere.
Ellipses and its changings
top Ellipse All points P, for which the distances of two fixed points or foci F1
and F2 have a constant sum, form an ellipse.
The ellipse in the centre position has the following cartesian equation.
The parameters a and b are called lengths of axis.
The ellipse is the formula of a relation.
......
The ellipse on the left has the equation
The constant sum is 2a=6.
...
You can add two halves of different ellipses to form a chicken egg.
A Gardener's Construction
You can draw an egg curve, if you wrap a rope (green) around an isosceles
triangle and draw with taut rope a closed line (1). The rope must be a
little bit longer than the circumference of the triangle. Elllipse arcs
develop, which together form an egg shaped curve (2).
The three main ellipse are totally drawn in a computer simulation (2,black,
red, blue, book 9). You are exacter, if you draw three more ellipses in
the sector of the vertical angles of the triangle angles to the sides AB,
AC und BC (3,4) .
Super Ellipse
......
If you take the exponent 2.5 instead of 2 in the equation (x/a)²+(y/b)²=1,
you get the equation of a super ellipse:
The modulus | | makes sure that the roots are defined.
In the drawing there is a=3 and b=2.
The Danish author and scientist Piet Hein (1905-1996) dealt with the super
ellipse in great detail (book 4). In particular that the shape made by
rotation around the x-axis can stand on the top, if it is made from wood.
You don't have to use power in contrast to the Columbus' egg.
The super ellipse belongs to the Lamé curves. They have the
equations
......
In the drawing there is a=3, b=2 and you substitute n with 1(parallelogram,
blue), 1.5(green), 2(Ellipse, bright red), 2.5 (super ellipse, red), and
3 (black).
From the Oval to
the Egg Shape You can develop the shape of a hen egg, if you change the equation
Question: What is the equation of a super ellipse? Answer: The equation of a super ellipse is (x/a)² + (y/b)² = 1, where the exponent of the terms is 2.5 instead of 2.
Question: How can you develop the shape of a hen egg from an oval? Answer: By changing the equation of an oval and adding more ellipses in the sector of the vertical angles of the triangle, you can create a hen egg shape.
Question: What is the mathematical formula for an ellipse? Answer: The standard equation of an ellipse with its center at the origin is (x/a)² + (y/b)² = 1, where a and b are the lengths of the semi-major and semi-minor axes, respectively. | 677.169 | 1 |
We have four rods of equal lengths hinged at their endpoints to
form a rhombus ABCD. Keeping AB fixed we allow CD to take all
possible positions in the plane. What is the locus (or path) of the
point D?
Four rods of equal length are hinged at their endpoints to form a
rhombus. The diagonals meet at X. One edge is fixed, the opposite
edge is allowed to move in the plane. Describe the locus of the
point X and prove your assertion.
LOGO Challenge 2 - Diamonds Are Forever
Stage: 2, 3 and 4 Challenge Level:
There are many ways to tackle the challenge.
You might like to consider returning to this challenge when you
have found out about procedures. For now you might like to see how
you might use the REPEAT command.
Here are some ideas:
Can you draw one rhombus?
Can you write a single line of code that draws the two rhombi
joined together?
You can change the pencolour by using the command SETPC
followed by a number between 0 and 15.
Try drawing pairs of rhombi in different colours and in
different arrangements. For example, what does the following line
of code produce?:
Question: What command can be used to change the pencil color in the LOGO programming language? Answer: SETPC | 677.169 | 1 |
There are
two pictures to keep in mind here, the generic right triangle:
First of
all, all physical angles have some size. We cannot visualize
an angle with negative physical size. They do not exist in anything
sufficiently similar to the physical space we live in.
However,
(especially when dealing with the unit circle), it is often
convenient to measure angles in a specific direction: counterclockwise.
[I'm not going to digress into "stratosphere" math now, but
that is dictating the convention here.]
Then, a
negative sign means we are measuring the angle "unconventionally",
i.e. clockwise.
This will
simplify the use of some of the trigonometric identities we
are going to look at.
Why
should I know the generic right triangle?
Given
a right triangle, the trig function values for the two acute
angles [angles smaller than a right angle] can be computed without
knowing the angles. I prefer to remember the formulae this way
[X is an angle]:
A mnemonic
for the formulae for sin(X), cos(X), and tan(X) is the (fictitious)
Indian Chief Soh-cah-toa, who had no problems with this part
of trigonometry. [I got this from Mr. Coole, a long time ago
-- I was in grade school then.]
The way
I wrote the formulae, above, emphasizes the following identities:
sin(X)csc(X)=1
cos(X)sec(X)=1
tan(X)cot(X)=1
These identities
do work hold when both functions involved are defined for the
angle X, regardless of size.
Another
fact, of some use, is the Pythagorean theorem: H²=A²+O². If
we relabel A as a, O as b, and H as c, we get the familiar form
of the Pythagorean theorem: c²=a²+b². This *only* works for
right triangles. The generalization of the Pythagorean theorem to non-right
triangles is called the law of cosines.
A common
strategy is to memorize how to compute tan(X), cot(X), sec(X),
and csc(X) in terms of sin(X) and cos(X), and then reduce everything
to this. This is not necessarily the least painful way to do
a trig problem, but it is often more important to start the
problem, than figure out how to do it elegantly. [EXERCISE:
derive this from the A, H, O formulation for acute angles. The
formulae work for arbitrary angles.]
The generic
right triangle also motivates some terminology [which we inherited
from twelfth century Arabian mathematics].
First of
all, we say two angles are complementary if they
add up to a right angle. That is, for an angle X, we say -X
is complementary to X. For instance,
the angle
in the upper right-hand corner is complementary to, i.e. the
Question: What are the trigonometric identities mentioned in the text? Answer: sin(X)csc(X)=1, cos(X)sec(X)=1, tan(X)cot(X)=1
Question: What does it mean for two angles to be complementary? Answer: Two angles are complementary if they add up to a right angle
Question: What is the direction convention used to measure angles in the unit circle? Answer: Counterclockwise | 677.169 | 1 |
Since 30°
is the complementary angle to 60°, we also have computed sin(30°)
and cos(30°).
How
does "wrapping the real number line around the unit circle"
work?
The unit
circle has a circumference of 2p. "Thus", all trig functions
will have the same value when evaluated 2p radians apart. We
say that all trig functions have a period. [In "stratosphere"
math, this period is arrived at in a very different way.]
The trig
function period identities are [X is an angle, n is a positive
integer]:
These identities
hold even in the undefined case [if one side is undefined, they
both are.]
To formally
demonstrate the n part, I would use natural induction, which
should be buried somewhere in College Algebra. If you don't
recall this term clearly, don't worry about it. However, I'm
only going to explain it for n=1.
We read
these from the unit circle immediately:
sin(X+[circle])
= sin(X)
cos(X+[circle]) = cos(X)
We use the
rewrite of sec(X) and csc(X) in cos(X) and sin(X) to derive
these:
sec(X+[circle])
= sec(X)
csc(X+[circle]) = csc(X)
Now, to
deal with tan(X) and cot(X), we have to be a little more clever.
For tan(X), write:
i.e. tan(X+[½][circle])=tan(X).
The identity cot(X+[½][circle]) = cot(X) is similar, but works
with the multiplicative reciprocals throughout.
What
are the triangle area formulae?
There are
two basic formulae, and one "impractical" one.
"½
base times height"
(Area)=(½)(length
of base)(length of height)
To use this
formula, pick one side of the triangle as the "base". Note its
length. Then, draw a perpendicular line segment from the vertex
of the triangle not in the "base", to the "base", and note this
line segment's length (this is the height). [This formula does
work for obtuse triangles. Extend the base to where it would
hit the perpendicular line.].
This formula
is easily visualized for a right triangle [a rectangle with
a line segment between two opposite vertices gives two right
triangles, both clearly with half the original area since they
are congruent]. It behaves reasonably for line segments [it
gives zero area, which is correct for a line segment; either
base or height is zero for a line segment.]
EXERCISE:
learn to use this formula by applying it to the reference triangles.
The 45-45-90 triangle with hypotenuse should
have area ½, and the 30-60-90 triangle with hypotenuse length
2 should have area .
Question: What is the period of all trigonometric functions as mentioned in the text? Answer: 2π radians
Question: How can you find the area of a triangle using the "½ base times height" formula for an obtuse triangle? Answer: Extend the base to where it would hit the perpendicular line.
Question: How does "wrapping the real number line around the unit circle" affect the trigonometric functions? Answer: It causes all trigonometric functions to have the same value when evaluated 2π radians apart, demonstrating their periodicity.
Question: Which of the following is NOT a triangle area formula mentioned in the text? A) ½ base times height B) base times height C) ½ base squared Answer: C) ½ base squared | 677.169 | 1 |
3D Shape Properties. Game.
A simple poster with a range of shapes listing their properties. Also, a "cut-out-and-use-as-you-want" sheet with outlined images of various 3D shapes. I've used the first in an introductory lesson and the cut-out sheet for a range of starter activities.
Very helpful resource, thank you. By the way, you were correct. A cone doesn't have a vertex. A vertex is only when two lines meet or intersect. Although in saying this, having looked on the internet - this seems to be an area of much discussion!
Question: What is a vertex according to the text? Answer: A vertex is only when two lines meet or intersect. | 677.169 | 1 |
7 is such a sharp number, pushed in between the sensuous curves of 6 and 8. All awkward angles and points, 7 is not a graceful number. It's odd. And sticks out in all the wrong places. 6 and 8 bend like dancers, Swaying or flowing as natural as the breeze. And poor 7 sandwiched in between them, like a middle school kid, all unsure and out of place.
Question: Which number is described as having angles? Answer: 7. | 677.169 | 1 |
Checkpoint
- Course 2, Unit 2
Patterns of Location, Shape, and Size
In each unit, there is a final lesson and Checkpoint that helps students
summarize the key ideas in the unit. The final Checkpoint will generally
be discussed in class, with the teacher facilitating the summarizing,
and students making notes in their Math Toolkits (teachers
may just refer to this as "notes") of any points they need to remember,
adding illustrative examples as needed. If your student is having difficulty
with any investigation in this unit, this Checkpoint and the accompanying
answers may help you recall the concepts involved, and give you the big
picture of what the entire unit is about. If your student has completed
the unit, then a version of this should be in his or her notes or toolkit.
Students should also have Technology Tips in their toolkits, which may
be useful for this unit.
Possible
Responses to Unit Summary Checkpoint
In this unit, students investigated how coordinates
and matrices can be used to model geometric ideas. The bolded words
are vocabulary and concepts your student should be familiar with.
a.
Describe several different
ways that coordinates are used to model geometric ideas. Illustrate
with examples. Coordinates can be used to model points, lines, shapes, distance, slope,
and transformations. Examples: points could be named (2, 1),
(-3, 0.5), (a, b), (c, d), (x, y); lines can
be created by joining points, such as (2, 1) and (-3, 0.5)
and the equation of the line can be written by finding the slope
between 2 points, and substituting the slope and a point into
the equation y = mx + b; lines can be named y = -2x + 3,
y = 4, x = -2.5; shapes can be created by connecting
in order (0, 0), (2, 3), (1, 5), (-4, 1), and
back to (0, 0); the distance from (2, 3) to (5, -4)
is the square root of (5 - 2)2 + (-4 - 3)2,
a result that becomes an obvious application of the Pythagorean Theorem
if the two points are used to create the hypotenuse of a right triangle;
the slope from (a, b) to (c, d) is (d - b)/(c - a),
a result familiar to students from their work in Course 1; patterns
of change from preimage points (2, 1) and (-3, 0.5) to
image points (4, 2) and (-6, 1) can be discerned from the
coordinates (size transformation in this case since both x-
and y-coordinates are multiplied by the same scale factor, 2). Other
kinds of transformations are as follows: translations, reflections, rotations.
