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Circle Properties Study Guide: Angles, Arcs, and Segments color coded study guide helps students learn and recall the properties of circles. One side covers the relationships of angle measures and arc measures. The other side illustrates properties of segment lengths.
If color printing is not an option, print it in BW and show the color version as a slideshow. Students can use highlighters or colored pencils to highlight the appropriate regions.
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
638.19So easy to cut and paste these in my smartboard notebook notes. Love the resource and next year, I may incorporate this in an interactive notebook page for some of the less advanced Geo classes that struggle getting the notes down accurately. Thank you!
Excellent reference for my kiddos. Wish I had the resources available to me to print it in color for them... we spent a bit too much time coloring in class today, I'm afraid. (But enjoyed it - and with state testing, the kids needed to just color.)
Exactly what I was looking for. I needed to teach my mathcounts team the basics of circle properties and this notes page had it all in a quick reference colorful page that my students could understand right away.
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QUESTIONS AND ANSWERS:
sberry90
This will be great as a visual organizer. Do you have anything available that can help "push" the formulas in? The students (10th graders) have a hard time keeping them all straight, and their long term retention (until the end-of-year test) is minimal. Would love to find some activities that will help with that.
September 12, 2012
Showing 1-1 of 1
TEACHING EXPERIENCE
Question: What should students do if they cannot print the study guide in color? Answer: They should print it in black and white and use highlighters or colored pencils to highlight the appropriate regions. | 677.169 | 1 |
Definitions 18 through 20
Def. 18. When a right triangle with one side of those about the right angle remains fixed is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cone. And, if the straight line which remains fixed equals the remaining side about the right angle which is carried round, the cone will be right-angled; if less, obtuse-angled; and if greater, acute-angled.
Def. 19. The axis of the cone is the straight line which remains fixed and about which the triangle is turned.
Def. 20. And the base is the circle described by the straight line which is carried round.
The right triangle ABC with right angle at A is rotated about the side AC to produce a cone. The axis of the cone is AC, and its base is the circle with center at A and radius AB.
The three different kinds of cone are not used by Euclid in the Elements, but they were important in the theory of conic sections until Apollonius' work Conics. In Euclid's time conic sections were taken as the intersections of a plane at right angles to an edge (straight line from the vertex) of a cone. When the cone is acute-angled, the section is an ellipse; when right-angled, a parabola; and when obtuse-angle, a hyperbola. Even the names of these three curves were given by the kind of angle, so, for instance, Euclid knew a parabola as a "section of a right-angled cone." It was Apollonius who named them ellipse, parabola, and hyperbola.
Question: In which work did Apollonius define the conic sections as ellipse, parabola, and hyperbola? Answer: In his work "Conics". | 677.169 | 1 |
In the system shown in Fig. 6-3B, we again start with Cartesian xyz -space. The xy -plane corresponds to the surface of the earth in the vicinity of the origin, and the z axis runs straight up (positive z values) and down (negative z values). The angle θ is defined in the xy -plane in degrees (but never radians) clockwise from the positive y axis, which corresponds to geographic north), where θ represents the angle measured clockwise between P′ and geographic north, r represents the distance or radius rom P′ to the origin, and h represents the distance (altitude or height) of P above the xy -plane. This scheme is preferred by navigators and astronomers.
Spherical Coordinates
Figure 6-4 shows three systems of spherical coordinates for defining points in space. The first two are used by astronomers and aerospace scientists, while the third one is preferred by navigators and surveyors.
In the scheme shown in Fig. 6-4A, the location of a point P is defined by the ordered triple ( θ,ø,r ) such that θ represents the declination of P, ø represents the right ascension of P, and r represents the radius from P to the origin, also called the range. In this example, angles are specified in degrees (except in the case of the astronomer's version of right ascension, which is expressed in hours, minutes, and seconds as defined earlier in this chapter). Alternatively, the angles can be expressed in radians. This system is fixed relative to the stars.
Instead of declination and right ascension, the variables θ and ø can represent celestial latitude and celestial longitude respectively, as shown in Fig. 6-4B. This system is fixed relative to the earth, rather than relative to the stars.
Fig. 6-4. (A) Spherical coordinates for defining points in three-space, where the angles θ and ø represent declination and right ascension, and r represents radius or range. (B) Spherical coordinates for defining points in three-space, where the angles θ and ø represent latitude and longitude, and r represents radius or range.
There's yet another alternative: θ can represent elevation (the angle above the horizon) and ø can represent the azimuth (bearing or heading), measured clockwise from geographic north. In this case, the reference plane corresponds to the horizon, not the equator, and the elevation can cover the span of values between, and including, –90° (the nadir, or the point directly underfoot) and +90° (the zenith). This is shown in Fig. 6-4C. In a variant of this system used by mathematicians, the angle θ is measured with respect to the zenith (or the positive z axis), rather than the plane of the horizon. Then the range for this angle is 0° ≤ θ ≤ 180°.
Fig. 6-4. (C) Spherical coordinates for defining points in three-space, where the angles θ and ø represent elevation (angle above the horizon) and azimuth (also called bearing or heading), and r represents radius or range.
Question: What is the fixed reference for the system shown in Fig. 6-4B? Answer: The earth | 677.169 | 1 |
Spatial Coordinates Practice Problems
Practice 1
Suppose you fly a kite above a perfectly flat, level field. The wind is out of the east-southeast, or azimuth 120°. Thus, the kite flies in a west-northwesterly direction, at azimuth 300°. Suppose the kite flies at an elevation angle of 50° above the horizon, and the kite line is 100 meters long. Imagine that it is a sunny day, and the sun is exactly overhead, so the kite's shadow falls directly underneath it. How far from you is the shadow of the kite? How high is the kite above the ground? Express your answers to the nearest meter.
Solution 1
Let's work in navigator's cylindrical coordinates. The important factors are the length of the kite line (100 meters) and the angle at which the kite flies (50°). Figure 6-5 shows the scenario. Let r be the distance of the shadow from you, as expressed in meters. Let h be the height of the kite above the ground, also in meters.
First, let's find the ratio of h to the length of the kite line, that is, h /100. The line segment whose length is h, the line segment whose length is r, and the kite line form a right triangle with the hypotenuse corresponding to the kite line. From basic circular trigonometry, we can surmise the following:
sin 50° = h/100
Fig. 6-5. Illustration for Solution 1.
sing a calculator, we derive h as follows:
We also know, from basic circular trigonometry, this:
cos 50° = r /100
sing a calculator, we derive r as follows:
In this situation, the wind direction is irrelevant. But if the sun were not directly overhead, the wind direction would make a difference. It would also make the problem a lot more complicated. If you like difficult problems, try this one again, but imagine that the sun is shining from the southern sky (azimuth 180°) and is at an angle of 35° above the horizon.
Question: How far from you is the shadow of the kite? Answer: 86.6 meters (rounded to the nearest meter)
Question: At what angle does the kite fly above the horizon? Answer: 50° | 677.169 | 1 |
You all know the fuselage-sections of a plane and probably also how to construct such a cone on a piece of paper.
In Dutch it's called a 'uitslag', a mathematical expression, so in case of a cone it's called an 'uitslag' of a cone. But of course it does not have to be cone at all. You also can make an 'uitslag' from a box, resulting in 6 squares/rectangles. How do you call an 'uitslag' in English ?
Even Wikipedia does not give an answer.
Question: What is the resulting shape when you construct an 'uitslag' of a cone? Answer: A cone | 677.169 | 1 |
5.3 Law of Sines
Introduction In Section 5.1 we saw how to solve right triangles. In this and the next section we consider two techniques for solving general triangles.
Law of Sines Consider the triangle ABC, shown in FIGURE 5.3.1, with angles a, ß, and ?, and corresponding opposite sides BC, AC, and AB. If we know the length of one side and two other parts of the triangle, we can then find the remaining three parts. One way of doing this is by the Law of Sines.
FIGURE 5.3.1 General triangle
THEOREM 5.3.1 The Law of Sines
Suppose angles a, ß, and ?, and opposite sides of length a, b, and c are as shown in FIGURE 5.3.1. Then
PROOF: Although the Law of Sines is valid for any triangle, we will prove it only for acute triangles—that is, a triangle in which all three angles a, ß, and ?, are less than 90°. As shown in FIGURE 5.3.2, let h be the length of the altitude from vertex A to side BC. Since the altitude is perpendicular to the base BC it determines two right triangles. Consequently, we can write
Question: Which figure is used to illustrate the general triangle? Answer: FIGURE 5.3.1 | 677.169 | 1 |
Loci: Convergence
Eratosthenes and the Mystery of the Stades
by Newlyn Walkup
Eratosthenes' Argument
Eratosthenes uses these five main assumptions as hypotheses for his famous geometric approximation of the Earth's circumference. His approximation would not be surpassed for centuries to come. The method devised by Eratosthenes is the basis for the complex "astrogeodetic method" which is used to measure the Earth today [5, p.153 ]. His elegant geometric argument, illustrated below, is sound and simple.
Claim: The circumference of the Earth is approximately 250,000 stades.
Proof:
Given 1. That Alexandria and Syene lie on the same meridian.
2. That light rays from the Sun which strike the Earth are parallel.
3. That the distance between Alexandria and Syene is 5000 stades.
4. That the angle formed by the shadow and the staff in Alexandria at the
summer solstice is equal to 1/50th of a circle.
5. That the Earth is a sphere.
By construction, the staff in Alexandria is perpendicular to the ground, so in the plane of the meridian it is orthogonal to the cross-sectional circle of the Earth.
By definition of orthogonal, the staff in Alexandria is perpendicular to a line m which is tangent to the Earth at the base of the staff.
Likewise, the staff in Syene is perpendicular to a line n tangent to the Earth at the staff's base.
Euclid III-19: If a straight line touches a circle, and from the point of contact a straight line is drawn at right angles to the tangent, the center of the circle will be on the straight line so drawn [10, p.45 ].
Therefore, since the staffs are perpendicular to tangents m and n, if the staffs are extended toward the Earth, their lines intersect at the center of the Earth.
By hypothesis, the light rays striking the Earth are parallel.
Since the staff in Syene casts no shadow, no angle is formed by the intersection of the light rays and the staff, thus the line of the staff is parallel to the light rays.