Question: What are the Technology Tips that students should have in their toolkits? Answer: The Technology Tips are not explicitly mentioned in the text, but they are referred to as something that may be useful for this unit.
Question: What kind of transformation can be observed in the pattern of change from preimage points (2, 1) and (-3, 0.5) to image points (4, 2) and (-6, 1)? Answer: A size transformation, as both x- and y-coordinates are multiplied by the same scale factor, 2.
Question: How can the distance between two points be calculated using coordinates? Answer: The distance from (2, 3) to (5, -4) can be calculated using the formula √[(x2 - x1)² + (y2 - y1)²], which is an application of the Pythagorean Theorem.
Question: What is the main purpose of the Checkpoint in each unit? Answer: The main purpose of the Checkpoint is to help students summarize the key ideas in the unit and recall concepts if they're having difficulty with any investigation. | 677.169 | 1 |
e)If it is a star, then it is a small pinprick of light in the night sky.
f) A really cool sneaker.
Question 65974: These two definationes are agreed upon:
a vertical lline is a line containing the center of the earth
a horizontal line is a line perpendicular to some vertical line
Which of the following would be true:
a. Could two horizontal lines be parallel
b. could two vertical lines be parallel
c. could two horzontal lines be perpendicular
d. could two vertical lines be perpendicular
e. could every horzontal line be a vertical line
f. could everty vertical line be a horizontal line
g. could a horizontal line be parallel to a vertical line
h. would every line be horizontal Click here to see answer by uma(370)
Question 68261: Please help me solve the following proof:
The directions are:
Draw and label a diagram. List, in terms of the diagram, what is given and what is to be proved. Then write a proof.
The problem states:
If the vertex angle of one isosceles triangle is congruent to the vertex angle of another isosceles triangle, then the triangles are similar.
I made two triangles, triangle ABC and triangle DEF
I think the given is supposed to be: angle A is congruent to angle D
I think we need to prove: triangle ABC is similar to triangle DEF
What are the steps to the rest of this proof? Click here to see answer by stanbon(57274)
Question 73426: Please help i dont understand proofs:
Given: MI=LD
Prove:ML=ID
for reasons name the propety that makes the statement is true
Statements Reasons
MI=LD Given
IL=IL Reflexive
MI+IL=LD+IL I think addition
MI+IL=ML;LD+IL=ID I have no clue
ML=ID I think substitution Click here to see answer by stanbon(57274)
Question 75900: Select the best description for the curve given by the parametric equation
x = t-5, y = 2t-1, t in R.
Choices for answers are:
a.) a straight line with slope
b.) a straight line through the point (5,1)
c.) a parabola with vertex (-5,1)
d.) a circle centered at (5, -1) Click here to see answer by stanbon(57274)
Question 81613: Write a proof argument for part (b) of the sufficient conditions for a parallelogram theorem: if both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram.
given: that the opposite sides are congruent.
prove: that the figure ABCD is a parallelogram
I drew in diagonals in the figure ABCD and labeled the point where they intersect E.
Question: Could two horizontal lines be parallel? Answer: Yes, two horizontal lines could be parallel.
Question: What is the best description for the curve given by the parametric equation x = t-5, y = 2t-1, t in R? Answer: A straight line with slope.
Question: What is the given in the proof about isosceles triangles? Answer: The vertex angle of one isosceles triangle is congruent to the vertex angle of another isosceles triangle.
Question: What property is used to prove ML=ID in the given proof? Answer: The property of addition of lengths. | 677.169 | 1 |
Since the earliest times, mankind has employed the simple geometric forms of straight line and circle, in art, architecture, and mathematics. Originally marked out by eye and later using a stretched cord, in time these came to be made with the simple tools of ruler and compass. This valuable reference book introduces the origins and principles of geometry using these basic tools, and shows some of the geometric constructions used by artists, architects, and mathematicians of old.
You can earn a 5% commission by selling Ruler and Compass: Practical Geometric Constructions (Wooden Books
Question: What were the initial methods used to mark out these shapes? Answer: Marked out by eye and later using a stretched cord | 677.169 | 1 |
Physics suppose that the separation between speakers A and B is 5.80 m and the speakers are vibrating in phase. They are playing identical 135 Hz tones, and the speed of sound is 343 m/s. What is the largest possible distance between speaker B and the observer at C, such that he ...
Tuesday, May 1, 2007 at 9:05pm by Mary
MATH A is the midpoint of BC. BA=2x+5 and BC=5x-3. solve for x and find the length of BC
Sunday, January 16, 2011 at 7:05pm by BOB
chemistry from UCI!!!!! 1 is equal bc of equilibrium 2 is endothermic bc of absorbing heat 3 is absorb....bc it is endothermic yay all finish!
Monday, January 25, 2010 at 11:01pm by Alex Le
solid mensuration The circle radius is 5 cm. That comes from the area. The triangle ACB is a right triangle with AB as a diameter. The length of AB is 10 cm, since it is a diameter (AC)^2 + (BC)^2 = (AB)^2 = 100 Area = (1/2)(AB)*(BC) = 11 (AC)*(BC) = 22 (BC)=22/(AC) (AC)^2 + 484/(AC)^2 = 100 ...
Saturday, December 1, 2012 at 7:13am by drwls
math Assume the 2x2 matrix to be A= \ a b c d Do the matrix multiplication AA and equate each element to A, a=a²+bc b=b(a+d) c=c(a+d) d=bc+d² From which we conclude a+d=1 or a=1-d and bc = a-a² = d-d² Take a=4, then d=-3 bc=4-4²=-12, If ...
Thursday, November 25, 2010 at 3:56pm by MathMate
11th grade Angle A=45, AB=10, and BC=8. This is an SSA configuration which could result in 0, 1 or two distinct triangles, depending on the length of the "dangling" leg BC, i.e. the side which does not touch the given angle A. To determine which case applies, we construct a ...
Thursday, June 3, 2010 at 10:08pm by MathMate
Math - explanation This is an example in the text book. Using vectors, demonstrate that the three points A(5, -1), B(-3,4) and C(13,-6) are collinear. Solution AB = (-8, 5) BC = (16, -10) Then BC = 2AB AB and BC have the opposite direction, so the points A, B, and C must be collinear. I don'...
Question: What is the length of AC in the fourth text? Answer: 4 cm (since (AC) * (BC) = 22 and BC = 22/(AC), solving for AC gives 4 cm)
Question: Who is the author of the second text? Answer: BOB
Question: Who is the author of the first text? Answer: Mary
Question: What is the speed of sound given in the first text? Answer: 343 m/s | 677.169 | 1 |
Showing 1 comment
A Parallelogram is a quadrilateral which has two parallel pair of sides.
parallelograms are:
Properties of
--- The opposite sides of a parallelogram are equal in length.
--- The opposite angles are ...
First we will see what the meaning of antiderivation is? An antiderivative of a
function f can be thought as a function F whose derivative is equal to f or we
can say that F' = f.
The method used to degree of polynomial is the greatest exponent of a term. The greatest exponent should
have a non-zero coefficient in a polynomial expressed as a sum or difference of terms which
is commonly knownUnderstanding of Median is important before knowing Centroid . Median is the line joining the
vertex with the midpoint of the opposite side. Centroid of a Triangle is Point of intersection of
all its ...
b. Periodontal ligaments. (1) Dense connective fibrous tissues that connect teeth to the alveolar bone. (2) One end is embedded in cementum and other end in bone. (3) Supports and protects the tooth from normal shock. Dental Anatomy
Question: What is an antiderivative of a function? Answer: An antiderivative of a function is a function whose derivative is equal to the original function. | 677.169 | 1 |
Since its a regular polygon its one vertex will be at the mid point of the side of the square by symmetry.
Therefore,
1/4+((1-x)/2)^2=x^2
x= (sqrt(28)-2)/6
if we anlyze dis problem then we are getting four 30-60-90 triangles on four peripherals so as we know 1:2:rt3 ratios we will end up wid 1/rt3 as the side of the hexagon but still am in dilemma as then i guess square would end it up as rectangle as one side would be unity and other would be 2/rt3.....pls throw some light
since, requirement is of smallest no. let us take 3 odd divisors as- 1,3,5 and even divisors as - 2 4 6, which when multiplied by the odd divisors will give 9 even divisors as 2*1, 2*3, 2*5, 4*1, 4*3, 4*5, 6*1, 6*3, 6*5
so we have 3 odd divisors 1, 3, 5
and 9 even divisors as 2,6,10,4,12,20,6,18,30
here we see, 30 is the highest no. but 4,12,18 and 20 are not its multiple..so the required number should be a multiple of 30..next highest number is 60...which serves are purpose
hello sir,
My ans before was incorrect. as the ans is 72. Since it has 3 odd factors, hence the no should contain an odd square number. The smallest of such no is. 9=3*3.
Total no of factors is =12
so the remaining factors should be even and should contain 2^3.
Hence the no is= 3^2*2^3=72
ans would be 72 as we know no. of factors needed is 12 so start with 2 and 3,let say 2^a *3^b is the required format of the number's factors,on listing all the combination of a and b as (a+1)*(b+1)=12,we will come to a conclusion ,the required a and b are 3 and 2 resp.thereby 2^3*3^2=72
Question: What is the side length of the hexagon? Answer: 1/sqrt(3)
Question: What are the odd divisors of 72? Answer: 1, 3, 5, 9
Question: What is the value of 'x' in the given equation? Answer: x = (sqrt(28) - 2) / 6 | 677.169 | 1 |
A hemisphere of radius 8 is inscribed in a cylinder of radius 8 and height 8. The figure shows top and side views of the hemisphere, the cylinder, and a cone whose radius and height are both 8, and whose base and vertex are coplanar with the bases of the cylinder.
Consider that part of the cylinder that is outside (above) the hemisphere. Slice this region by a plane that is parallel to the cylinder base and 5 inches above the equator of the hemisphere. The intersection is a ring between two concentric circles. The same plane slices the cone, creating a disk. Show that the ring and the disk have the same area.
(Continuation) Suppose now that the three radii and two heights are all r. Show that the ring and the disk have the same area, no matter what the height of the slicing plane.
(Continuation) If the cone were filled with liquid, it could be poured into the cylinder, which still has the hemisphere stuck in the bottom. Will all the liquid fit? Expressed in terms of r, what is the volume of the cone? of the empty cylinder? of the hemisphere?
(Continuation) Show that a sphere of radius r encloses a volume of (4/3)πr3.
----
The rectangle shown has been formed by fitting together four right triangles. As marked, the sizes of two of the angles are α and β (Greek "alpha" and "beta"), and the length of one segment is 1. Find the two unmarked angles whose sizes are α and α + β. By labeling all the segments of the diagram, discover formulas for sin(α+β) and cos(α + β), written in terms of sin α, cosα, sinβ, and cosβ.
----
Imagine covering an unlimited plane surface with a single layer of pennies, arranged so that each penny touches six others tangentially. What percentage of the plane is covered?
----
The table shows the results of a probability experiment. Forty-eight dice were rolled, and each die that showed "2" on top (a deuce) was removed. The remaining dice were rolled again, and deuces were removed. This procedure was repeated until all the dice were gone.
Each entry in the left column is a roll number. The corresponding entry in the right column is the number of dice that had not yet turned up deuces after this roll.
These variables are not linearly related. What happens if you try to apply logarithms to straighten this data?
(Continuation) If forty-eight dice are rolled once, how many do you expect to remain after removing the deuces? How many dice do you expect to remain after two applications of the above procedure? After ten applications?
(Continuation) If a prize were offered for predicting how many rolls will be needed to remove all 48 dice, what would your guess be? Explain.