Question: What is Eratosthenes' approximate value for the Earth's circumference in stades? Answer: 250,000 stades | 677.169 | 1 |
NWN: 0071: Algebra Mind Map 6:
3.5.1.1 (Bedroom one):
There's a lot of new stuff here. There are two equations to remember, the first deals with finding the distance between two points on a coordinate plane and the second deals with graphing circles. To find the distance between two points we use the distance formula: d = [(x2-x1)^2 + (y2-y1)^2]^0.5, which means that for two points on a coordinate system if you take the square root of the sum of the quantity difference of the x coordinates squared and the quantity difference of the y coordinates squared, the result you get is the distance between those two points.
The equation for the circle is derived from the distance equation and is r^2 = (x-a)^2 + (y-b)^2, where r is the radius of the circle and a and b are the coordinates of the centre of the circle. The OPPOSITE values of a and b in the equation give the point on the coordinate plane at the centre of the graphed circle. Also r will always be positive, so when working with the circle equation you can disregard any negative values resulting from the use of radicals when solving for r. When calculating a and b for given values of r, x and y it's often necessary to complete the square (i.e. (a=1 and 1/2b)^2) for either the x coordinate or y coordinate or both to get your results for a and b in the circle equation. Note than when a and b are zero the centre of the circle will be at the origin of the x and y axes. Circles will always have a coefficient of 1 for the x and y coordinates of the circle equation.
Neverwinter Nights:
The first of two bedrooms in the reception area, and Zanzi looks at the familiar objects in this room but in the context of the Moonstone Mask she sees them in a totally new light. Paradoxically, in some ways she has come full circle (with her familiarity of the bed and desk) yet there is a massive distance from the feelings originally evoked by these objects.
Question: What is Zanzi seeing the objects in the context of? Answer: The Moonstone Mask | 677.169 | 1 |
A secant line of a curve is a line that (locally) intersects two points on the curve. The word secant comes from the Latinsecare, to cut.
It can be used to approximate the tangent to a curve, at some point P. If the secant to a curve is defined by two points, P and Q, with P fixed and Q variable, as Q approaches P along the curve, the direction of the secant approaches that of the tangent at P, (assuming that the first-derivative of the curve is continuous at point P so that there is only one tangent). As a consequence, one could say that the limit as Q approaches P of the secant's slope, or direction, is that of the tangent. In calculus, this idea is the basis of the geometric definition of the derivative. A chord is the portion of a secant that lies within the curve.
A secant line on a map is a line where the projection is without distortion.
Question: What is the definition of a secant line in the context of a curve? Answer: A secant line is a line that locally intersects two points on the curve. | 677.169 | 1 |
Circumscribing Shapes When you circumscribe a shape, you generally draw it such that it contains a circle using the least bounding area. Every edge of the circumscribed shape is drawn tangent to the circle so that it brushes it at a single point. You might think this process is hard, but it is actually quite easy if you recall that any regular shape can be quickly and easily drawn within a circle. That means we simply need the radius of the inscribing circle for the circumscribing shape. Confused?
Don't be. Making this realization only makes our life easier. It means that if we find a single point for the circumscribing shape, we can use that point to find the rest of them with ease. Finding a single point involves making use of the trigonemtric sine rules that allow us to take any two angles and a side and get the length of the remaining sides. Take a look at the figure for getting the first point of a circumscribing square. We are using a square for now because it is pretty easy.
We start by making the following assertions. The angle from the center of the circle to two points on the circumscribing square is going to be 360 / n. This is called the internal angle and is the same angle we used earlier for inscribing shapes as well. Now, we extend the legs of a triangle beyond the radius of the circle until the third leg drawn opposte our internal angle touches the circle at only one point. The distance from the center to where the point touches is the radius or Cr. This will be important given information in a second.
Because the far leg is drawn tangent to the circle, the radius is perfectly perpindicular. That means the radius forms a right angle with the tangent line breaking our larger triangle T(C, P1, P2) into two triangles T(C, P1, T1) and T(C, P2, T1) where C is the center point, P1 and P2 are the points on the circumscribing square we are creating and T1 is the tangent point. By definition the radius also becomes a perpindicular bisector splitting that internal angle of ours from 360 / n in half to 360 / n / 2. The third angle or cross side angle is whatever is left of the 180 degrees allowed in a triangle. We now have three angles and a side. The following sin rules now apply:
sideA / sin(angleA) = sideB / sin(angleB) = sideC / sin(angleC)
In the example, sideA will be Cr, angleA will be the cross side angle and angle b will be 90 degrees. This reduces the sideB equation to:
Frigin sweet. We have a basic equation that gives us the distance between the center of our circle to the first point in the circumscribing square. Use this distance as the radius of a new circle given the center of the original. Now, inscribe the rest of the square within this larger circle and you are done. You've just circumscribed the circle within a square. You could just as easily have used another shape though.
Question: What is the role of the radius (Cr) in the process? Answer: It's the distance from the center of the circle to the tangent point, which is used to find the distance between the center and the first point in the circumscribing square.
Question: Why is finding a single point for the circumscribing shape helpful? Answer: It allows us to find the rest of the points with ease using trigonometric sine rules.
Question: What does it mean to circumscribe a shape? Answer: To draw a shape such that it contains a circle using the least bounding area, with every edge tangent to the circle at a single point.
Question: What is the final step to circumscribe a circle within a square? Answer: Inscribe the rest of the square within the larger circle using the found distance as the radius. | 677.169 | 1 |
Author:
David Annal
This site provides information about all aspects of tessellations, from their history and development to...
Type: Reference Material
Date Added: Feb 19, 2005
Date Modified: Nov 22, 2011
Author:
Rudy Lopes
A conic section is a curve formed by the intersection of a cone with a plane. Use this printable study...
Type: Reference Material
Date Added: Dec 18, 2007
Date Modified: Dec 18, 2007
Question: Which material was added to the site earlier, tessellations or conic sections? Answer: tessellations | 677.169 | 1 |
Joseph Malkevitch
Department of Mathematics and Computer Studies
York College (CUNY)
Jamaica, New York 11451
Session 1
Note: Geometry is a vast subject and it "sits" within the vaster subject of mathematics. We must begin our discussions somewhere so I will use "common language" for a variety of mathematical and geometrical terms which will be made more precise (or fuzzy) as we go along.
b. We live on a large surface which can be crudely approximated by a sphere. Should we know something about the "spherical geometry?"
c. Can we tell what space we live in by computing the angle sum for a triangle in the space? If one has a plane triangle, then its angle sum is exactly 180 degrees, but on the surface of a Euclidean sphere, the angle sum would be greater than 180 degrees, while in the Bolyai-Lobachevsky Plane the angle sum would be less than 180 degrees.
The study of visual phenomena
a. If we had to describe the shape of a maple leaf to an intelligent English- speaking alien over the phone, what would we say?
b. Why do the paintings of the Renaissance look more realistic than 2-dimensional representational art of earlier cultures (Egyptian, Greek, Assyrian, Roman, etc.)? Hint: Railroad tracks, which we know to be parallel, appear to meet in the distance.
c. Which pixels on a digital screen (200x200) should be lit up to represent a Euclidean circle? Which pixels for the line y = 3x + 2?
d. Young children can tell a maple leaf from an oak leaf but what is the difference in geometrical terms?
Geometry as a branch of mathematics and geometry as a branch of physics
Physics has theories. Theories can be falsified. New information can force one to realize that a current theory is not correct. Thus, peculiarities in the orbit of the planet Mercury, noticed at the turn of the 20th century, showed that Newton's Laws were not sufficient to explain Mercury's orbit. Einstein's Theory of General Relativity involving the "curvature" of space was involved in getting a richer understanding of what was going on.
Question: What is one reason why Renaissance paintings look more realistic than earlier 2D representational art? Answer: The use of linear perspective, which creates the illusion of depth and distance. | 677.169 | 1 |
What name would a young child who has a pet dog give to a kind of dog that the child has not seen before? What name would the child give to a raccoon?
b. Mathematicians give a name to an interesting "property" that some collection of objects possesses (e.g. continuous function; planar graph; rational number, etc.)
c. Imagine trying to deal with the world without words for colors. Suppose there were words for colors but not for blue. Remember the difficulty that some people are color blind or or have color perception problems. Thus, I may be able to make the distinction between blue, green and violet, but to someone else they would all look alike or be only subtly different.
d. What name would you give the shape below:
p
What name would you give the shape below:
d
What name would you give the shape below
What name would you give the shape below:
If you turn this page upside down, what names would the shapes be given?
This example shows that there are situations where we use different names for shapes that have been rotated 180 degrees while in other cases the name would not change.
Question: In the given example, what is the difference in names given to shapes that have been rotated 180 degrees compared to those that have not? Answer: The names do not change for shapes rotated 180 degrees in this example. | 677.169 | 1 |
Exponential roots in .NET framework
Printer PDF417 in .NET framework Exponential roots
Using Barcode printer for .NET framework Control to generate, create PDF417 image in VS .NET applications.
The wonderful world of math is also home to concepts like cube roots, fourth roots, fifth roots, and so on These roots are a factor of a number, which, when cubed (multiplied by itself three times), taken to the fourth power (multiplied by itself four times), and so on, produce the original number A couple of examples seem to be in order: The cube root of 27 is 3 If you cube 3 (also known as raising it to the third power or multiplying 3 3 3), the product is 27 The fourth root of 16 is that number which, when multiplied by itself four times, equals 16 Any guesses Drumroll, please: 2 is the fourth root of 16 because 2 2 2 2 = 16
Geometry is the branch of mathematics that makes grown adults cry end of discussion What You want a more specific explanation of geometry than that Okay, geometry is the branch of mathematics concerned with measuring things and defining the properties of and relationships between and among shapes, lines, points, angles, and other such objects Hey, don t blame us; you asked for it Before you read any further, you should note a few things to remember: Arcs, circles, triangles, and angles are measured in degrees and (not very often) in minutes (which are smaller than degrees) A circle has 360 degrees (360 ) A quadrilateral (shapes with four sides like a square or rectangle) has 360 Any arc or angle that isn t a complete circle or quadrilateral measures less than 360
Angles are formed when two lines intersect at a point Angles are measured in degrees The greater the number of degrees, the wider the angle is: A straight line is 180 A right angle is exactly 90 An acute angle is more than 0 and less than 90 An obtuse angle is more than 90 but less than 180 Complementary angles are two angles that equal 90 when added together Supplementary angles are two angles that equal 180 when added together Take a look at the different types of angles in Figure 8-1
A triangle consists of three straight lines whose three angles always add up to 180 The sides of a triangle are called legs Triangles can be classified according to the relationship between their angles or the relationship between their sides or some combination of these relationships: Isosceles triangle: Has two equal sides, and the angles opposite the equal sides are also equal Equilateral triangle: Has three equal sides, and all the angles measure 60 Right triangle: Has one right angle (90 ); therefore, the remaining two angles are complementary (add up to 90 ) The side opposite the right angle is called the hypotenuse, which is the longest side of a right triangle Check out Figure 8-2 to see what these triangles look like
Question: What are the two types of angles that add up to 90 degrees? Answer: Complementary angles
Question: Which branch of mathematics is geometry a part of? Answer: Mathematics
Question: What is the sum of the angles in a triangle? Answer: 180 degrees
Question: What is the measure of a right angle in degrees? Answer: 90 | 677.169 | 1 |
Philosophy of Math Lecture 30- Non-Euclidean Geometry—History and Examples - Part 3 of 4 This is the Part 3FINDING THE (GCF) USING THE EUCLIDEAN ALGORITHM Using the Euclidean Algorithm (contnouos division in finding the GCF of a certain set of numbers.
Non-Euclidean Geometry This video discusses elliptical and hyperbolic geometries. It may be interesting but only if you are a math nerd!
Some Non-Euclidean Geometry from Thinkwell Calculus - Preview Do you wish that Professor Burger was your teacher? Click the link to learn more about Thinkwell's Online Video Calculus Course.
Euclidean & Non-Euclidean Geometries Part 5: Axioms (Cont.) Continued from Part 4. I knock a glass candleholder off the shelf during the video, and the sound, while not very loud, might surprise you or your cat. I also knock something else off the ledge, but I don't remember what it was. EUCLID'S POSTULATE III. For every point O and every point A not equal to O there exists a circle with center O and radius OA. DEFINITION. The ray AB is the following set of points lying on the line AB: those points that belong to the segment AB and all points C such that B is between A and C. The ray AB is said to emanate from A and to be part of line AB. DEFINITION. Rays AB and AC are opposite if they are distinct, if they emanate from the same point A, and if they are part of the same line AB = AC. DEFINITION. An "angle with vertex A" is a point A together with two nonopposite rays AB and AC (called the sides of the angel) emanating from A. DEFINITION. If two angles BAD and CAD have a common side AD and the other two sides AB and AC form opposite rays, the angles are supplements of each other, or supplementary angles. DEFINITION. An angle BAD is a right angle if it has a supplementary angle to which it is congruent. EUCLID'S POSTULATE IV. All right angles are congruent to each other. DEFINITION. Two lines m and n are parallel if they do not intersect, ie, if no point lies on both of them. EUCLID'S POSTULATE V. (THE PARALLEL POSTULATE) For every line l (el) and for every point P that does not lie on l (el) there exists and unique line m through P ...
Euclidean Crisis Demo
DystopiaGround - Love for Sail - Euclidean Track #3 from the album "Euclidean" composed by "Kenji Ito", arranged by "Shinji Hosoe" and performed by "nao (former vocalist for fripSide)".
Non-Euclidean Level Design This is an old version. Please view the new version here:
Question: True or False: Professor Burger is mentioned as a teacher in the text. Answer: True.
Question: What is the main topic of this text? Answer: The text discusses Euclidean and Non-Euclidean geometries, with a focus on the Euclidean algorithm for finding the greatest common factor (GCF) and the history and examples of non-Euclidean geometries.
Question: What is the fifth postulate of Euclidean geometry, also known as the parallel postulate? Answer: The fifth postulate states that for every line l and for every point P that does not lie on l, there exists a unique line m through P that does not intersect l. | 677.169 | 1 |
vector
Representation and Reference Systems
The simplest representation of a vector is as an arrow connecting two points. Thus, AB is used to designate the vector represented by an arrow from point A to point B, while BA designates a vector of equal magnitude in the opposite direction, from B to A. In order to compare vectors and to operate on them mathematically, however, it is necessary to have some reference system that determines scale and direction. Cartesian coordinates are often used for this purpose. In the plane, two axes and unit lengths along each axis serve to determine magnitude and direction throughout the plane. For example, if the point A mentioned above has coordinates (2,3) and the point B coordinates (5,7), the size and position of the vector are thus determined. The size of the vector in the x -direction is found by projecting the vector onto the x -axis, i.e., by dropping perpendicular line segments to the x -axis. The length of this projection is simply the difference between the x -coordinates of the two points A and B, or 5 - 2 = 3. This is called the x -component of the vector. Similarly, the y -component of the vector is found to be 7 - 3 = 4. A vector is frequently expressed by giving its components with respect to the coordinate axes; thus, our vector becomes [3,4].;g666;none;1;g666;;;block;;;;no;1;4224n;178277n;;;;;vector-ab666;;;left;stack;;;;;;;;left;stack;2745n;;;The components of the vector AB → are given by its projections on each of the coordinate axes.CE5
Knowledge of the components of a vector enables one to compute its magnitude—in this case, 5, from the Pythagorean theorem [(32 + 42)1/2 = 5)]—and its direction from trigonometry, once the lengths of the sides of the right triangle formed by the vector and its components are known. (Trigonometry can also be used to find the component of the vector as projected in some direction other than the x -axis or y -axis.) Since the vector points from A to B, both its components are positive; if it pointed from B to A, its components would be [ - 3, - 4] but its magnitude and orientation would be the same.
It is obvious that an infinite number of vectors can have the same components [3,4], since there are an infinite number of pairs of points in the plane with x - and y -coordinates whose respective differences are 3 and 4. All these vectors have the same magnitude and direction, being parallel to one another, and are considered equal. Thus, any vector with components a and b can be considered as equal to the vector [ a,b ] directed from the origin (0,0) to the point ( a,b ). The concept of a vector can be extended to three or more dimensions.
Question: What is the magnitude of the vector AB →? Answer: 5
Question: What is the concept of a vector extended to? Answer: Three or more dimensions.
Question: How many vectors can have the same components [3,4]? Answer: An infinite number. | 677.169 | 1 |
The denition indicates that when a gure is shrunk or enlarged from a center O, by a factor r, the image of each point P of the gure is determined by multiplying OP by r to produce OP on OP . P is the image of P. With this notation, we can write SO,r(P) = P to say that P is the image of P under the size transformation with center O and scale factor r. The enlarging and shrinking lines are determined by a point called the center of the size transformation. The multiplier that enlarges or shrinks the lengths is called the scale factor. For example, the blue gure, A B C P , in Figure 11.9 is the image of parallelogram ABCP when it is enlarged by using point O as the center and 2 as the scale factor. Note that OA is 2 times OA, OB is 2 times OB, OC is 2 times OC, and OP is 2 times OP. The yellow gure, A B C P , is the shrunken image of ABCP when 1 is 2 the scale factor instead of 2. In this case, OA is one-half OA, OB is one-half OB, OC is one-half OC, and OP is one-half OP. Size transformations can also be carried out with a computer drawing program, as illustrated in Figure 11.10.
ISBN: 0-536-08809-8
Buddy Mays/CORBIS
Steve Kaufman/CORBIS666
C H A P T E R 11
EXTENDING GEOMETRY
Original image
Transformed image
F I G U R E 11 .1 0
Effecting a size transformation by using a computer drawing program.
Source: 2006 Corel Corporation Ltd. Box shot(s) reprinted with permission from Corel Corporation.
Size Transformations and Similarity. To relate size transformations to similarity, we rst consider some of the size properties of transformations. Measuring the sides BC B of the parallelograms in Figure 11.9 reveals that AB = AC . This result suggests that B the ratios of lengths of sides of a gure and the ratios of the lengths of corresponding sides in its image are equal. Because of this property and the fact that size transformations also preserve betweenness and angle measure, size transformations also preserve shape. Sometimes a gure can be made to correspond to a gure with the same shape by a single size transformationbut not always. In Figure 11.11 on p. 671, the original triangleon the left in part (a)was rst reected in line r, then this reected image was subjected to a size transformation to produce the blue image of the triangle shown in part (b). Similar gures were described, with focus on proportional variation, on p. 384 and on p. 621. The idea that a combination of transformations is needed to transform a gure into a gure of different size and shape is used in the following denition of similarity.
Question: Which figure in the text is the shrunken image of ABCP with a scale factor of 1/2? Answer: The yellow figure, A B C P
Question: What is the notation used to represent the image of a point P under a size transformation with center O and scale factor r? Answer: SO,r(P)
Question: What is the name of the transformation that combines size transformations and other transformations to create similar figures? Answer: Similarity | 677.169 | 1 |
In the pentagram shown in Mini-Investigation 11.6, an isosceles triangle that forms one of the points of the star, such as ^BFG, is called a golden triangle because the ratio of its longer side to its shorter side is the golden ratio, f = (1 + 15)> 2 L 1.618. We know that the measure of an interior angle of the pentagram, such as ABC, is 108, so we can conclude from the equations 2x + a = 108, 4x + a = 180, and b = 2x, that x = a = 36 and b = 72. Using this information, try to identify 9 other golden triangles in the pentagram. The golden triangle is interesting but hasnt received as much attention as the often-discussed and used golden rectangle. Mathematicians, artists, architects, and others have long considered the golden rectangle an especially pleasing formin nature, in buildings, and in works of art. To construct a golden rectangle, we start with a unit square, as in Figure 11.27(a), and nd the midpoint M of one side. Next, we consider MB. Then, as shown in Figure 11.27(b), we use M as the center and MB as the radius and strike an arc on the extension of DC at E. Finally, we construct EF DE and complete the golden rectangle AFED. To verify that the ratio of the length to the width of rectangle AFED is the golden ratio, and that AFED is a golden rectangle, we use the Pythagorean Theorem and show that d = 15 . It then follows that DE = 1 + 15 = (1 +2 15) . 2 2 2 This length of DE is approximately (1 + 22.236) , or 1.618, and AFED is a golden rectangle. Note that 1.618 is an approximation of the golden ratio, referred to on p. 563.
1
D
M (a) 1 B
C
A
F
1
d
1
D
1 2
M
1 2 (b)
C d
E
ISBN: 0-536-08809-8
F I G U R E 11 . 2 7
Construction of a golden rectangle692
C H A P T E R 11
EXTENDING GEOMETRY
V1
Star Polygons
Question: What is the length of DE in the golden rectangle AFED? Answer: DE = 1 + 15 = 1.618.