----
Question: What is the volume of the hemisphere in terms of r? Answer: (2/3)πr³
Question: What is the height of the cylinder? Answer: 8 inches
Question: What is the radius of the hemisphere and the cylinder? Answer: 8 inches | 677.169 | 1 |
The figure at right shows an outermost 1 × 1 square, within which appears an inscribed circle, within which appears an inscribed square, within which appears another inscribed circle, within which appears another inscribed square. Although the figure does not show it, this process can be continued indefinitely. Let L1 = 1 be the length of a side of the first (largest) square, L2 be the length of a side of the second square, L3 be the length of a side of the third square, and so on. Show that the numbers L1, L2, L3, ... form a geometric sequence, and calculate L20.
(Continuation) Let An be the area of the nth square. Is the sequence A1, A2, A3, ... geometric? Explain.
----
When the note middle C is struck on a piano, it makes a string vibrate at 262 cycles per second. When the corresponding note one octave higher (denoted C') is struck, it makes a string vibrate at 524 cycles per second (twice as fast as the first string). These two numbers form part of a geometric sequence of frequencies
. . ., C, C#, D, D#, E, F, F#, G, G#, A, A#, B, C', . . .
known as equal-tempered tuning. Given C = 262 and C'= 524, find the frequency of G, and the frequency of the note that is n steps above middle C. (For example, F is 5 steps above C.)
(Continuation) Without calculating any of the indicated frequencies, explain why the ratio G:C is the same as the ratio A:D, which is the same as the ratio B:E. Now calculate the ratio G:C and show that it is approximately equal to 3:2, the musical interval known as a fifth. The disagreement between the two ratios is why some violinists do not like to make music with pianists.
----
On the same system of coordinate axes, graph the circle x2 + y2 = 25 and the ellipse 9x2 + 25y2 = 225. Draw the vertical line x = 2, which intersects the circle at two points, called A and B, and which intersects the ellipse at two points, called C and D. Show that the ratio AB :CD of chord lengths is 5:3. Choose a different vertical line and repeat the calculation of the ratio of chord lengths. Finally, using the line x = k (with |k| < 5, of course), find expressions for the chord lengths and show that their ratio is 5:3. Where in the diagram does the ratio 5:3 appear most conspicuously? Because the area enclosed by the circle is known to be 25π, you can now deduce the area enclosed by the ellipse.
(Continuation) What is the area enclosed by the ellipse x2/a2 + y2/b2 = 1?
----
Question: What is the frequency of the note that is n steps above middle C? Answer: The frequency of the note that is n steps above middle C is 262 * (2^(n/12)) cycles per second.
Question: Is the sequence A1, A2, A3,... also a geometric sequence? Answer: No, the sequence A1, A2, A3,... is not a geometric sequence because the ratio of consecutive terms is not constant. For example, A2/A1 = (1/2)^2 = 1/4, but A3/A2 = (1/2)^2 * (1/4) = 1/16, which is not equal to 1/4.
Question: Is the sequence L1, L2, L3,... a geometric sequence? Answer: Yes, the sequence L1, L2, L3,... is a geometric sequence with the first term (L1) being 1 and the common ratio (r) being 1/2. | 677.169 | 1 |
Two circles of radii 9 and 17 centimetres are enclosed within a
rectangle with one side of length 50 cm. The two circles touch each
other, and each touches two adjacent sides of the rectangle. Find the
perimeter of the rectangle.
Given a 3,4,5 triangle, and inside it, inscribed, two circles of equal
radii. Both circles touch one leg and the hypotenuse, and both are
tangent to each other at one point. Find the radius of the circles.
Dr. Ian replies to a question about motivating students to learn about
special quadrilaterals with some excellent general advice on
motivation, math education, and real life application of mathematics.
Question: What is the type of triangle mentioned in the text? Answer: 3,4,5 triangle | 677.169 | 1 |
Its C....its necessary to know the angle to be able to draw a circle around a quadrilateral.The sides of a quadrilateral may be equal but if the angles of the sides are obtuse or acute we can't draw a circle.
Vivek.
_________________
"Start By Doing What Is Necessary ,Then What Is Possible & Suddenly You Will Realise That You Are Doing The Impossible"
well I understand the question now. All the vertices of Quadrilateral should lie on the circumference.
I am also modifying my post. An acute rohmbus will not fit in a circle such that all vertices lie on the circumference. Looks like B is the answer But I have a following doubt.
If you make a rectangle very thin then the circle has to have an infinite radius. Limit of width of rectangle -> 0 R -> infinity
Whereas if you make all sides equal and all the angles = 90 degrees then definitely you can draw a circle around it.
Question: What happens to the radius of the circle around a rectangle as its width approaches zero? Answer: The radius approaches infinity. | 677.169 | 1 |
to follow, and hundreds of others have been given. Here's a nice one given by
Thabit ibn-Qurra (826-901).
Proof: Start with the right triangle ABC with right angle at C. Draw a square
on the hypotenuse AB, and translate the original triangle ABC along this square
to get a congruent triangle A'B'C' so that its hypotenuse A'B' is the other
side of the square (but the triangle A'B'C' lies inside the square). Draw perpendiculars
A'E and B'F from the points A' and B' down to the line BC. Draw a
line AG to complete the square ACEG.
Note that ACEG is a square on the leg AC of the original triangle. Also, the
square EFB'C' has side B'C' which is equal to BC, so it equals a square
on the leg BC. Thus, what we need to show is that the square ABB'A' is equal to
the sum of the squares ACEG and EFB'C'.
But that's pretty easy by cutting and pasting. Start with the big square ABB'A'.
Translate the triangle A'B'C' back across the square to triangle ABC, and
translate the triangle AA'G across the square to the congruent triangle BB'F.
Paste the pieces back together, and you see you've filled up the squares ACEG and
EFB'C'. Therefore, ABB'A' = ACEG + EFB'C',
as required.
Q.E.D.
Similar triangles
Two triangles ABC and DEF are similar if (1) their corresponding
angles are equal, that is, angle
A equals angle D, angle B equals angle E, and angle C equals angle
F, and (2) their sides are proportional, that is, the ratios of the three corresponding sides
are equal:
ABDE
=
BCEF
=
CAFD
In fact, as Euclid showed, each of these two conditions implies the others. That is to say,
if corresponding angles are equal, then the three ratios are equal
(Prop. VI.4), but if
the three ratios are equal, then corresponding angles are equal
(Prop. VI.5).
Thus, it is enough to know either that their corresponding angles are equal or that
their sides are proportional in order to conclude that they are similar triangles.
Typically, the smaller of the two similar triangles is part of the larger. For
example, in the diagram to the left, triangle AEF is part of the triangle ABC,
and they share the angle A. When this happens, the opposite sides, namely
BC and
EF, are parallel lines.
This situation frequently occurs in trigonometry applications, and for many of
those, one of the three angles A, B, or C is a right angle.
Question: Which of the following is NOT a part of the proof given by Thabit ibn-Qurra? A) Drawing a square on the hypotenuse B) Translating the triangle along the square C) Showing that the sum of the squares is equal to the square on the hypotenuse Answer: C) Showing that the sum of the squares is equal to the square on the hypotenuse (This is the conclusion, not a part of the proof)
Question: What is the name of the shape formed by drawing a square on the hypotenuse of a right triangle and translating it? Answer: A square | 677.169 | 1 |
Because the two triangles are congruent, this means that their corresponding parts are equal. So their two longer legs are equal which means that AB = DE = 24 cm
This means that AB = 24 cm
Also, their two hypotenuses are equal. So AC = DF = 25 cm which means that DF = 25 cm
----------------------------
Also, congruent triangles have equal corresponding angles. Since angle BAC is the angle between the longer leg and the hypotenuse for the first triangle and angle EFD is the same for the second triangle, this makes
angle BAC = angle EFD = 16 degrees
So angle BAC = 16 degrees
The same applies to angles ACB and EFD. These angles are also corresponding angles. So this means that
angle ACB = angle EFD = 74 degrees
which means that angle EFD = 74 degrees
Now from before, it was stated that the sum of all angles in a triangle add to 180 degrees. Since we know two angles of the first triangle, this means that we can find the remaining angle by first adding the two angles to get: 74+16=90
Now subtract this result from 180 to get 180 - 90 = 90
So the remaining angle ABC is 90 degrees which means angle ABC = 90
Because angles ABC and DEF are corresponding angles, this tells us that angle ABC = angle DEF = 90
Now simply connect every vertex with a line. Make sure that EVERY point has two lines connecting to it (so a bridge doesn't form). In addition to connecting every vertex, draw a line from any one vertex to itself to form a loop (in green):
Question: What is the measure of angle DEF? Answer: 90 degrees | 677.169 | 1 |
No Need to change the TriangleError code for either challenge. You just need to check for invalid triangles and raise the error if the triangle isn't.
def triangle(a, b, c)
if a==0 && b==0 && c==0
raise TriangleError, "This isn't a triangle"
end
if a <0 or b < 0 or c <0
raise TriangleError, "Negative length - thats not right"
end
if a + b <= c or a + c <= b or b + c <= a
raise TriangleError, "One length can't be more (or the same as) than the other two added together. If it was the same, the whole thing would be a line. If more, it wouldn't reach. "
end
# WRITE THIS CODE
if a == b and b == c
return :equilateral
end
if (a==b or b == c or a == c)
return :isosceles
end
:scalene
end
There are some absolutely brilliant people on StackOverflow...I'm reminded of that every time I visit :D
Just to contribute to the conversation, here's the solution I came up with:
def triangle(a, b, c)
raise TriangleError if [a,b,c].min <= 0
x,y,z = [a,b,c].sort
raise TriangleError if x + y <= z
equal_sides = 0
equal_sides +=1 if a == b
equal_sides +=1 if a == c
equal_sides +=1 if b == c
# Note that equal_sides will never be 2. If it hits 2
# of the conditions, it will have to hit all 3 by the law
# of associativity
return [:scalene, :isosceles, nil, :equilateral][equal_sides]
end
Question: What is the error message raised when the triangle is invalid due to negative side lengths? Answer: The error message raised when the triangle is invalid due to negative side lengths is "Negative length - thats not right".
Question: What are the three types of triangles that the code can identify? Answer: The three types of triangles that the code can identify are equilateral, isosceles, and scalene. | 677.169 | 1 |
reflection with mirrors
reflection with mirrors
You stand 1.80 m in front of a wall and gaze downward at a small vertical mirror mounted on it. In this mirror you can see the reflection of your shoes. If your eyes are 1.95 m above your feet, through what angle should the mirror be tilted for you to see your eyes reflected in the mirror? (The location of the mirror remains the same, only its angle to the vertical is changed.)
I know that the angle of incidence is equal to the angle of refraction and I tried creating triangles to solve this problem. For the first case, where you can see your shoes, when I drew the triangle, it was isoceles, so I split it up into to equal right triangles and found that the angles of incidence and refraction were about 62 degrees, but I did not know where to go from there or if I was even on the right track.
Question: What is the relationship between the angle of incidence and the angle of refraction in this scenario? Answer: They are equal. | 677.169 | 1 |
Oblique projection
Graphical projection is a protocol by which an image of a three-dimensional object is projected onto a planar surface without the aid of mathematical calculation, used in technical drawing.- Overview :...
Graphical projection is a protocol by which an image of a three-dimensional object is projected onto a planar surface without the aid of mathematical calculation, used in technical drawing.- Overview :...
Overview:
it projects an image by intersecting parallel rays (projectors)
from the three-dimensional source object with the drawing surface (projection plane). parallel lines of the source object produce parallel lines in the projected image. The projectors in oblique projection intersect the projection plane at an oblique angle to produce the projected image, as opposed to the perpendicular angle used in orthographic projection.
Mathematically, the parallel projection of the point on the -plane gives . The constants and uniquely specify a parallel projection. When , the projection is said to be "orthographic" or "orthogonal". Otherwise, it is "oblique".
The constants and are not necessarily less than 1, and as a consequence lengths measured on an oblique projection may be either larger or shorter than they were in space. In a general oblique projection, spheres of the space are projected as ellipses on the drawing plane, and not as circles as you would expect them from an orthogonal projection.
Oblique drawing is also the crudest "3D" drawing method but the easiest to master. Oblique is not really a 3D system but a 2 dimensional view of an object with 'forced depth'. One way to draw using an oblique view is to draw the side of the object you are looking at in two dimensions, i.e. flat, and then draw the other sides at an angle of 45 degrees, but instead of drawing the sides full size they are only drawn with half the depth creating 'forced depth' - adding an element of realism to the object. Even with this 'forced depth', oblique drawings look very unconvincing to the eye. For this reason oblique is rarely used by professional designers and engineers.
Oblique pictorial
In an oblique pictorial drawing, the angles displayed among the axes, as well as the foreshortening factors (scale) are arbitrary.
More precisely, any given set of three coplanar segments originating from the same point may be construed as forming some oblique perspective of three sides of a cube. This result is known as Pohlke's theorem, from the German mathematician Pohlke, who published it in the early 19th century.
The resulting distortions make the technique unsuitable for formal, working drawings. Nevertheless, the distortions are partially overcome by aligning one plane of the image parallel to the plane of projection. Doing so creates a true shape image of the chosen plane.
This specific category of oblique projections, whereby lengths along the directions and are preserved, but lengths along direction are drawn at angle using a reduction factor is very much in use for industrial drawings.
Cavalier projection is the name of such a projection, where the length along the axis remains unscaled.
Question: What is the mathematical representation of the parallel projection of a point on the -plane? Answer: The parallel projection of the point on the -plane gives.
Question: What is the name of the specific category of oblique projections where lengths along the directions and are preserved, but lengths along direction are drawn at an angle using a reduction factor? Answer: Cavalier projection
Question: What are the constants that uniquely specify a parallel projection? Answer: and
Question: Which drawing method is oblique drawing often compared to in terms of ease of mastery? Answer: The crudest "3D" drawing method | 677.169 | 1 |
Looking at functions via multiple representations (graphical, numeric, algebraic, verbal) has always served me well. Some representations shine a different light on the function. Putting 3.5 into the form 7/2, a hint that Dan gave, opened up a new door for making sense of the shape.
I played with the different applets that were provided overnight by four readers of Dan's post…another positive aspect of the MTB…great minds just waiting to use their talent for the betterment of mankind! The 7 was pretty easy to see since there were 7 line segments. The 2 took awhile for me to see a pattern. What I finally noticed by trying out different fractions was that the denominator indicated how many "rotations" around the shape it took to "close up" the polygon. These terms are used loosely, since a full rotation would take one back to the "starting point", which would "close up" the polygon.
3 – gon, 3/1 – gon, or you may know it as a triangle
For example, in a whole number-gon, such as a 3-gon (triangle), which, by the way, could also be called a 3/1 – gon, follow this procedure:
Start at A. Heading counter-clockwise and draw segment AA'. Construct a 60 degree angle degrees at A'. Draw A'B the same length as AA'. Construct a 60 degree angle at B. Draw BB' and B' will coincide with A. You have closed the 3/1 -gon after 1 rotation.
m/n – gons and partial sweeps or revolutions
When dealing with a m/n-gon, when n is not 1, it will take more than one sweep to get exactly back to the starting point I am calling each "sweep around" a "partial revolution", meaning that it gets close to the starting point, but doesn't quite make it that time around.
Although my explanation was not mathematical precise…it was more like hand-waving and a "kinda" explanation… I posted my comment to Dan's post, putting that interpretation out there.
Here is part of my comment to Dan's post:
Wow! I love math! I was blown away by this post and the idea of thinking of a 3.5 gon as a 7/2 "gon" that means that it has seven "sides" and takes two rotations to complete.
Out of the denizens of the MTB, a comment appeared from none other than Michael Serra, author of one of my favorite Geometry books, Discovering Geometry: An Investigative Approach. Of course, he has a better and more mathematically succinct way to describe how the "n" behaves. Before I explain his way, which is skipping vertices, and much less wishy-washy than my explanation, I will share the following comment that he made:
Question: How did putting 3.5 into the form 7/2 help understand the function's shape? Answer: It hinted at the number of sides and rotations needed to close the polygon, providing a new perspective on the function's shape.
Question: What is the procedure to draw a 3/1-gon (triangle)? Answer: Start at A, draw AA', construct a 60-degree angle at A', draw A'B, construct a 60-degree angle at B, and draw BB' until B' coincides with A.
Question: Who is the author of one of the user's favorite Geometry books mentioned in the text? Answer: Michael Serra. | 677.169 | 1 |
Given three points, if they are non-collinear, there are three pairs of parallel lines passing through them – choose two to define one line, and the third for the parallel line to pass through, by the parallel.
Given two distinct points, there is a unique double line through them.
Degenerate ellipse with semiminor axis of zero
Another type of degeneration occurs when an ellipse, rotated and translated to its simplest form , has its semiminor axis b go to zero and thus has its eccentricity go to one. The result is a(degenerate because the ellipse is not differentiable at the endpoints) with its foci
Focus (geometry)
Question: What happens to an ellipse when its semiminor axis becomes zero? Answer: It becomes a degenerate line | 677.169 | 1 |
No...Angle AOB and BOC...are congruent.....So how do you find the answer?? Thanks
stapel
09-10-2005, 06:20 PM
Yes, AOB and BOC are congruent. And since you say that OC divides BOD "in half", BOC and COD are also congruent.
Eliz.
ToOtSiE_PoP
09-10-2005, 06:46 PM
Oh Okay...Duh..So can you please explain how you got this answer?
stapel
09-10-2005, 06:56 PM
I didn't give the answer; I told you how to find it.
Eliz.
ToOtSiE_PoP
09-10-2005, 07:09 PM
Ok...so would the answer for m(BOC) be 45.6..but that makes no sense does it! Im so confused....I dont understand how to solve for x....
This is how I got 45.6....
13x-9=45
+9 +9
13x=54
x=4.2 (rounded)..and I plugged it into the eq...
stapel
09-10-2005, 07:13 PM
You said that BOC was half of a right angle. Where is "45.6°" coming from?
Eliz.
ToOtSiE_PoP
09-10-2005, 07:18 PM
Correct, but inside angle BOC I am given an equation (13x-9) degrees....and b/c BOC is congruent to COD, I wrote 13x-9=45...to solve for x..but I got 4.2 (rounded to nearest tenth)and plugged it into the equation 13(4.2)-9...
stapel
09-10-2005, 07:34 PM
When you round, you're not going to get an exact value. That's the point (and downside) of rounding. And plugging the rounded value back into the angle-measure expression doesn't change the actual angle measure. If the angle measure is 45°, then it's 45°, not 45.6°.
Eliz.
ToOtSiE_PoP
09-10-2005, 07:52 PM
I am still confused about how I am to find each measure....How do I find the answer for m(BOC) using the equation.....I understand m(cod) is 45 degr....but I need to show steps showing that m(BOC) =45 degr...(since its = to COD) but I dont get how I am to get 45 degr..using the equation...
stapel
Question: How can you show that m(BOC) equals 45 degrees, given that it is congruent to COD? Answer: Since COD is given as 45 degrees and BOC is congruent to COD, you can directly state that m(BOC) is also 45 degrees. | 677.169 | 1 |
09-10-2005, 08:14 PM
You said that BOD was split in half to form BOC and COD. Don't they have the same angle measure then, since they're each one half of the original angle?
And you said that BOD was a right angle. What is the measure of a right angle? What is half of that?
Eliz.
ToOtSiE_PoP
09-10-2005, 08:22 PM
45 degrees..so why is this equation inside BOC..(13x-9)? Inside AOB..there is also an eqaution inside that (12x-4) degrees...and it is congruent to BOC which is 45 degrees...but arent I supposed to use these equations to find each measure???? How do I solve for x to get each measure?? Am I supposed to use each equation like 12x-4=180...I am confused...
stapel
09-10-2005, 08:39 PM
The angle measure is being given as "13x - 9". Set this expression equal to the known value, and solve for x. Don't round the answer.
The problem comes, of course, when you don't get "45" as the answer for the measure of the angle that is supposed to be congruent, which leads me to wonder if you have yet presented the exercise accurately. Does OC really split BOD "in half"? Or just "into two angles"? Is there any other information in the drawing?
Eliz.
ToOtSiE_PoP
09-10-2005, 10:11 PM
"Angles AOB AND BOC are congruent"...Only these angles are congruent...The drawing doesnt state that angles BOC and COD are congruent....So the right angle is bisected into 2 differnet angles...
stapel
09-10-2005, 10:23 PM
the right angle is bisected into 2 differnet angles...
"Bisected" means "split exactly in half". You can't cut something into to equal but not equal parts. Is the right angle bisected, or is it cut into two different angles?
If the angle is bisected, then the algebraic expressions don't work out. If it is not, then solve for x and see where that takes you.
I don't understand this drawing at all...It needs to specify the measurement of the whole angle...AOD...Oh wait I am supposed to find the measurement of that angle too...I was trying to draw something like the drawing Iam referring to....no such luck...If you have time..can you draw what you think looks like the drawing I am referring to? Its really frustrating!!
That's IT ; OVER AND OUT : those 2 angles both equal 56 degrees...
if that changes anything in your problem, then go see your $#%@#$@!@
Question: What is half of the measure of a right angle? Answer: 45 degrees
Question: To find the measure of each angle, should you use the equations to solve for x? Answer: Yes | 677.169 | 1 |
As a concrete example, consider a sphere of radius a, where x = a sin u cos v, y = a sin u sin v, z = a cos u. Here, u and v are the polar angles. In this case, it is easy to see that the normal vector is (sin u cos v, sin u sin v, cos u). The first derivatives are xu = (a cos u cos v, a cos u sin v, -a sin u) and xv = (-a sin u sin v, a sin u cos v, 0). If we had not realized that the normal could be obtained easily, it could be found from the cross product of these tangent vectors. This gives E = a2, F = 0, G = a2 sin2u, so ds2 = a2du2 + a2sin2u dv2. We can see by inspection that this result is valid, considering displacements along a meridian and along a circle of latitude: ds2 = (adu)2 + (a sin u dv)2. If u and v are orthogonal coordinates, then F = 0.
Since the metric form is ds2 = a2dφ2 + a2cos2φdθ2 for a sphere where we have now taken φ = latitude and θ = longitude, the ratio of the north-south distance dy to an east-west distance dx in a small displacement is dy/dx = dφ/cos φ dθ. This is the slope of the displacement, or the tangent of its azimuth from north. Whenever F = 0, the coordinates are rectangular and we can form this ratio simply by looking at the metric form. Let's now consider a cylinder wrapped around the sphere, where z replaces the latitude φ, and ds2 = a2dθ2 + dz2. Here, dy/dx = dz/adθ.
Now suppose we map the sphere on the cylinder, so that the point φ,θ on the sphere is represented by the point z,θ on the cylinder. This can be done by a arbitrary z = z(φ), but a very useful mapping results when angles are preserved. That is, two displacements that make an angle w on the sphere are to make the same angle when mapped onto the cylinder. This will be true when the angle between the meridian and a displacement is preserved, since we only have to refer the displacements to the meridian. This means that the slopes of the displacements are to be the same. We know the slopes, so we have the requirement that dφ/cos φ dθ = dz/adθ, or dz/dφ = a/cos φ. This integrates to z = a ln tan(φ/2 + π/4) + C. If z = 0 when φ = 0, then C = 0. This is the desired relation between z and φ, which gives a Mercator map. The scale is determined by the factor a. Distance representing longitude is given by x = a θ.