Question: What is the measure of an interior angle of the pentagram? Answer: 108 degrees. | 677.169 | 1 |
If we stop after only considering star polygons that can be produced by sequentially connecting points on a circle with line segments, we miss a lot of very interesting star-shaped gures. For example, a quiltmaker might want to use star-shaped gures like those shown in Figure 11.29. To compare these six-pointed gures with six-pointed star polygons, we observe that E 6 F and E 6 F produce regular hexagons, E 6 F and E 6 F produce triangles, and E 6 F 1 5 2 4 3 produces a straight line. Thus the polygons shown in Figure 11.29 are not star polygons, nor can they be produced by connecting points on a circle with line segments and then erasing the interior parts of the segments. However, we can use the term star-shaped polygon to describe a nonconvex symmetric gure like the one shown in Figure 11.29(a) or (b) that isnt a star polygon. Star-shaped polygons have n star-tip points, 2n congruent sides, n congruent point angles with measure a, and n congruent dent angles b such that b = A 360 B + a, or a = b - A 360 B . Exercise n n 48 at the end of this section asks you to verify this relationship. A six-pointed, starshaped polygon with point angle 30 is denoted by 630 . Figure 11.30 compares a star polygon and star-shaped polygon. It illustrates the importance of precise denitions, which are required to differentiate two different but closely related ideas. Although the basic shape of the ve-pointed stars shown are the same here, that isnt always the case when an n-pointed star polygon and an n-pointed star-shaped polygon are compared. This information about the relationship between dent angles and point angles is useful in constructing star-shaped polygons that meet our specications. For example, if you want to produce a star-shaped polygon with a specic point angle for a special quilt, you can quickly calculate the measure of the dent angle for that gure. Conversely, if you want to make a star-shaped polygon with a specic dent angle for tessellation or other695
5 (a) Star polygon 2 Nonsimple, nonconvex 5 vertices 5 sides 5 point angles F I G U R E 11 . 3 0
(b) Star-shaped polygon 5 36 Simple, nonconvex 10 vertices 10 sides 5 point angles
Comparison of a star polygon and a star-shaped polygon.
design purposes, you can quickly calculate the measure of the point angle for that gure. Example 11.8 demonstrates how to construct a desired star-shaped polygon.
Example
11.8
Problem Solving: The Star Design
An artist wants to make a painting that includes a ve-pointed star-shaped polygon with a fairly thin point angle of 18. How could the artist accurately construct the star-shaped polygon 518 ?
SOLUTION
Question: Can a star-shaped polygon be produced by connecting points on a circle with line segments? Answer: No. | 677.169 | 1 |
360 n x
52. Making Connections. Show a connection between geometry and art by making and coloring a tessellation that uses the following polygons:
x
x
x {418} Regular pentagon
53. Making Connections. Connect algebra and geometry by nding an algebraic expression for the ratio of the length to width of golden rectangle AHCD. Do not use approximate values for square roots.
D. Communicating and Connecting Ideas
49. Discuss how you would produce an accurate drawing of a six-pointed star-shaped polygon with a point angle of 60. Describe your procedure in detail. 50. Work in a small group and devise a procedure to test whether people really do think that the golden rectangle is the most aesthetically pleasing rectangle. Write a paragraph explaining your procedure and the results.
A
B
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1 2
M
1 2
F
C701
Section
11.4
The Regular Polyhedra Prisms and Pyramids Cylinders, Cones, and Spheres Symmetry in Three Dimensions Visualizing Three-Dimensional Figures
In this section, we classify and dene various types of three-dimensional gures. We also explore properties of these gures, such as symmetry, and relationships involving them. Finally, we describe and analyze different ways to view three-dimensional gures.
Three-Dimensional Figures
Essential Understandings for Section 11.4
All polyhedra can be described completely by their faces, edges, and vertices. There is more than one way to classify most solids. Polyhedra can have reectional or rotational symmetry. Solid gures can be viewed from different perspectives.
The Regular Polyhedra
Vertex Face
Edge F I G U R E 11 . 31
Parts of a polyhedron.
Question: Which of the following is NOT a part of a polyhedron? A) Faces B) Edges C) Colors D) Vertices Answer: C) Colors | 677.169 | 1 |
As shown in Figure 11.37, a pyramid is named by the shape of its base. When the base of a pyramid is a regular polygon, the lateral faces are isosceles triangles, and the altitude is perpendicular to the base at its center, the pyramid is called a right regular pyramid. In such a pyramid, the height of an isosceles triangular lateral face is called the slant height. In contrast to the pyramid shown in the cartoon on p. 706, the pyramids built by the ancient Egyptians were spectacular. Mini-Investigation 11.9 suggests that you broaden your perspective on the applicability on Eulers formula.
M I N I - I N V E S T I G A T I O N 11 . 9
Draw a square pyramid and show that Eulers formula holds for it.
Using Mathematical Reasoning
Do you think that Eulers formula holds for prisms and pyramids? Copy and complete a chart like the following to help you investigate this question and support your conclusion.
Name of Polyhedron Number of Number of Number of Base Edges Vertices V Faces F Number of V F E 2? Edges E (Yes or No)
ISBN: 0-536-08809-8
Triangular prism Square prism n-gon prism Triangular pyramid Square pyramid n-gon pyramid
3 ? n 3 ? n
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?706
C H A P T E R 11
EXTENDING GEOMETRY
The New Yorker Collection 1963 Robert Weber from cartoonbank.com. All Rights Reserved.
Cylinders, Cones, and Spheres
If you imagine prisms, pyramids, and polyhedra with thousands of faces, these solid gures come very close to the solids with curved surfaces shown in Figure 11.38. In the cylinder and cone, the shaded circles are called bases. The radius, r, of the cylinder or cone is the radius of its base. The height, h, of a cylinder is the perpendicular distance from one base to the other. The line through the centers of the bases of a cylinder is called the axis of the cylinder. The height, h, of the cone is the perpendicular distance from the vertex, or apex, V, to the base. The line through the center of the base and the vertex is called the axis of the cone. Point O is the center of the sphere, and r is the radius of the sphere. From ice cream cones to cans to architecture
r
V
h
h
O
r
r (a) Circular cylinder F I G U R E 11 . 3 8
r (b) Circular cone (c) Sphere
ISBN: 0-536-08809-8
Cylinders, cones, and spheres707
Question: What is the radius (r) of a sphere? Answer: The radius of a sphere is the distance from its center to any point on its surface.
Question: What is the name of a pyramid with a regular polygon base, isosceles triangular lateral faces, and a perpendicular altitude from the base's center? Answer: A right regular pyramid.
Question: What is the height of an isosceles triangular lateral face of a right regular pyramid called? Answer: The slant height.
Question: What is the topic of Mini-Investigation 11.9? Answer: To draw a square pyramid and show that Euler's formula holds for it. | 677.169 | 1 |
10. A truncated cube 11. A truncated rectangular prism 12. Show that Eulers formula holds for the truncated octahedron and cube octahedron in Figure 11.35(b) and (c). 13. Describe the different planes of reectional symmetry for each of the gures shown in Exercise 7. Assume that the prism in (a) is a right prism with equilateral triangular bases and that the pyramids in (b) and (c) are right regular pyramids. 14. Describe the different axes of rotational symmetry for each of the gures shown in Exercise 7. Assume that the triangular faces of gure (a) are equilateral triangles and the nontriangular face of gure (c) is a square. 15. One type of axis of rotational symmetry and one type of plane of reectional symmetry are shown in the following square prism. How many different planes of symmetry and different axes of rotational symmetry does the square prism have?
A. Reinforcing Concepts and Practicing Skills
1. a. What is the difference between a regular tetrahedron and a nonregular tetrahedron? b. Describe or sketch a nonregular tetrahedron. 2. Complete Exercise 1 for an octahedron. 3. Name some real-world objects that can be modeled by the following geometric gures: a. a cube b. a rectangular prism other than a cube c. a triangular prism d. a pyramid e. a cylinder f. a cone g. a sphere 4. Name or describe some other hexahedrons besides a cube. 5. a. How many pairs of bases does a right rectangular prism have? b. Is there any other prism with more than one pair of faces that can be identied as bases? 6. Make a sketch of the following: a. a circular cylinder with radius of approximately 3 cm and height of approximately 20 cm b. a circular cylinder with radius of approximately 10 cm and height of approximately 3 cm c. a circular cone with radius of approximately 3 cm and height of approximately 20 cm d. a circular cone with radius of approximately 10 cm and height of approximately 3 cm 7. Show that Eulers formula holds for each of the following polyhedra:
(a)
(b)
(c)
Copy and complete the following table:
Polyhedron F V F V E
a. b. c.
? ? ?
? ? ?
? ? ?
? ? ?
Show that Eulers formula holds for the polyhedra described in Exercises 811. 8. A right pentagonal prism 9. A hexagonal pyramid
Question: What is the total number of planes of reflectional symmetry in a square prism? Answer: A square prism has 11 planes of reflectional symmetry: 2 vertical planes through each of the 4 edges and 1 horizontal plane through the center of each base.
Question: What is the relationship between the number of faces (F), vertices (V), and edges (E) in a polyhedron, according to Euler's formula? Answer: Euler's formula states that for any convex polyhedron, F - V + E = 2.
Question: How many different axes of rotational symmetry does a square prism have? Answer: A square prism has an infinite number of axes of rotational symmetry, as it can rotate around any line passing through its center and perpendicular to its bases. | 677.169 | 1 |
1. Describe the ve regular polyhedra and explain how they are named. 2. Give the dimensions of a picture frame that is a golden rectangle. 3. Show that Eulers formula holds for a square pyramid. 4. Draw two patterns of squares that can be used to form cubes. 5. Draw the front, side, and top views of the following object: 15. Consider the following gure:
a. If it is drawn on a balloon that can be stretched, shrunk, and twisted, draw two different shapes that could be produced. b. If it is made of wire, draw two different shapes that its shadow could take.
Reasoning and Problem Solving
6. Draw a cube in perspective. 7. Draw a tetrahedron in perspective. 8. Draw a sketch of the star polygon {8}. Give another sym3 bold for this same star polygon. Is it a regular polygon? 9. An eight-point star-shaped polygon has a point angle of 36. What is the measure of its dent angle? 10. Describe the symmetry properties of the following gures: 16. If you marked the midpoints of the edges of a cube and sliced off all its corners through the midpoints of its edges, how many and what type of faces would the truncated gure have? 17. Do you believe the following generalization about the measure of a point angle of a star polygon is true? The measure of a point angle of the star polygon E n F is d (|n - 2d|180) . Support your belief in writing, giving n evidence. 18. How do the axes of rotational symmetry of an octahedron compare to the axes of rotational symmetry of a cube?
(a)
(b)
11. How many planes and axes of rotational symmetry does a right rectangular prism have? 12. Name three regular polygons that will tessellate the plane. 13. Draw a small portion of a semiregular tessellation. a. Tell why it is semiregular. b. What is true about a tessellation that utilizes a combination of regular polygons but isnt semiregular? 14. Use tracing paper to nd the image of the quadrilateral shown for the following transformations:
B E D A C
a. TAB 4 c. MAB e. TAB*RB,180
b. R A,90 d. SE,2
19. The Fancy Quilt Problem. Stacy wants to make a quilt that uses a 12-pointed star-shaped polygon that tessellates with equilateral triangles. To make the star, she needs to know the measure in degrees of the point angles and the dent angles of the star-shaped polygon. If possible, supply this information and sketch the quilt pattern. If not, explain what additional information you need in order to do so REVIEW
721
Question: What is the measure of a point angle of an eight-point star-shaped polygon with a point angle of 36 degrees? Answer: The measure of a dent angle in an eight-point star-shaped polygon is 180 - 36 = 144 degrees. | 677.169 | 1 |
To determine the trigonometric functions for angles of π/3 radians (60 degrees) and π/6 radians (30 degrees), we start with an equilateral triangle of side length 1. All its angles are π/3 radians (60 degrees). By dividing it into two, we obtain a right triangle with π/6 radians (30 degrees) and π/3 radians (60 degrees) angles. For this triangle, the shortest side = 1/2, the next largest side =(√3)/2 and the hypotenuse = 1. This yields:
,
,
.
Inverse functions
The trigonometric functions are periodic, and hence not injective, so strictly they do not have an inverse function. Therefore to define an inverse function we must restrict their domains so that the trigonometric function is bijective. In the following, the functions on the left are defined by the equation on the right; these are not proved identities. The principal inverses are usually defined as:
For inverse trigonometric functions, the notations sin−1 and cos−1 are often used for arcsin and arccos, etc. When this notation is used, the inverse functions could be confused with the multiplicative inverses of the functions. The notation using the "arc-" prefix avoids such confusion, though "arcsec" can be confused with " arcsecond".
Just like the sine and cosine, the inverse trigonometric functions can also be defined in terms of infinite series. For example,
These functions may also be defined by proving that they are antiderivatives of other functions. The arcsine, for example, can be written as the following integral:
Analogous formulas for the other functions can be found at Inverse trigonometric function. Using the complex logarithm, one can generalize all these functions to complex arguments:
Properties and applications
The trigonometric functions, as the name suggests, are of crucial importance in trigonometry, mainly because of the following two results.
Law of sines
The law of sines states that for an arbitrary triangle with sides a, b, and c and angles opposite those sides A, B and C:
also known as:
where R is the radius of the triangle's circumcircle.
A Lissajous curve, a figure formed with a trigonometry-based function.
It can be proven by dividing the triangle into two right ones and using the above definition of sine. The law of sines is useful for computing the lengths of the unknown sides in a triangle if two angles and one side are known. This is a common situation occurring in triangulation, a technique to determine unknown distances by measuring two angles and an accessible enclosed distance.
Law of cosines
The law of cosines (also known as the cosine formula) is an extension of the Pythagorean theorem:
also known as:
In this formula the angle at C is opposite to the side c. This theorem can be proven by dividing the triangle into two right ones and using the Pythagorean theorem.
Question: Are the trigonometric functions injective? Answer: No, they are not injective.
Question: Which of the following is NOT a method to define inverse trigonometric functions? A) Infinite series B) Antiderivatives C) Complex logarithm D) Integration Answer: D) Integration
Question: What is the law of sines used for in a triangle? Answer: It is used for computing the lengths of the unknown sides in a triangle if two angles and one side are known.
Question: What is the length of the hypotenuse in the right triangle? Answer: 1 | 677.169 | 1 |
Parabola Simulation
Move the Point (C) on the Directrix and Notice how the Parabola Point traces the Parabola Locus.
Try moving the Focus point and note the shape of the parabola depending on the distance from Directrix
Definition of a parabola: A Parabola is the collection (locus) of all points. That are equidistant from a line (The Directrix) and a point not on the line (The Focus).
Does this definition stand up to the measurements shown in the applet?
QUESTIONS:
What happens to the parabola as you move the focus closer to the directrix?
Question: According to the text, does the definition of a parabola hold true to the measurements shown in the applet? Answer: The text does not provide a definitive answer to this, it only asks the question. | 677.169 | 1 |
I tried it over and over, and again and again I got the answer for number 10 on the Episode Quiz to be "A", although the site says the correct answer is "C". Could somebody please explain?
Answers
This is a tough one!Try following these steps:1. Make a perpendicular from the top two points of the trapezoid down to the bottom base. You'll then have 2 triangles and a rectangle.2. You have a 45-45-90 triangle on the right. The hypotenuse is 6 * sqrt 2. This means that the two sides are a length of 6 given the 45-45-90 ratio. The side that you drew from the top point is therefore 6.3. You know that the sides that you drew are the same length given the properties of a trapezoid (top and base are parallel). 4. In the 30-60-90 triangle on the left, you now know the side you drew has a length of 6. 5. Use the 30-60-90 ratio of a-2a-a*sq root 3, you find that a=2sqrt3. Since you're looking for the side that is 2a, you get 4sqrt3, which is answer "C"Make sense?
Question: What is the length of the hypotenuse of the 45-45-90 triangle in the solution? Answer: 6 * sqrt 2 | 677.169 | 1 |
In a parallelogram, where both pairs of opposite sides and angles are equal, this formula reduces to
Alternatively, we can write the area in terms of the sides and the intersection angle θ of the diagonals, so long as this angle is not 90°:[10]
The area of a quadrilateral ABCD can be calculated using vectors. Let vectors AC and BD form the diagonals from A to C and from B to D. The area of the quadrilateral is then
which is half the magnitude of the cross product of vectors AC and BD. In two-dimensional Euclidean space, expressing vector AC as a free vector in Cartesian space equal to (x1,y1) and BD as (x2,y2), this can be rewritten as:
In the following table it is listed if the diagonals in some of the most basic quadrilaterals bisect each other, if their diagonals are perpendicular, and if their diagonals have equal length.[18] The list applies to the most general cases, and excludes named subsets.
Note 1: The most general trapezoids and isosceles trapezoids do not have perpendicular diagonals, but there are infinite numbers of (non-similar) trapezoids and isosceles trapezoids that do have perpendicular diagonals and are not any other named quadrilateral.
Note 2: In a kite, one diagonal bisects the other. The most general kite has unequal diagonals, but there is an infinite number of (non-similar) kites in which the diagonals are equal in length (and the kites are not any other named quadrilateral).
In any convex quadrilateral ABCD, the sum of the squares of the four sides is equal to the sum of the squares of the two diagonals plus four times the square of the line segment connecting the midpoints of the diagonals. Thus
where x is the distance between the midpoints of the diagonals.[20]:p.126 This is sometimes known as Euler's quadrilateral theorem and is a generalization of the parallelogram law. A corollary is the inequality
where equality holds if and only if the quadrilateral is a parallelogram.
This relation can be considered to be a law of cosines for a quadrilateral. In a cyclic quadrilateral, where A + C = 180°, it reduces to pq = ac + bd. Since cos (A + C) ≥ -1, it also gives a proof of Ptolemy's inequality.
The shape of a convex quadrilateral is fully determined by the lengths of its sides in sequence and of one diagonal between two specified vertices. The two diagonals p, q and the four side lengths a, b, c, d of a quadrilateral are related[24] by the Cayley-Mengerdeterminant, as follows:
Each pair of opposite sides of the Varignon parallelogram are parallel to a diagonal in the original quadrilateral.
Question: Is a parallelogram a special case of a quadrilateral? Answer: Yes.
Question: What is the area of a quadrilateral ABCD using vectors? Answer: Area = 1/2 * |AC x BD|.
Question: In which case does the area formula reduce to base * height? Answer: When the diagonals intersect at a 90° angle. | 677.169 | 1 |
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You have 3 points labelled A, B and C. You then have another 3 points labelled
0
You have 3 points labelled A, B and C. You then have another 3 points labelled
1, 2 and 3. The aim of the puzzle is to connect point A with point 1, 2 and 3. Point
B with point 1, 2 and 3 and point C with point 1, 2 and 3.
Now while connecting the points you have to follow one rule - the lines cannot
cross over each other.
1 Answer
0
There is no solution to it, if you consider 2 dimensions. It is impossible to join
each of points A, B and C with points 1, 2 and 3 without lines crossing each
other..
The other possible 3D structure is Pyramid. Take points 1, 2 and 3 as a vertices of the triangular base and points A, B and C along the height of the Pyramid which is perpendicular to the triangular base and passing through the apex.
Question: Can each of points A, B, and C be connected to points 1, 2, and 3 without lines crossing each other in a 2D plane? Answer: No. | 677.169 | 1 |
Parallelogram vs Rectangle
ParallelogramParallelogram vs Rhombus
Parallelogram and rhombusMedian vs Average (Mean)
Median and mean are measures of central tendency in descriptive statistics. Often Arithmetic mean is considered as the average of a set of observations. Therefore, here mean is considered as the average. However, average is not the arithmetic mean at all the times.
Average
The arithmetic...
Mathematics vs Applied Mathematics
Mathematics first emerged from the daily necessity of the ancient people to count. Trading, referring to time, and measuring the crop or land required numbers and values to represent them. Search of creative ways of solving above problems resulted in the basic form of...
Ellipse vs Oval
Ellipse and ovals are similar looking geometrical figures; therefore, their appropriate meanings are sometimes confusing. Both being planar shapes with similar looks, such as an elongated nature and the smooth curves make them almost identical. However, they are different, and their subtle...
Arithmetic Sequence vs Geometric Sequence
The study of patterns of numbers and their behaviour is an important study in the field of mathematics. Often these patterns can be seen in nature and helps us to explain their behaviour in a scientific point of view. Arithmetic sequences and Geometric sequences are two...