Question: What is the radius of the sphere mentioned in the text? Answer: The radius of the sphere is 'a'. | 677.169 | 1 |
5/7 Muilti Purpose Ruler
The circle is divided by 72 degree lines into 5 equal segments (yellow lines) and by 52 degree lines into 7 equal segments (blue lines). Repositioning marks enable further division of the circle. At each intersection of lines and circles is a small hole for marking with a fine tip marker.
Question: Are there any markings on the intersections of lines and circles? Answer: Yes, there are small holes for marking with a fine tip marker. | 677.169 | 1 |
falls back to earth, the curved path it follows is a
parabola. It is an open plane curve formed by the
intersection of a cone with a plane parallel to its side.
Hyperbola: It is the curve
produced when a cone is cut by a plane that makes a larger
angle with the base than the side of the cone does.
Mathematics Q.
What is triangulation and how do you calculate it?
A. Triangulation is a
geometrical method to measure the height of a distant object.
Let AB be an object whose height h is to be measured. Let O
be the observation point. By joining AOB, we form a triangle.
Using an apparatus called sextant at O, we measure angle AOB
= Θ . This is called angle of elevation of the object.
In triangle AOB, tan Θ =
AB/OB = h/x. Thus, h = x tan Θ . Here, x is the distance of
the object from the observation point. If x is known then h
can be calculated.
In certain cases, x is not
known. The object is said to be inaccessible. In such a case,
we hold the sextant at any point C and measure angle ACB = Θ
. The sextant is then moved to any other point D where CD = x
is measured and angle ADB = Θ' is measured. Geometrically,
one can show
h = x / (cot Θ - cot Θ ' ).
Thus, h can be calculated.
Misc.
Misc. Q Hello,
I was wondering about this problem, and expecting you to
tell me if my ideas about it may be right (even if only
theoretically, in the hypothesis of being real an absolute
unit), if you please:
The problem:
If we have to go from one point to other that is ten meters
away, we must pass first by the 5 meters to the goal mark;
then by the 2.5,then 1.25 and so on:
If we always must cross the middle point of what is left, we
ever advance less than the total remaining distance...
Possibility 2:There is not, and my question is about this
hypothesis; in the abstract world of mathematics, we can
solve the contradiction between an infinitely divisible
quantity (let s say 1) and its finiteness by convention: we
can have 0.999...and we say that is less than 1 although is
an infinitely extensive number, and we can only achieve 1
when we establish by convention the indivisibility in some
point of the scale (lets state from 0.9 the indivisibility in
0.01 and we do 0.9 + 0.01 x 10=1 and we got to the goal)
But as I see it, as long as the basic function of numbers is
to describe the relation between singularity and plurality,
to divide a number is in a sense the same as to multiply it,
Question: What is the hypothesis in the miscellaneous section about the mathematical world? Answer: That in the abstract world of mathematics, an infinitely divisible quantity can be considered finite by convention.
Question: What is the shape of the path followed by an object when it is thrown horizontally and falls back to earth? Answer: A parabola. | 677.169 | 1 |
Probability and Geometry
The activity and two discussions of this lesson connect probability and geometry. The Polyhedra discussion leads to platonic solids, and the Probability and Geometry discussion leads to connections between angles, areas and probability. The subtle difference between defining probability by counting outcomes and defining probability by measuring proportions of geometrical characteristics is brought to light.
Question: What is the main difference highlighted between two ways of defining probability? Answer: The difference between defining probability by counting outcomes and defining probability by measuring proportions of geometrical characteristics. | 677.169 | 1 |
In the Pentomino problem, we worked with rearranging the shapes on a two-dimensional plane, and we needed to consider how the shapes were alike or different. In this section, we now consider a third dimension of depth as we look for the various possibilities.
The mathematical concepts include visualizing the relationship between the angles of an actual cube and an isometric drawing of a cube, recognizing the possible views of a three-dimensional figure, using two-dimensional drawings to record three-dimensional figures, and using reflections and rotations.
Question: How do these activities connect to subjects other than mathematics?
In solving these problems, it is helpful for you to communicate the process(es) used in analyzing the various viewpoints. Working in a group or with partners will help you find various strategies for reaching the solution. Looking at the three-dimensional model from different views and recording those views on isometric models are means of representation. Reasoning about the relationships between the views will help you eliminate duplicates. Several problem-solving strategies are useful in exploring these problems, including (1) solve a simpler problem, (2) account for all possibilities systematically, (3) make a model, and (4) change your point of view.
Question: True or False: Making a model is not a useful strategy for exploring these problems. Answer: False. Making a model is one of the problem-solving strategies mentioned. | 677.169 | 1 |
Question 32750
Since you are given two sides of the triangle and you know that neither of the two are the hypotenuse.
You can say that either a=3 and b=4 or the other way where a=4 and b=3. Then by using the pythagorean theorem which states a^2+b^2=c^2 you will be able to get your answer.
So
(3)^2+(4)^2=c^2
9+16=c^2
25=c^2
Now by taking the square root of each side you get that c=+ - 5, but since you are dealing with lengths and you can't have a negative length, then c=5.
Question: What is the value of c^2 when a=3 and b=4? Answer: 25 | 677.169 | 1 |
Vectors from 444
We assume vector addition and multiplication of a vector by a scalar (i.e.,
a real number) are well-understood.
Notation: Let I, J, K be (1, 0, 0), (0, 1, 0), (0, 0, 1).
Center of Mass for equal masses
The center of mass of a set of points (with equal masses at each point) is
the mean, or average. Thus for two points, AB, the midpoint is (1/2)(A+B), for
a triangle ABC, the center of mass (the centroid) is (1/3)(A + B + C). For a
set of four points ABCD, the center of mass is (1/4)(A+B+C+D).
Applications and exercises
Find the midpoint of (1, 1, 1) and (1, 0, 2).
For an equilateral triangle, all the usual centers (centroid, circumcenter,
incenter, orthocenter) are all the same point.
Unequal Masses, Barycenters and parametric equation of a line and a plane
Given two points A and B and two masses a, b at those points, the center of
mass is the weighted average:
M = (a/(a+b))A + (b/(a+b))B.
If a+b = 1, then M = aA + bB. In the case a+b = 1, ( if we set b = t, then
a = 1 - t, so the formula for the center of mass is simpler:
P(t) = (1-t)A + t.B.
The formula P(t) = (1-t)A + t.B can be written in coordinates as, e.g., (x(t),
y(t), z(t)). This is the (affine) parametrization of the line AB. The
parameter t provides a coordinate system on the line AB.
When both masses are nonnegative, then P is on the segment AB. If one of the
masses is negative, then P is on line AB but not on the segment.
For 3 points A, B, C, with masses a, b, c, the center of mass, or barycenter
is
M = (1/(a+b+c))(aA+bB+cC).
Again this is simpler if a+b+c = 1, so the point M = aA+bB+cC. Finally, if
b = s and c = t, then if a = 1 - s - t the sum = 1.
P(s,t) = (1 - s - t)A + sB + tC
The point P(s,t) = (1 - s - t)A + sB + tC ranges over the whole plane ABC as
s and t take on all real values, with P inside triangle ABC if all 3 masses
Question: In an equilateral triangle, where are the centroid, circumcenter, incenter, and orthocenter located? Answer: They are all at the same point.
Question: Can the point P be on the segment AB if one of the masses is negative? Answer: No, P will be on line AB but not on the segment. | 677.169 | 1 |
Triangle An emblem of the triad or three-in-one, expressing more than the three dots alone: the points, lines, and the whole figure give a septenate composed of two triads and a monad. The triangle also symbolizes twin rays proceeding from a central point, and when the other ends of these lines are joined, the base line signifies that which is produced by the interaction and interblending of the two formative rays. The apex, the side lines, and the base thus represent the three chief stages of cosmic evolution. The idea is further elaborated in the square pyramid. The Pythagoreans recognized the triangle as the first regular rectilinear figure, as three is the first odd number -- the one being considered as the origin and unit, out of which all subsequent parts flow. The usual form of the triangle in symbology is equilateral, with the apex up or down. The circle, triangle, and square form another important triad representing stages in evolution. For interlaced triangles, see also SIX-POINTED STAR
Question: What is the usual form of the triangle in symbology? Answer: Equilateral, with the apex up or down. | 677.169 | 1 |
Concurrence Ml
concurrent, lines, geometry and triangle
CONCURRENCE (ML. concurrentia, con currence, from concurrcre, to run together, from con-, together + currcrc. to run) AND COL LINEARITY (from Lat. corn-, together + linca, line). If several lines have a point in common they arc said to be concurrent. The common point is called the focus or vertex of the pencil of lines. if several points lie on one straight line they are said to be collinear. The line is called the base of the range of points. That portion of geometry concerned with concurrent lines and collinear points is called the theory of concur rence and collinearity. Some of its fundamental propogitions are: If a transversal intersects the sides of a tri angle ABC in the points X, Y, Z, the segments of the sides of the triangle are connected by the relation (AZ:ZB) • (BX:XC) • (CY:YA)= — 1. Conversely, if the points be so taken that the rela tion holds, then the three points arc collinear. (This relation is known as Menelaus's theorem.) lf the three lines AO, BO. CO drawn from the vertices of the triangle ABC are concurrent in 0 and meet the opposite sides in X. I', Z. then BX • CY'AZ =-- —CX•AY•BZ, and conversely (Ceva's theorem).
If three lines perpendicular to the sides of a triangle ABC at X. Y. Z are concurrent. then
-h
= 0.
Conversely, if this relation holds, the per pendiculars are concurrent.
If the lines joining the vertices of two tri angles are concurrent, their corresponding sides intersect in three collinear points. (This proposi tion, known as Desargues's theorem, is true for any rectilinear figures.) The opposite pairs of sides of a hexagon in scribed in a conic intersect in three collinear points (Pascal's theorem).
The lines joining the opposite vertices of a hexagon circumscribed about a conic are con current (Brianchon's theorem).
The polars of a range of points with respect to a circle (q.v.) are concurrent. and conversely.
Question: What does it mean for lines to be concurrent? Answer: If several lines have a point in common, they are said to be concurrent. | 677.169 | 1 |
Loci: Convergence
Mathematical Treasures
by Frank J. Swetz and Victor J. Katz
Charles Bossut's Traite elementairede geometrie
This is the title page of the Traité élémentaire de géométrie et de la maniere d'appliquer l'algébre a la géométrie (1775), written by Charles Bossut (1730-1814). This work was one of many texts written by Bossut in connection with his teaching at several different institutions in France. It was one of the first books to give a detailed explanation of analytic geometry.
On page 397 and the following pages, Bossut discusses the equation of the circle. Note that he does not assume that the two coordinate axes are perpendicular.
On pages 398-399, Bossut carefully works out the equation of the circle, displaying it as equation (A) on p. 399. Note that because his axes are not assumed to be perpendicular, the equation has an xy term.
On page 400, Bossut shows various simplifications of the equation of the circle. He derives equation (B) in the case where the two axes are perpendicular. In the next two equations, the position of the origin is simplified.
On pages 424-425, Bossut discusses certain properties of tangents to parabolas and deals with them algebraicially.
On pages 426-427, Bossut proves the familiar property of a paraboloa, that the line from the focus to a point on the parabola makes the same angle with the tangent as the line from that point parallel to the axis.
Question: What is the relationship between the line from the focus to a point on the parabola and the line from that point parallel to the axis, according to Bossut's proof? Answer: They make the same angle with the tangent
Question: On which pages does Bossut prove the familiar property of a parabola about the angle made by the line from the focus to a point on the parabola with the tangent? Answer: 426-427 | 677.169 | 1 |
i am confused in one problem and please help me,i will give picture from problem below
and question says : $AB$ is a diameter of the circle. All triangles above the diameter in the diagram are equal in area. All triangles below the diameter are equal in area.compare total area of triangles above $AB$ and total are below $AB$
i have chosen that this can't be determined by given information,because in spite of we have fact that below $AB$ we have more triangles then above $AB$,we dont know length of bases of each triangles,all sides except bases are radius so equal to each other,but in question answer is different,and explanation is following:
The total area of the upper triangles is less than the area of the lower triangles. The more triangles that you cut the semicircle into, the more of the circle that is occupied.
is this right?it is test taken from GRE test,i am preparing for passing it
So just by taking more triangles, they will necessarily be closer to circumference and hence their total area is larger. GRE is about intuitiveness, so this reasoning is mainly intuitive but based on some knowledge you need to have.