Question: What is one key difference between a parallelogram and a rhombus? Answer: A rhombus is a special type of parallelogram where all sides are equal in length. | 677.169 | 1 |
If you point your "gun" straight ahead, stick out your middle finger so that it points left and all your fingers are at right angles to each other.
If you have two vectors that you want to cross multiply, you can figure out the direction of the vector that comes out by pointing your thumb in the direction of the first vector and your pointer in the direction of the second vector. Your middle finger will point the direction of the cross product.
Remember that when you change the order that vectors get cross multiplied, the result goes in the opposite direction. So it's important to make sure that you go in the order of .
Question: Which two fingers are used to represent the directions of the two vectors being cross multiplied? Answer: Thumb and pointer finger. | 677.169 | 1 |
The Sine Ratio Lesson Plan introducing the sine ratio and guiding students to find the length of a side of a right triangle using sine. The inverse sine ratio is also covered. This lesson follows my lesson plan "Introduction to Trigonometry" which is also posted in my TPT Store.
I also have lesson plans on the cosine and tangent ratios posted. If you prefer to teach the three ratios in one lesson, I have the lesson "Sine, Cosine & Tangent" posted at:
Word Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
153
Question: Are there lesson plans available for the cosine and tangent ratios as well? Answer: Yes. | 677.169 | 1 |
Basic Geometric Shapes
In the kindergarten the young kids need to be introduced with the basic shapes. The main basic shapes a kindergarten student should learn are triangles, rectangles, squares, diamonds and circles.
There are three dimensional basic shapes also, like rectangular prisms, or cones or cylinders. The KG students should be able to recognized these basic shapes by looking at daily life objects such as pop cans, ice cream cones and ball.
Question: Which of the following is a three-dimensional shape but not mentioned in the text?
Answer: Sphere | 677.169 | 1 |
Re: Cool problem of the day!
The lengths of the sides of the octagon are 1, 2, 3, 4, 5, 6, 7 and 8 units in some
order. Find the maximum area of the hexagon (square units).
This reminds me of a really neat (but actually pretty hard) problem:
Given a polygon, we define a "flip-stick" to be the following process: take a 'cut' of the polygon (i.e. a line segment that cuts the polygon into exactly two new polygons), then take one of the polygons you produced, reflect it and stick it back on: as long as the flipped thing doesnt overlap with the rest of the polygon, in which case the move is illegal. So essentially you cut of a bit, flip it and stick it back on if its allowed. Does there exist a sequence of "flip-sticks" that can transform a square into an equilateral triangle?
Question: What shape is the target of the transformation in the second problem? Answer: An equilateral triangle | 677.169 | 1 |
I would like to coin together certain properties of [#permalink]
28 Dec 2006, 21:27
6
This post received KUDOS
I would like to coin together certain properties of Coordinate Geometry (from the basics) that I learnt from various sources, some even from the forum. Would appreciate your help in adding more or correcting these.
SLOPE:
slope m=(y1-y2)/(x1-x2)
*A straight line with a -ve slope passes through II and IV quadrants
A straight line with a +ve slope passes through I and III quadrants
*If the slope is 1 the angle formed by the line is 45 degrees
*If the slope of a line is n, the slope of a line perpendicular to it is its -ve reciprocal, -1/n
*If a line is horizontal, slope=0, equation is y=b
*If a line is vertical, slope is not defined, equation is x=a, where a is x-intercept.
*Parallel lines have same slope.
EQUATION:
Equation of a circle is (x - a)^2 + (y-b)^2 = r^2, where (a,b) is the center and r is the radius
Equation of a circle is x^2+y^2=r^2 if (0,0) is the center
Equation of a line is y=mx+b, where m=slope and b=y intercept
* y intercept is the value of y when x is 0
x intercept is the value of x when y is 0
* Points that solve the equation of a line are in the same line
*Given a point and slope, equation of the line can be found
*Given the equation, x and y intercepts can be found
DISTANCE between two points = sqrt[(x1-x2)^2+(y1-y2)^2]
MIDPOINT: xm=(x1+x2)/2 ym=(y1+y2)/2
I know, the properties do not end here. Hope you guys join in.
Last edited by Sumithra on 16 Jan 2007, 13:46, edited 3 times in total.
Question: How can you find the equation of a line given a point and its slope? Answer: Given a point (x1, y1) and a slope m, the equation of the line can be found using the point-slope form: y - y1 = m(x - x1).
Question: What is the equation of a horizontal line? Answer: The equation of a horizontal line is y = b, where b is the y-intercept.
Question: What are the coordinates of the midpoint of a line segment with endpoints (x1, y1) and (x2, y2)? Answer: The coordinates of the midpoint are (xm, ym), where xm = (x1 + x2) / 2 and ym = (y1 + y2) / 2. | 677.169 | 1 |
Serendipity is wonderful …. The first lesson I need to teach next week is the Law of Sines in my Math Studies class and I also need a second blog post for my current COETAIL course about using creative commons images for teaching students. One of the new capabilities of Geometer's Sketchpad Version 5 lets users import photos into the software and to create geometric constructions on top of them. Images can also be modified using transformations. Texas Instruments thought the ability to do math on images was such a good idea that they added it to the new operating system for the TI nSpire calculator.
I have also been wanting to try and use Dan Meyer's Levels of Abstraction idea to encourage students to be what Dan calls "patient problem solvers" The lesson will start with a warmup activity where I show the tree on the left and ask what measurements would you need to find the height of the tree without climbing it? (They have already solved other IB right triangle application problems in previous classes.) I increase the level of abstraction by adding a coordinate grid in Geometer's Sketchpad. Can we still use right triangle trig for the left tree on the slanted hill?
After students decide that the left tree's height can be measured with a right triangle, I have them go up another level of abstraction by having students construct a right triangle on top of the image. See left image below. (Good questions here are; What are the ¨top¨and ¨base¨of the tree? What angles could be measured? What triangle lengths could be measured without climbing the tree? If Sketchpad measures the base of the triangle constructed over the photo in centimeters, what else do you need to know about the photo to get the tree's real world height in meters?)
After using the tangent ratio to find the height of the tree in the photos on the left, I show them the tree in the right photo at the top of this post. Can we use right triangle trig? I increase the level of abstraction by adding a coordinate grid in Geometer's Sketchpad and ask how can we tell that right triangle trig will not help in finding the height of the tree? Students then increase the level of abstraction by constructing an acute triangle on top of the image. For this tree what angles and distances could we measure without climbing it? Does right triangle trig give us the correct answer? (Students can use Sketchpad's measurement ability to check it themselves.)
Students will then complete a Geometer's Sketchpad Lesson Link activity to discover the formula for the Law of Sines. As a lesson wrap-up students are asked to use the Law of Sines formula that they have discovered to find the height of the tree in the photos on the right of this blog post.
Note: Part of this post was written with the help of a Mac OS blog editor called MarsEdit. If you have Geometer's Sketchpad you can download the GSP file here.
Last week I had an experience of how a math lesson could be so much more authentic than a traditional textbook math lesson as I taught my robotics class.
Question: What is the name of the concept used by Dan Meyer to encourage students to be patient problem solvers? Answer: Levels of Abstraction
Question: What is the second blog post topic the author is working on for their COETAIL course? Answer: Using creative commons images for teaching students
Question: Which trigonometric ratio is used to find the height of the tree in the first photo? Answer: Tangent ratio
Question: What is the first step in the warm-up activity for the lesson? Answer: Showing a tree on the left and asking students what measurements they would need to find the height of the tree without climbing it | 677.169 | 1 |
Radii to Tangents
Explanation
When a radius is drawn to a point of tangency, the angle formed is always a right (90 degree) angle. This fact is commonly applied in problems with two tangent segments drawn to a circle from a point. If two radii to tangents are drawn in, a kite with two right angles is formed and the missing angles or sides can be found. Related topics include central angles, tangent segments to a circle, and chords.
Transcript
When we have a tangent to a circle which means it's a line or line segment or even a ray that intersects the circle on only one point if we had to draw a radius to that point, something special happens. Over here I've written a tangent to a circle is and I left it blank purposely perpendicular to the radius drawn at the point of tangency so I can label this as a 90 degree angle. You're going to apply this in lots of different problems where you see a tangent in some form being drawn where it's connected to a radius.
Question: Which special angle is formed when a radius is drawn to the point of tangency in a tangent to a circle? Answer: A 90 degree angle. | 677.169 | 1 |
In
the figure given here, if ABCD, ABFE and CDEF are parallelograms,
then prove that ar(ΔADE) = ar(ΔBCF).
Diagonals AC and BD of
quadrilateral ABCD intersect at O such that ar (ΔBOC) = ar
(ΔAOD). Then show that ABCD is a trapezium.
Prove that equal chords
of a circle subtend equal angles at its centre.
Find the area of a
triangle whose sides measure 20 cm, 30 cm and 40 cm.
A rectangular sheet of
paper 44 cm × 18 cm is rolled along its length and a cylinder
is formed. Find the volume of the cylinder.
A dice is
thrown once. Find the probability of getting
(i) a number greater than 2 (ii) a number less than
4.
SECTION C
Draw the
graph of the equation 3x – 2y = 4. From the graph, find the
coordinates of the point when:
(i) x = 2
(ii) y = 7
A railway half ticket costs half the full fare. The
reservation charges are the same for both half and full
tickets. A family of three adults and five children pay Rs 1200 for
their travel from Hyderabad to Bengaluru. If the basic fare is Rs x
and reservation charge is Rs y, then find the linear equation that
represents the given information.
In the
figure given here, ABCD is a parallelogram. Compute the values of x
and y.
Question: What is the relationship between the areas of triangles ΔADE and ΔBCF? Answer: They are equal (ar(ΔADE) = ar(ΔBCF))
Question: What is the linear equation that represents the cost of travel for a family of three adults and five children from Hyderabad to Bengaluru, given that a railway half ticket costs half the full fare and the reservation charges are the same for both? Answer: 3x + 5y = 1200 (where x is the basic fare for an adult and y is the basic fare for a child) | 677.169 | 1 |
In the Appendix, we give discovery activities that could be used in the classroom or as an assignment for students; in fact the author has used them at the end of a non-Euclidean Geometry course. Although they are targeted to post-calculus geometry students, the activities are gradual, and could be made accessible to students at lower levels by cutting some of the last activities from each non-Euclidean geometry, and/or cutting one or more of the non-Euclidean geometries from the activity. Indeed, the weaker students in the author's classes usually do not complete the hyperbolic elements of the discovery activity. The activities also seem appropriate as extra-credit problems for calculus students wanting to peek at more advanced material.
We use Geometry Playground [1], as the basis for our discovery activities. Although there are other packages that allow discovery in non-Euclidean geometries (Cinderella [2], Spherical Easel [3], NonEuclid [4], Geometry Explorer [5]), Geometry Playground combines all of the "standard" geometries (and a few unusual ones) in one package, and is available freely to anyone with an internet connection, allowing students to use it in a take-home activity.
We would like to thank the referees for many useful suggestions.
2. The Plane Truth
We begin our explorations in Euclidean geometry, the most familiar geometry to most of us. Euclidean geometry might best be explored via the wonderful Euclid's Elements [6], a reproduction of Euclid's own work, written in English, and published with an applet accompanying each theorem to aid in visualization. Wikipedia's Euclidean Geometry page [7] and Wolfram MathWorld's Euclidean Geometry page [8] are also good references.
Most students have seen the Pythagorean Theorem so many times that they have lost count. Many have seen at least two or three proofs of it as well, though most would be hard pressed to recall one, much less elucidate what, exactly, constitutes a proof. While dynamical software allows discovery of concepts for making conjectures, many students are more convinced by their "discoveries" via dynamic software than by mathematically rigorous proofs. We hope to use that surety students have with their own discoveries to convince them of the need for actual proof.
We first consider Figure 1b in the Euclidean plane. Although it is easy to evaluate the area of a circle using the well known formula \(A=\pi r^2\) (where \(A\) represents the area and \(r\) the length of a radius of the circle, both as measured in Euclidean geometry), finding the formula in the first place requires some knowledge of calculus. (It is true that the formula was discovered long before calculus, but the methods involved used limits, a concept that is now taught as part of calculus.)
Question: Can these activities be adapted for lower-level students? Answer: Yes, by removing some of the last activities or reducing the number of non-Euclidean geometries in the activity. | 677.169 | 1 |
We define distances and circles in hyperbolic space analogously to how we did on the sphere, and assume that the reader is familiar with the Poincaré Disc Model of hyperbolic geometry, in particular, its (distance-inducing) differential \(ds=2(dx^2+dy^2)^{1/2}/(1-x^2-y^2)\). For those who need a refresher, Wikipedia has a reference page for the Poincaré Disc Model [13], Wolfram Math World [14] has another, and the Geometry Center [15] has an archive containing descriptions, pictures, equations, and an applet concerning it.
4a It is harder to conjecture a formula for circumference in hyperbolic geometry. Using Geometry Playground, we note that for small \(r\), \(A\approx \pi r^2\) and \(C\approx \pi r^2\), but for large \(r\), both \(A\) and \(C\) seem to approach \(\pi e^r\). Those familiar with hyperbolic trigonometric functions are probably already beginning to see the "parallel," and are conjecturing that \(C=2\pi\sinh r\), where \(\sinh r=(e^r-e^{-r})/2\) (and similarly \(\cosh r=(e^r+e^{-r})/2\)). If this is true, then a quick repeat of the steps of Subsection 3.1 shows that \(A=2\pi\cosh r-2\pi\), a remarkable resemblance to the spherical case.a.2 Using Integral Calculus
We take a circle in the Poincaré Disc model, and for convenience center it at the origin. In the model, we assume that it passes through the point \((x_0,0)\). We can find its area by integrating the square of the line element, \(ds^2=4(dx^2+dy^2)/(1-x^2-y^2)^2\) over the region inside the circle. This is easier in polar coordinates, where
\[A=\int_0^{2\pi}\int_0^{x_0} \frac4{(1-r^2)^2} rdrd\theta\]
This is relatively simple as double integrals go. After substitution \(u=(1-r^2\), \(du=-2rdr\), we can move quickly to the solution
(3)\[A=(4\pi x_0^2)/(1-x_0^2)\]
However, we want to know the area in terms of the radius. To find the radius, we need to calculate the distance between the origin and the point \((x_0,0)\). This requires a hyperbolic arclength integral. Thus:
\[r=\int_0^{x_0} ds\]
\[=\int_0^{x_0} \frac{2(dx^2+dy^2)}{(1-x^2-y^2)}\]
\[=\int_0^{x_0}\frac{2dx}{1-x^2}\]
Question: What is the formula for the line element in the Poincaré Disc Model? Answer: \(ds=2(dx^2+dy^2)^{1/2}/(1-x^2-y^2)\) | 677.169 | 1 |
The following problem comes directly from a sample problem found at yourteacher.com [21], and was meant for a high school algebra class. "Raul is 6 feet tall, and notices that his shadow is 5 feet long. The shadow of his school building is 30 feet long. How tall is his school building?"
In fact, the problem assumes knowledge of geometric ideas. Solve this problem, explaining at which steps you are using algebra and at which steps you are using geometry.
One student gave the following response, with a sketch of two appropriate triangles (parenthetical clarification added).
We know angles with the ground and [angles with the] sun are the same. Since AA [two angles are congruent], the triangles are similar. The similar triangles theorem says 6/5 = x/30, so x=36. The building is 36 feet. Except we live on a sphere, so we'd have to use a spherical similar triangles theorem, which is probably different. But since we don't care about hundredths of an inch, I guess Euclidean rules will give a good approximation. 36 feet is about right.
Appendix: Discovering Pythagoras
Here we give a discovery activity that we have used, in its entirety, with undergraduate students studying non-Euclidean geometry. We use freeware Geometry Playground [1] for these investigations; simply select the main link on the Geometry Playground page to run it. Geometry Playground allows straightedge and compass construction in Euclidean and non-Euclidean geometries such as spherical and hyperbolic. We let \(r\) represent the radius of a circle as measured in each geometry, while \(C\) and \(A\) represent the circumference and area of that circle, again as measured within that geometry.
This link will open Geometry Playground in Euclidean geometry with a pre-constructed right triangle having vertices \(A\), \(B\), and \(C\), the last corresponding to the right angle, and a constant term of \(\pi\) that can be used in forming sums, differences, products, and ratios. Construct three circles, one with center \(A\) and containing point \(B\), another with center \(C\) containing \(A\), and the third with center \(B\) containing point \(C\). Measure the areas of these circles.
Denote the length of the side with vertices \(B\) and \(C\) by \(a\), the side with vertices \(A\) and \(C\) by \(b\), and the third, the hypotenuse, by \(c\). Write the areas of the circles in terms of \(a\), \(b\), and \(c\). How do you know that the expressions you wrote down are correct?
Find a relationship between the areas of the circles. Make sure that the relationship you find remains after moving points in the triangle about. Write the relationship in terms of \(a\), \(b\), and \(c\). Hint: Sum the areas of the two smaller circles (you can do this by selecting Measure → Sum) and compare the sum to the area of the larger circle, perhaps by finding the ratio or the difference.
Question: What is the length of the school building's shadow? Answer: 30 feet
Question: Is Raul's shadow longer or shorter than him? Answer: Shorter. Raul is 6 feet tall, and his shadow is 5 feet long. | 677.169 | 1 |
In the diagram to the right, triangle PQR has a right angle [#permalink]
13 Apr 2007, 10:12- If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other
- In a right triangle, the length of the altitude from the right angle to the hypotenuse is the geometric mean of the lengths of the two segments of the hypotenuse
- In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. Each leg of the right triangle is the geometric mean of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg
- The areas of two similar triangles are proportional to the squares of any two homologous sides.
However, I still can't solve the question. I just run around in circles. Maybe these can help jar something loose in someone else...
Question: What is the relationship between the length of the altitude and the two segments of the hypotenuse in a right triangle? Answer: The length of the altitude is the geometric mean of the lengths of the two segments of the hypotenuse. | 677.169 | 1 |
Since then a straight line AD touches the circle ABE, and from the point of contact at A a straight line AB has been drawn across in the circle ABE, the angle DAB equals the angle AEB in the alternate segment of the circle.
But the angle BAD also equals the angle at C, therefore the angle AEB also equals the angle at C.
Therefore again the segment AEB of a circle has been described on AB admitting an angle equal to the angle at C.
Next, let the angle at C be obtuse.
Construct the angle BAD equal to C on the straight line AB and at the point A as is the case in the third figure. Draw AE at right angles to AD. Bisect AB again at F. Draw FG at right angles to AB, and join GB.
Question: What is the description given for the segment AEB of a circle? Answer: It admits an angle equal to the angle at C. | 677.169 | 1 |
Question 135888: Please help me solve this problem. The problem is that there are two ladder leaned up against a wall. One is 20m and the other is 15m. They both reach the same height up the wall. The bottom of the 20m ladder is 7m farther from the building than the 15m ladder. Click here to see answer by checkley77(12569)
Question 13 Click here to see answer by checkley77(12569)
Question 139343: Ok, I have a right triangle with an altitude of 8, the shorter side of the hypotenuse divided by the altitude is 4 and the longer side is represented by a. The shortest side of the triangle is 8.9 and the last side is represented by b. I need help finding a and b. Click here to see answer by scott8148(6628)
Question 140716: My problem is C squared equals 8x squared(substituted for A) plus 15x squared(once again substituted)
Since the question is an odd one, i got my answer of 23x squared but the book got something different...
The book got 17x
i do not understand how they got that. Click here to see answer by oscargut(891)
Question 143529: I just want to make sure I have the right answer, if I don't could you please explain the steps to me.
Problem---You are given a right triangle with the following lengths. Leg 1 = 2, Leg 2 = x, Hypotenuse = 8. Use the Pythagorean Theorem to find x. Make sure you include an equation.
Here is my answer
c^2 - a^2 = b^2
8^2 - 2^2 =
sqrt64 - sqrt4 =
sqrt60 =
Sqrt4 X 15 =
2 sqrt15 = X
Answer is 2 sqrt15 = x
Question 144665: Triangles HIJ and MNO are similar. The length of the sides of HIJ are 111, 96, and 108 centimeters. The length of the smallest side of MNO is 288 centimeters, what is the length of the longest side on MNO?
manual responses are 315 cm , 324 cm, 333cm 945cm. I don't even know where to start with only one side of MNO given. Click here to see answer by scott8148(6628)
Question: In the similar triangles problem, what is the length of the longest side of triangle MNO?
Answer: The length of the longest side of MNO can be found by setting up a proportion using the corresponding sides of the similar triangles: (longest side of MNO / 288) = (108 / 111). Solving for the longest side of MNO gives ≈ 324 cm.