HINT You need to use the fact that although the triangles in each half of the diagram look to be different sizes, triangles in the same half all have the same area. Express the area of such a triangle in terms of the angle at the centre (knowing that two sides in each triangle have length equal to the radius). You should be able to prove that you have two sets of congruent triangles.
Then you could use the fact that the pentagon is constructible to compute the sine of the central angles in the bottom half. The sine of the angles in the top half should be known to you. And that should be enough information to give the ratio explicitly (if this is what is requited) - else use a calculator or tables to give a numerical value.
Question: Is the total area of triangles above AB less than the total area of triangles below AB? Answer: Yes.
Question: Is the GRE test mainly about intuitive reasoning or factual knowledge? Answer: The GRE test is mainly about intuitive reasoning but based on some factual knowledge. | 677.169 | 1 |
Did you try the experiment? If you did not, go do the experiment - we will wait for you!
Welcome back. Did you find a pattern? Good!
OK, here is the pattern between the angle of the mirrors in the Mirror Multiplier and the number of candles you see.
A full circle has a measure of 360 degrees. A single mirror has an angle of 180 degrees and you see two candles
If you do the math of 360 divided by 180 you get 2, or two images. Do the math of 360 divided by 45 and you get 8, or eight images, like the first picture. Try 360 divided by 90 - go to your Mirror Multiplier and make the angle between the two mirrors a 90-degree angle, place in a candle. How many candles do you see? Right, 360 divided by 90 equals 4; you see four candles.
Your Mirror Multiplier with the help of a protractor also works in reverse. Move the mirrors in your Mirror Multiplier to make 12 candles. What angle is this? Well 360 divided by 12 equals 30. So the angle needed to make 12 candles is 30 degrees.
Question: What is the mathematical relationship between the number of candles seen and the angle of the mirrors? Answer: The number of candles seen is equal to 360 degrees divided by the angle between the mirrors. | 677.169 | 1 |
My Geometry Lesson Today
We talked about a family today, the Quadrilateral family. It starts off with Big Momma. Big Momma had 4 sides. Big Momma was ugly and boring, so her husband left her. But, before he left her they had 3 babies. Parallelogram, Trapezoid, and Kite. Kite was ugly like her mom. She had 0 pairs of parallel sides and 2 pairs of congruent adjacent sides. Since she was ugly and boring like her mom she died alone with 25 cats. The second child was the Trapezoid. She was average looking and had one pair of parallel sides. Then there was the most beautiful child, the Parallelogram. She had 2 pairs of parallel sides.
Since the Trapezoid wasn't ugly (like Kite), she had a baby. The Isosceles Trapezoid. She has one pair of congruent sides and one pair of parallel sides. And this is where this family ends because the Isosceles Trapezoid became a Nun.
The Parallelogram had two babies. The Rectangle and the Rhombus. The Rectangle has two pairs of parallel side and four right angles. The Rhombus, as my teacher would say, "the girl with the most beautiful sides". She had two pairs of parallel sides and four congruent sides. Now Rhombus learned her true identity. She wasn't meant to be a girl, she was meant to be a guy. So she became a he. Rectangle and Rhombus got so close that they had a kid! They made the Square. She had four right angles and four congruent sides. Boy! was she beautiful!!
And that is where this family tree ends because no guy wanted to deal with a wreck like Square. She tried to explain to them that her parents never thought of each other like brother and sister but no one listened to her. So, Square died in a mental hospital because everyone thought she was crazy.
This was the story my teacher told us. I will never look at figures the same. Never. Don't even get me started on Triangles. The main point I think my teacher was trying to tell us is… If you're ugly, you're going to die alone .. or… If you have a screwed up family, you're going to die alone.
Question: Which two quadrilaterals had a child together? Answer: Rectangle and Rhombus
Question: What was the profession of the Isosceles Trapezoid? Answer: Nun
Question: What was the main point or moral of the story according to the narrator? Answer: If you're ugly, you're going to die alone, or if you have a screwed up family, you're going to die alone. | 677.169 | 1 |
FREE Lesson Plan - 1st of 3 Free Items
Estimating Angles, Area, and Length
Grade Levels: 3 - 7
Objective
Math students in middle school will use estimation to approximate values, angle, and area measurements of a triangle.
Materials
paper
pencil
string
Procedure
Demonstration
Explain to students that they are going to work as a class to estimate the measurements of several angles and compare the estimates with measured values. Then, students will work in groups of four to estimate a traingle's angles and area. Explain that this lesson covers two benchmark units, degrees and centimeters.
Draw two triangles on the chalkboard, and write the base and height for each: first triangle, height = 732 and base = 1239; second triangle, height = 128 and base = 985. Have students select an acute angle from the first triangle, and show them that they can visualize whether the angle is less than or greater than 90 degrees. Then have them determine if the angle is less than or greater than 45 degrees. This will help them narrow the angle's range to 45 degrees (0-45 or 45-90). If the angle is less than 45 degrees, students can determine whether the angle is closer to 0 or 45 degrees. Guide them through this process for the first triangle, and then repeat the process for the second triangle. Prompt them with questions about the angle's relation to 0, 45, 90, 135, and 180 degrees to help them narrow the acceptable range, and then have them make their estimate. Finally, have students measure the actual angles and compare the estimates with measured values.
Have students estimate the area for each triangle by estimating the product dictated by the formula for the area of a triangle (area = [base/2] x height) and document their process in their notebooks. Explain that they should choose numbers that are close to the originals, but are easier to work with. For example, with the parameters given for the first triangle, a student might say,
"The base is 1239, which is very close to 1200, so I will divide that by 2 to get 600. 600 x 732 is difficult to calculate, but 732 is very close to 700, and 600 x 700 = 420,000."
Have students calculate the actual area with the exact measurements and compare these measurements to their estimates. The actual area of the first triangle is 453,474, so the estimate is only off by 7% (453,474 - 420,000/453,474 = .07 = 7%).
Guided Practice
Question: In the example given for the first triangle, what is the estimated area calculated by the student? Answer: 420,000 square units
Question: What are the materials needed for this lesson? Answer: Paper, pencil, and string.
Question: Which angles should students consider when estimating the angle of a triangle? Answer: Angles less than or greater than 0, 45, 90, 135, and 180 degrees. | 677.169 | 1 |
Polygons/336504: Suppose that the interior angles of a convex pentagon are five consecutive numbers. What is the measure of the largest angle? 1 solutions Answer 241203 by edjones(7569) on 2010-08-30 00:55:27 (Show Source):
You can put this solution on YOUR website! 180(n-2)=sum of the interior angles of a polygon. n=number of sides
Let the consecutive numbers be x, x+1, x+2, x+3, x+4.
5x+10=180(5-2)
...=180*3
...=540
5x+10=540
5x=530
x=106
x+4=110 degrees the largest angle.
.
Ed
Mixture_Word_Problems/336495: A hospital needs to dilute a 60% boric acid solution to a 10% solution. If it needs 40 liters of the 10% solution, how much of the 60% solution and how much water should it use? 1 solutions Answer 241201 by edjones(7569) on 2010-08-30 00:40:26 (Show Source):
You can put this solution on YOUR website! 1/(√2) + x = x/(√2)
1+sqrt(2)x=x multiply each side by sqrt(2)
sqrt(2)x=x-1
2x^2=x^2-2x+1
x^2+2x-1=0
x^2+2x =1
x^2+2x+1=1+1 complete the square.
(x+1)^2=2
x+1=+-sqrt(2) sqrt of each side.
x=-1-sqrt(2) the only answer.[x=-1+sqrt(2) is extraneous]
.
Ed
Probability-and-statistics/332409: Find the indicated probability.A study conducted at a certain college shows that 54% of the school's graduates find a job in their chosen field within a year after graduation. Find the probability that among 9 randomly selected graduates, at least one finds a job in his or her chosen field within a year of graduating. 1 solutions Answer 238248 by edjones(7569) on 2010-08-17 00:29:41 (Show Source):
You can put this solution on YOUR website! 46% don't of the school's graduates don't find a job in their chosen field within a year after graduation.
.46^9=.000922 probability that none of the 9 get a job.
Question: What is the measure of the largest interior angle in a convex pentagon where the angles are five consecutive numbers? Answer: 110 degrees
Question: What is the probability that none of the 9 randomly selected graduates find a job in their chosen field within a year of graduating? Answer: 0.000922
Question: If the hospital needs 40 liters of a 10% boric acid solution, how much of the 60% solution should it use? (Answer in liters) Answer: 10 liters | 677.169 | 1 |
A Little Bit of Trigonometry
Some tasks.
1. Measure side a and side c, and compute a/c. Use a protractor to measure angle A,
and find sin A with a calculator. Does a/c = sin A?
2. Measure side b, and compute b/c. Find cos A with a calculator. Are they
the same?
3. Do the same for angle B: measure b, and compute b/c. Measure angle B, and
find sin B with a calculator. Are they the same?
4. You can also compute the tangents of the two angles, by forming ratios
of sides, and by using a calculator.
5. How could you find the measure of angle A, if you did not have a protractor?
Question: What is the first step in task 1?
Answer: The first step in task 1 is to measure side a and side c. | 677.169 | 1 |
I'll refer to this image:
Red vector (named direction) is direction
Two blue vectors represent range where I want to do something
α (alpha) is "falloff" of direction
Purple vector (v) is the vector which I need to check if it's between two blues.
In other words, I need to check if (v) is between (direction-α) and (direction+α)
Let a1 be the angle of the alpha vector on the left (- alpha) side of d, a2 be the angle of the alpha vector on the right (+ alpha) side.
Let v be the angle of the vector you're testing.
var/x = v - a1 while(x > 180) x -= 360 while(x < -180) x += 360
return x < (a2 - a1)
That should work (completely untested) assuming that alpha < 90, also assuming angles are in degrees. Conversion to radians is easy enough.
Basically it works by making a1 the zero point and then checking if we're within a2. Angles are fiddled with such that being pushed past -180 or 180 flips sign appropriately. The reason it doesn't work for cases where alpha is >= 90 is because it ends up taking the smaller angle.
Not at all sure if this is the fastest or most correct way to do it, but it feels functional. If you need alpha >= 90, comment again and I'll think about it some more.
All you should need is arccos(dot(v,d)), where dot() is the dot product (v.x*d.x + v.y*d.y). The range is only 0 to 180 degrees, but you don't seem to care about left or right so that's not much of an issue.
Question: What are the angles a1 and a2 representing? Answer: a1 is the angle of the alpha vector on the left (- alpha) side of d, and a2 is the angle of the alpha vector on the right (+ alpha) side. | 677.169 | 1 |
Quadratic_Equations/417959: One side of a triangle is half the longest side. The third side is 8 meters less than the longest side. The perimeter is 42 meters. Find all three sides. (Enter your answers from smallest to largest.)
? meters
? meters
? meters 1 solutions Answer 292675 by ewatrrr(10682) on 2011-03-06 11:50:45 (Show Source):
Hi
Using the vertex form of a parabola, where(h,k) is the vertex
y=2x^2-6x-14 |completing the square to put into the vertex form
y= 2(x-3/2)^2 -9/2 -14
y= 2(x-3/2)^2 -37/2 |vertex Pt(3/2,-37/2) x = 3/2 the line of symmetry
CHECKING our Answer***
Question: Is the triangle a right-angled triangle? Answer: No | 677.169 | 1 |
When the midpoints of the sides of a Quadrilateral are joined,
another quadrilateral results. using the diagram helps to see what is
special about this new quadrilateral. Try the Quadrilateral applet for yourself.
The Pythagorean Theorem can be proved in many ways. This
diagram uses shears to show a version of Euclid's proof. The applet is a
slight variation on the one provided by the publishers of GSP. Try the
Pythagorean Theorem applet for yourself.
Trig tracers show the paths of points on a circle, leading to
sine and cosine curves. The applet is a slight variation on the one
provided by the publishers of GSP. Try the Trig tracers applet for yourself.
Vertically opposite angles are formed by a pair of
intersecting lines. By manipulating the lines, the sizes of the angles
can be compared. Try one of the applets, JSP version or GeoGebra version for yourself.
When an object is subjected to Two reflections , there are
various possibilities for the composite transformation. These can be
seen by manipulating the reflections as well as the original object. Try
the Two reflections applet for
yourself.
Transformations
Transformations in the plane can be studied using dynamic geometry. The examples here have been made with GeoGebra software.
The isometries are rigid transformations that preserve distances. The three most important isometries are those that reflect, translate or rotate the plane. Objects and their images are congruent.
Enlargements (or dilations) are not isometries, but are examples of similitudes. Images are similar to the original objects; that is, they have the same shape, but are not necessarily the same size.
Three-dimensional geometry
Three-dimensional objects can also be manipulated. The examples here have been made with the remarkable Cabri 3D software. If you
do not have the necessary (free) Cabri 3D plug-in on your browser, you
will be directed to download it. There is more information from the software developers here, as well as many more examples.
The five Platonic solids comprise all the polyhedra with faces that are congruent regular polygons and for which each vertex is the same. There are only five of these: the tetrahedron, the cube (hexahedron), octahedron, dodecahedron and icosahedron.
The dual of a cube is an octahedron, a surprising link between these two Platonic solids.
The volumes of pyramids and prisms are related, as can be seen with this pentagonal example.
Other uses of dynamic geometry
Dynamic geometry software can be used for purposes other than directly geometric ones. Here are some examples, using GeoGebra, a free software package available here.
The normal distribution is of critical importance in studying sampling distributions as well as other purposes. (Download the original GeoGebra file by right-clicking here.)
Question: Which software was used to create the examples of transformations in the plane? Answer: GeoGebra software.
Question: What is the relationship between the volumes of pyramids and prisms, as shown in the text? Answer: They are related. | 677.169 | 1 |
New from the Blog
Geom(ag)etry: Dodecahedrons and Icosahedrons
What follows is an explanation very similar to the one about Cubes and Octahedrons. We are now starting to see how two other Platonic solids are tied together: the Dodecahedron with its 12 pentagonal faces and Icosahedron made up by 20 equilateral triangles (I remember Platonic solids are those made up by identical regular polygons, and with the same number of faces meeting at each vertex).
As for Cubes and Octahedrons, also Dodecahedrons and Icosahedrons share a reciprocal "duality" (in geometry, polyhedra are associated into pairs called duals, where the vertices of one correspond to the faces of the other).
This is the main reason Dodecahedrons and Icosahedrons are tied together. Well, if you already read the explanation about Cubes and Octahedrons, I guess you are expecting to find also another reason... so I invite you to follow reading!
Let's see some Dodecahedron related polyhedra. Each edge of a Dodecahedron can be divided in three parts (see the edge b-e in the drawing below on the left) so that the extremities b-c and d-e are equal each other, as are the segments a-c, c-d and d-f. Doing the same operation on every face, and cutting out the Dodecahedron vertices along the red lines, we obtain a "Truncated Dodecahedron":
After the truncation, each Dodecahedron face has turned into a regular decagon, and each vertex have turned into an equilateral triangle, where the length of the decagon and triangle sides are the same each other. The Truncated Dodecahedron is composed of 32 regular faces (12 decagons and 20 triangles), 60 edges, and 90 identical vertices.
Here is the Geomag Truncated Dodecahedron model, where the green rods are the polyhedron edges and the metal rods are only needed to keep the model rigid (this is valid for all the following models too):
There is another way to cut the Dodecahedron vertices: it's enough to divide each edge into two halves and join the obtained central points each other across the cube faces. The result is shown below on the right: it's name is... well, I'll tell you later!
Let's put aside the Dodecahedron for a while, considering now the Icosahedron instead. Each edge of an Icosahedron can be divided into three identical parts: in the drawing below on the left the red lines join these intermediate points. Cutting the vertices of the Icosahedron over these lines, we obtain one regular hexagon for each face, and one regular pentagon for each vertex. The length of the hexagon and pentagon sides are the same each other.
Question: What is the relationship between Dodecahedrons and Icosahedrons? Answer: They are duals, sharing a reciprocal relationship in geometry. | 677.169 | 1 |
Since the center of each square is the center of the sphere, therefore two sides, EF and FG, along with the one diameter EG of the octahedron form a 45°-45°-90° triangle. Thus, the square on the diameter of the sphere is twice the square on the side of the octahedron.
Coordinates for the vertices of the octahedron
If the sphere circumscribing the octahedron is the unit sphere, then a natural coordinate system to impose would have the three coordinate axes be the three perpendicular diameters. Then the points a unit distance from the origin are the six vertices of the octahedron, namely,
(1,0,0) (–1,0,0) (0,1,0) (0,–1,0) (0,0,1) (0,0,–1)
Duals of the regular polyhedra
As will be shown in proposition XIII.18, there are exactly five regular polyhedra. The accompanying table lists these five polyhedra along with the numbers of the their faces, edges, and vertices. Their names are taken from the number of their faces, except, of course, the cube, which otherwise would be called a hexahedron.
Polyhedron
Faces
Edges
Vertices
tetrahedron
4
6
4
octahedron
8
12
6
cube
6
12
8
icosahedron
20
30
12
dodecahedron
12
30
20
Note that there are two pairs of polyhedra in this table where the numbers are related. One pair is the octahedron and cube, the other is the icosahedron and dodecahedron. For these pairs the number of faces of one of the pair equals the number of vertices of the other, and both of the pair have the same number of edges. These are the pairs of "duals." The numbers for the tetrahedron indicate that it dual to itself.
We can see the correspondence between the parts of one of these polyhedra and the parts of its dual. Consider the octahedron. Place a point in the circumcenter of each of the eight faces. Connect two of these points if the faces that contain them share an edge. For each of the six vertices of the octahedron, connect the four circumcenters of the adjacent faces to make a square. What results is a cube with six vertices, 12 edges, and eight faces.
An analogous construction for the cube yields an octahedron. Likewise the constructions for the icosahedron and dodecahedron yield each other, and the construction for a tetrahedron yields another tetrahedron.
Use of this construction
Constructing this octahedron is an end in itself. The construction is also used in proposition XIII.18 where the five regular polyhedra are compared.
Question: What shape is the result of connecting the circumcenters of the faces of an octahedron that share an edge and connecting the circumcenters of the adjacent faces to each vertex? Answer: A cube
Question: What is the relationship between the square on the diameter of the sphere and the square on the side of the octahedron? Answer: The square on the diameter of the sphere is twice the square on the side of the octahedron. | 677.169 | 1 |
Second equation is x= 2y + 100
You can put this solution on YOUR website! Let M = Mary's age
then because "Susan is six years older than Mary."
M+6 = Susan's age
.
From:"She is twicw as old as Mary"
M+6 = 2M
Solve for M by subtracting M from both sides:
M+6-M = 2M-M
6 years old = M (Mary's age)
.
Susan's age:
M+6 = 6+6 = 12 years old (Susan's age)
Equations/162904: The length of a rectangle is 3 in. more than its width. The perimeter is 30 in. Write and solve the equation to find the width of the rectangle. Formula is P=2 x length + 2 x width
Please show all work.
Thanks 1 solutions Answer 120057 by nerdybill(6958) on 2008-10-19 11:32:11 (Show Source):sum of the measures of the angle, three times its complement, and four times its supplement is at most 480 degrees." we get
x + 4(180-x) + 3(90-x) <= 480
x + 720 - 4x + 270 - 3x <= 480
990 - 6x <= 480
-6x <= -510
x >= 85 degrees
.
Since they asked for the "smallest" angle, it would then be 85 degrees.Half the supplement of the angle is 12 degrees less than twice the complement of the angle."
.
(1/2)(180-x) = 2(90-x) - 12
180-x = 4(90-x) - 24
180-x = 360 -4x - 24
180-x = 336 -4x
180+3x = 336
3x = 156
x = 52 degrees
You can put this solution on YOUR website! You did not need to expand it. This, is enough:
(2x-5)(x+8)=0
.
Notice that if EITHER then contents of the left parenthesis or the right parenthesis, is zero, then the ENTIRE left side is zero.
.
Therefore, the solutions are found by setting the two term on the left to zero.
Solution 1:
(2x-5)=0
2x = 5
x = 5/2
.
Solution 2:
(x+8)=0
x = -8
.
Therefore,
x = {-8, 5/2}
Question: What is the formula for the perimeter of a rectangle in the given text? Answer: P = 2 × length + 2 × width
Question: Who is 'M' in the context of the text? Answer: Mary's age
Question: What is the equation that can be used to find the width of a rectangle, given that its length is 3 inches more than its width and the perimeter is 30 inches? Answer: 3 × width + 2 × (width + 3) = 30
Question: What are the solutions to the equation (2x - 5)(x + 8) = 0? Answer: x = -8, 5/2 | 677.169 | 1 |
Course Title: Geometry
Grade: 8
Credits: 1
A. Course Description:
Geometry is a survey of Geometry, emphasizing the history of mathematics and proofs.
B. Course Objectives/Methods:
Geometry is a broad overview of the study of shape. It includes a general history of mathematics, the study of line, angle, and a variety of shapes. The derivation of theorems and justification of information is accomplished by proofs. This leads to a natural discussion of the nature of mathematics as a formal language. Emphasis will be placed on formal construction.
C. Course Goals:
Students will be able to do the Following:
Recognize the order, design, and beauty in the world around us, both in nature and in man-made inventions.
Recognize uses of geometry in man made construction
Construct a variety of shapes and lines with given criteria
Develop construction skills to the extent that new constructions can be made based on previous knowledge
Question: How many credits is this course worth? Answer: 1 | 677.169 | 1 |
From the slope and the coordinates of one of the points, we can get the equation of the line. If the coordinates of the third point satisfy that equation, the third point lies on that same line.
The slope of the line connecting (a,0) to (0,b) is
Since the y-intercept is at (0,b), b is the intercept, and we can write as the slope-intercept form of the equation of the line connecting (a,0) to (0,b).
Substituting the x-coordinate of (3a,-2b) into the equation we can find if that point lies on the same line.
For , the point on the line has so point (3a,-2b) lies on the same line as the other two points.
If we were not asked for the equation of the line, we could calculate the slopes for two different pairs of points. If we found the same slopes connecting two of the points with the other point, that would mean they all lie on the same line.
Question 749486: consider triangle ABC, where the picture is here
1. let ADF be the area of the triangle ADF, then
2. when BD=4 and CF=2, then BC=(B) and x satisfies the equation
solving this equation , we have
AD = (E)
solve for A,B,C,D, and E Answer by lynnlo(4155) (Show Source):
Question 749435:Thanks! Answer by Alan3354(30993) (Show Source):
You can put this solution on YOUR website!--------------
12 by 12 must mean the floor of the stall.
What part is in the frame depends on the location relative to the floor and the focal length.
If it's above the center of the floor 9' up, and 63 degs is the included angle, it will cover a circle 5.52' in diameter. Not all of the floor.
------------
To view all of the floor, it would have to be 13.85' above it.
The area of a triangle is the base times the height divided by 2, hence
Solve the quadratic. You will need the quadratic formula or, if you are a glutton for punishment, you can complete the square.
John Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
Question 749071: in a right-angled triangle one of the acute angles is 20 greater than then the other.fine the angles of the triangle Found 2 solutions by Cromlix, solver91311:Answer by Cromlix(307) (Show Source):
The sum of the measures of the interior angles of any triangle is 180 degrees. Since a right triangle has one angle that measures 90 degrees, the sum of the measures of the two acute angles must be 180 minus 90 equals 90 degrees.
Solve for
Question: What is the value of E when AD = 5 and x = 2? Answer: E = 1. | 677.169 | 1 |
mathAngle-angle-side congruence between two (or more) triangles. Congruent triangles have sides and angles of identical measure.
Abscissa
The horizontal axis, or the first coordinate in an ordered pair.
Absolute Maximum
The highest point on a graph, especially over a specified domain. It is the greatest value of f(x) over a defined interval of x, provided y=f(x).
Absolute Minimum
The lowest point on a graph, especially over a specified domain. It is the least value of f(x) over a defined interval of x, provided y=f(x).
Absolute Value
The distance on the real number line between an value and zero. It applies best to things for which negative values have no meaning, such as mass or length.
Accuracy
The quality of approaching an exact value. Distinct from precision, accuracy means to approach correctness, to tend toward an established value.
Acute Angle
An angle whose measure is less than 90 degrees.
Acute Triangle
A triangle whose interior angles are each acute, that is, less than 90 degrees (or π/2 radians).
Additive Inverse for Arithmetic
The opposite of a given number. Change the sign of a number to have its additive inverse. The sum of a number and its additive inverse is always zero.
Additive Inverse for Matrices
Mr. X takes the mystery out of Additive Inverse for Matrices, a matrix when added to another equals the Zero Matrix. Subscribe to my youtube channel for more instructional math videos.
Additive Property of Equality
This property allows us to add equals to equals to stay equal. Given two equal values, we may add the same quantity to both values and retain an equality.
Adjacent
Next to each other. The idea is especially important in geometry, as with adjacent angles that share a common ray.
Adjacent Angles
Next to each other. Adjacent angles share a common ray and subsequently have a common vertex.
Algebra
The branch of mathematics that allows manipulation of symbols and values to determine quantities that are not always fixed. Variables are essential to algebra.
Algorithm
A sequence of steps to accomplish a familiar task; a recipe.
Alpha
The first letter of the Greek alphabet.
Alternate Exterior Angles
Given two parallel lines cut by a transversal, angles exterior to the parallel lines and on opposite (alternate) sides of the transversal are congruent.
Alternate Interior Angles
Given two parallel lines cut by a transversal, angles interior to (between) the parallel lines and on opposite (alternate) sides of the transversal are congruent.
Alternating Series
A series in which successive terms have opposite signs. Every other term is positive; every other term is negative.
Altitude
Height. The perpendicular or orthogonal distance above a fixed reference, as height above mean sea level. In geometry, the shortest distance from the base of an object to its apex (or top).
Altitude of a Cone
The shortest line segment from the apex (tip) of a cone to the plane of its base.
Altitude of a Cylinder
Question: What are adjacent angles? Answer: Angles that share a common ray and have a common vertex.
Question: Which is the highest point on a graph, over a specified domain? Answer: The absolute maximum. | 677.169 | 1 |
A function is considered Continuous if its graph has no gaps, no holes, no steps, and no cusps or discontinuities.
Continuous CompoundingContinuous Function
When the graph of a function has no holes, no gaps, no steps, or no discontinuities, then it is considered Continuous. It may have cusps.
Continuously Differentiable
When a function is Continuously Differentiable it is both continuous and smooth.
ContractionContrapositive
Given a conditional statement, its Contrapositive is logically equivalent and is obtained by negating the original hypothesis and conclusion as well as reversing their order.
Convergence
To approach a limit is to experience Convergence. Mathematical series experience convergence when the sum of their expanded terms reaches a boundary or limit.
Convergent Series
A series is said to be Convergent when its sum approaches a limit.
ConverseConvex
When a geometric or physical entity has no indentations. Or, when a polygon has the property where no line segment across it leaves the interior of the polygon, the polygon is said to be Convex.
Coordinate
A value associated with the location of a point is a Coordinate. In one dimension a Coordinate is a single value. In two dimensions, a point is defined by two Coordinates as an ordered pair.
Coordinate Geometry
This branch of mathematics is a combination of algebra and geometry; it is analytic geometry.
Coordinate PlaneCoplanar
In the same plane; of the same plane. Most generally, points within the same plane are said to be Coplanar.
Corollary
A Corollary is like a baby theorem.
Correlation
When two variables have a strong linear relationship, either increasing proportionally or one variable decreasing as the other increases, we say there is (strong) Correlation between the variables.
Correlation CoefficientCorresponding AnglesCosecant
One of the six basic trig functions, the Cosecant function is the reciprocal of the sine function, and the cofunction of the secant. The Cosecant of theta can be expressed as (r/y) for an angle in standard position, or the ratio of hypotenuse over opposite side in a right triangle.
Cosine
One of the six basic trig functions, the Cosine is the cofunction of the sine function and the reciprocal of the secant function. In standard position the Cosine of theta is (x/r). In a right triangle the cosine of an angle is the ratio of the adjacent side to the hypotenuse.
Cotangent
One of the six basic trig function, Cotangent is both the reciprocal function and the cofunction of the tangent function. For a right triangle, the Cotangent of an angle is the ratio of adjacent side to the opposite. In standard position for angle theta, the Cotangent can be expressed as (x/y).
CoterminalCountableCounterclockwise
For angles in standard position, we use a Counterclockwise rotation for positive measurement of the angle's rotation. This is the direction opposite the traditional movement of analog clock hands.
Counting Numbers
Question: What does it mean for a function to be continuously differentiable? Answer: It means the function is both continuous and smooth.
Question: Which of the following is NOT a trigonometric function? A) Cosine B) Cosecant C) Square Root D) Sine Answer: C) Square Root | 677.169 | 1 |
A planar figure that neither crosses itself or contains a gap is a Simple Closed Curve; note that a curve can be "straight" according to the mathematicians.
A number associated with a line graphed in a plane, Slope is the ratio of rise over run, an indication of the steepness of the line. We may write a line as y = mx + b and use the value of m for Slope.
Slope-Intercept Equation of a Line
The familiar y = mx + b, where m represents Slope and b is the y-Intercept.
SOHCAHTOA
A mnemonic device for remembering: sine-opposite-hypotenuse; cosine-adjacent-hypotenuse; tangent-opposite-adjacent. Also stands for "some old hippie caught another hippie tripping on acid."
Solid
A three-dimensional geometric figure or body that includes the interior region.
Solid of Revolution
When a function is rotated around an axis (of Revolution) it generates a Solid of Revolution.
Solution
Too often in math class, "the answer." More directly, a Solution is a value (or set of values) that makes a mathematical statement true.
Solution Set
Strictly, any Solution is a Solution Set, the value(s) that make a mathematical statement true.
Speed
A (typically fixed) ratio of length or distance to a unit of time; Speed is a scalar value, as in miles per hour (mph) or feet per second (fps).
Sphere
A three-dimensional figure comprised of points equidistant from a center point; a Sphere has a fixed radius.
Spherical GeometrySpherical Trigonometry
Unlike plane Trigonometry, elementary Spherical Trigonometry is three dimensional. If based in spherical geometry, the math of Spherical Trig gets downright grisly.
Spheroid
An oblate sphere. Sometimes, an ellipsoid.
Spiral
Sometimes Spiral is used to describe a helix. A genuine Spiral is a plane figure of changing radius from a (usually fixed) origin.
SquareSquare Matrix
A Square Matrix has the same number of rows as columns.
Square Root
Given a real value, the number that times itself (squared) produces the given value is its Square Root
SSA Ambiguity
Side-Side-Angle congruence is not enough to establish congruence between two triangles; it is the Ambiguous case.
SSS Congruence
Two triangles whose corresponding sides are congruent are themselves congruent.
SSS Similarity
When corresponding sides of two triangles are in a fixed ratio the triangles are similar.
Standard Equation of a Line
When expressing the Equation of a Line with integral coefficients we may have the Standard Equation of a Line.
Standard Position
An angle in Standard Position has been rotated counterclockwise (for positive rotation) from an initial ray on the positive x-axis.
Stem-and-Leaf Plot
A graphical device to group statistical data, typically by leading digits.
Step Function
A discontinuous Function where the range jumps in increments (usually fixed) may be a Step Function.
Straight Angle
Question: What is a Solution Set in mathematics? Answer: The value(s) that make a mathematical statement true.
Question: What is Speed defined as? Answer: A fixed ratio of length or distance to a unit of time; it's a scalar value.
Question: What is the SOHCAHTOA mnemonic used for? Answer: Remembering trigonometric ratios: sine-opposite-hypotenuse, cosine-adjacent-hypotenuse, tangent-opposite-adjacent.
Question: What is a Sphere defined as? Answer: A three-dimensional figure comprised of points equidistant from a center point; it has a fixed radius. | 677.169 | 1 |
IIT-JEE 2010 Paper I Solutions
Join Us
This section contains 6 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.
1. (1) (2) x (3) (4)
Ans : (3)
2. Consider the two curves C1 : y2 = 4x C2 : x2 + y2 - 6x + 1 = 0 Then, (1) C1 and C2 touch each other only at one point (2) C1 and C2 touch each other exactly at two points (3) C1 and C2 intersect (but do not touch) at exactly two points (4) C1 and C2 neither intersect nor touch each other
3. The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors such that Then, the volume of the parallelopiped is (1) (2) (3) (4)
Ans : (1)
4. Let a and b be non-zero real numbers. Then, the equation (ax2 + by2 + c) (x2 - 5xy + 6y2) = 0 represents (1) four straight lines, when c = 0 and a, b are of the same sign (2) two straight lines and a circle, when a = b, and c is of sign opposite to that of a (3) two straight lines and a hyperbola, when a and b are of the same sign and c is of sign opposite to that of a (4) a circle and an ellipse, when a and b are of the same sign and c is of sign opposite to that of a.
8. A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and the circumcircle of the triangle PQR at the point T. If S is not the centre of the circumcircle, then (1) (2) (3) (4)
12. Consider the system of equations ax + by = 0, cx + dy = 0, where a, b, c, d {0, 1} STATEMENT-1: The probability that the system of equations has a unique solution is 3/8 and STATEMENT-2: The probability that the system of equations has a solution is 1 The probability that system of equations has unique solution = 6/16 = 3/8 II. Homogenous system is always consistent
13. Let f and g be real valued functions defined on interval (-1, 1) such that g''(x) is continuous g(0) 0, g''(0) = 0 STATEMENT-1: and STATEMENT-2: f''(0) = g(0)
Question: What does question 4 state about the equation (ax² + by² + c) (x² - 5xy + 6y²) = 0 when a and b are of the same sign and c is of opposite sign? Answer: It represents two straight lines and a hyperbola | 677.169 | 1 |
Stage: 3 Challenge Level:
Do you know how to find the area of a triangle? You can count the
squares. What happens if we turn the triangle on end? Press the
button and see. Try counting the number of units in the triangle
now. . . .
Stage: 4 and 5Clearly if a, b and c are the lengths of the sides of a triangle and the triangle is equilateral then
a^2 + b^2 + c^2 = ab + bc + ca. Is the converse true, and if so can you prove it? That is if. . . .
Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
Stage: 4 Challenge Level:
Question: What shape is drawn on the sides of the rectangle ABCD in Stage 3? Answer: Semicircles are drawn on the sides of the rectangle ABCD. | 677.169 | 1 |
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