Question: In the right triangle problem, what is the length of 'a'?
Answer: 'a' is the longer side of the hypotenuse and can be found using the formula: a = √(altitude^2 + shorter side^2) = √(8^2 + 8.9^2) ≈ 13.4
Question: What is the difference in length between the two ladders?
Answer: The 20m ladder is 5m longer than the 15m ladder. | 677.169 | 1 |
If instead we use polar coordinates with the origin at one focus, with the angular coordinate still measured from the major axis, the ellipse's equation is
where the sign in the denominator is negative if the reference direction points towards the center (as illustrated on the right), and positive if that direction points away from the center.
In the slightly more general case of an ellipse with one focus at the origin and the other focus at angular coordinate , the polar form is
The angle in these formulas is called the true anomaly of the point. The numerator of these formulas is the semi-latus rectum of the ellipse, usually denoted . It is the distance from a focus of the ellipse to the ellipse itself, measured along a line perpendicular to the major axis.
Semi-latus rectum.
General polar form
The following equation on the polar coordinates (r, θ) describes a general ellipse with semidiameters a and b, centered at a point (r0, θ0), with the a axis rotated by φ relative to the polar axis:[citation needed]
where
Angular eccentricity
Degrees of freedom
An ellipse in the plane has five degrees of freedom (the same as a general conic section), defining its position, orientation, shape, and scale. In comparison, circles have only three degrees of freedom (position and scale), while parabolae have four. Said another way, the set of all ellipses in the plane, with any natural metric (such as the Hausdorff distance) is a five-dimensional manifold. These degrees can be identified with, for example, the coefficients A,B,C,D,E of the implicit equation, or with the coefficients Xc, Yc, φ, a, b of the general parametric form.
Ellipses in physics
Elliptical reflectors and acoustics
If the water's surface is disturbed at one focus of an elliptical water tank, the circular waves created by that disturbance, after being reflected by the walls, will converge simultaneously to a single point — the second focus. This is a consequence of the total travel length being the same along any wall-bouncing path between the two foci.
Similarly, if a light source is placed at one focus of an elliptic mirror, all light rays on the plane of the ellipse are reflected to the second focus. Since no other smooth curve has such a property, it can be used as an alternative definition of an ellipse. (In the special case of a circle with a source at its center all light would be reflected back to the center.) If the ellipse is rotated along its major axis to produce an ellipsoidal mirror (specifically, a prolate spheroid), this property will hold for all rays out of the source. Alternatively, a cylindrical mirror with elliptical cross-section can be used to focus light from a linear fluorescent lamp along a line of the paper; such mirrors are used in some document scanners.
Question: What is the semi-latus rectum of an ellipse? Answer: It is the distance from a focus of the ellipse to the ellipse itself, measured along a line perpendicular to the major axis. | 677.169 | 1 |
Question 67951: I need help with this problem please! Two guy wires running from the top of the telephone pole down to the ground form an isosceles triangle. One of the two equal angles of the triangle is seven times the third angle (the vertex angle). Find the measure of the vertex angle. Thanks for any help! Click here to see answer by stanbon(57278)
Question 68052: Find the length of X. Express in simplified radical form.
I will try to draw this out so it makes it easier:
-
- -
- -
- -
51ft - - X
- -
- -----
- - -
----------------------------------
45 ft
How do I go about solving this type of equation? Thank you so very much! Click here to see answer by [email protected](15645)
Question 68245: Hi, could you please help me on how to solve this problem?
Determine the area of the triangle bounded by the following equations using Heron's formula: x-2y=12, x+3y=9, and 4x-y=-16. I know the Heron's formula but I do not know how to solve the problem. I first put the equations in function form but I don't know if that is how to solve the problem.
Please help me! I need this for homework that is due in three days!
Thank you. Click here to see answer by adamchapman(301)
Question 68325: Multi-Step Problem. You can use the method described below to approximate the distance across a stream without getting wet. As shown in the diagram you need a - Stand on the edge of the stream and look straight across to a point on the other edge of the stream. Adjust the visor of your cap so that its in the line with that point.
- Without changing the inclination of your neck and head, turn sideways until the visor is in line with a point on your side of the stream.
- Measure the distance BD between your feet and that point.
QUESTIONS
a.) From the description of the measuring method, what corresponding parts of the 2 triangles can you assume are congruent?
b.) What theorem or postulate can be used to show that the 2 triangles are congruent?
c.) Explain why the length of BD is also the distance across the stream. Click here to see answer by stanbon(57278)
Question 69010:
While you are in Washington, D.C., you visit the Washington Monument. In the scale diagram of a cross section of the monument you are given that AB= 8.4 meters, BC= 152.5 meters, ED= 17.6 meters, and CE= 5.25 meters.
Question: What is the length of X in the given problem, expressed in simplified radical form?
Answer: X = 3√13 ft | 677.169 | 1 |
Question 71294: If one leg is 1847 units long and the other leg is 3694 units long. What is the hypotenuse. The answer should be in simplified radical form. (No Decimals)
This is not from a textbook but it is for a problem set in Geometry. Click here to see answer by funmath(2925)
Question 72576: The measures of the angles of a triangle are 3x, 2x-10, and x+40 degrees. Find the number of degrees in the difference between the measures of largest and smallest angles.
I tried:
2x-10=0
2x=10
x=10/2
x=5
And after that I replaced x by 5. My answer is 3*5=15, (2*5)-10=0, and 5+40=45.
Then, the difference between 45 an 0 is 45°
Question 72631: Click here to see answer by bucky(2189 jim_thompson5910(28536 diluzion(5)
Question: What is the value of 'x' in the second question? Answer: 5 | 677.169 | 1 |
Hint
When a question asks "which of the following cannot ...," you must find the choice that does not satisfy the given situation. In any triangle, the sum of the lengths of any two sides must be greater than the length of the third side. Examine the answer choices to see which one does not satisfy this condition.
Answer
The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Four of the answer choices can be eliminated because each set of three numbers can be the lengths of three sides of a triangle. For example, in the third choice , the sum of any two numbers is greater than the third number. However, the second choice cannot be the lengths of the sides of a triangle, since is not greater than . This choice is the correct answer, since the question asked which set of numbers could NOT be lengths of the sides of a triangle.
Question: What is the relationship between the sum of the lengths of any two sides of a triangle and the length of the third side? Answer: The sum of the lengths of any two sides must be greater than the length of the third side. | 677.169 | 1 |
High Scohol Mathematics - Trignometry II
3. The measure of angle of depression of the bottom of a building on a level fround from the top of the tower is 60° . How far is the building from the tower?
4. A tower is 100/√3 meters high. Find the angle of elevation if the point of observation is 100 meters away from its foot.
5. As observed from the top of a light house, 100 meters high above sea level, the angle of depression of a ship, sailing directly towards it, changes from 30° to 45° . Determine the distance travelled by the ship during the period of observation.
Re: High Scohol Mathematics - Trignometry II
Re: High Scohol Mathematics - Trignometry II
OOps! I am sorry, I forgot to mention in #3 that the tower is 50 meters high. Regarding your solution to #5, the solution isn't right. I shall wait for any other membet to attempt, else I shall post all the solutions (sans diagams )
Question: What is the measure of the angle of depression in question 3? Answer: 60° | 677.169 | 1 |
Combinatorics and Graph Theory
Binomial Construction of the Trinomial Triangle
The trinomial triangle can be constructed in a binomial way using unit vectors of geometric algebra of quarks. This sheds some light on the question, how it is possible to transform mathematically entities of two elements into entities of three elements or vice versa.
Question: What does the text suggest about transforming mathematical entities of two elements into entities of three elements? Answer: The text suggests that it is possible to transform mathematically entities of two elements into entities of three elements, or vice versa. | 677.169 | 1 |
This project is an interesting interactive one.
The construction of the cricket reinforces geometric terminology, and
transformations. The cricket itself is an amusing small toy, and
students will enjoy making it hop. It takes a little practice to make
the cricket hop successfully, but students usually figure this out
faster than the teacher! It is best to make the origami cricket out
of a piece of bright green paper, and preferably a slightly heavier
weight than typing paper, although any paper will do. Some students
will enjoy decorating their cricket, with eyes, a mouth, and even
"whiskers"!
The worksheet that accompanies the Origami
Cricket reinforces and reviews many of the right triangle theorems.
Finding the measures of angles 1 and 2 provide a review of one of the
trickier theorems of Geometry: the Angle Bisector Proportionality
Theorem which states that the bisector of an acute angle of a right
triangle divides the opposite side into two segments which are
proportional to the remaining sides of the triangle:
To construct the
origami cricket, follow the instructions below:
Start with a
square with side x:
Fold on a diagonal of
the square:
Rotate the isosceles
right triangle 135° counter-clockwise:
Fold on the
perpendicular bisector of the base:
Fold on the
perpendicular bisector of the hypotenuse, FE. Do this separately for
each side as shown:
Fold on the angle
bisector of angle FBE twice, on the top portion and on the bottom
potion.
You are finished!
Turn your cricket over, strike the highest point with your finger,
and your cricket will jump!
I advise my students to listen carefully the
moment they decide to take no more mathematics courses. They might be
able to hear the sound of closing doors.
James CaballeroGo
back to Table of Contents
Question: What is the main objective of the project described? Answer: The main objective is to create an origami cricket toy while reinforcing geometric terminology and transformations, and reviewing right triangle theorems. | 677.169 | 1 |
impact of the sun's rays, ... showed that the pyramid has to the stick
the same ratio which the shadow [of the pyramid] has to the shadow [of
the stick].
Now of course Thales could have used these
geometrical methods for solving practical problems having merely observed
the properties and having no appreciation of what it means to prove a
geometrical theorem. This is in line with the views of Russell who writes
of Thales contributions to mathematics:
Thales is said to have travelled
in Egypt, and to have thence brought to the Greeks the science of geometry.
What Egyptians knew of geometry was mainly rules of thumb, and there
is no reason to believe that Thales arrived at deductive proofs, such
as later Greeks discovered.
On the other hand, there are claims that
Thales put geometry on a logical footing and was well aware of the notion
of proving a geometrical theorem. However, although there is much evidence
to suggest that Thales made some fundamental contributions to geometry,
it is easy to interpret his contributions in the light of our own knowledge,
thereby believing that Thales had a fuller appreciation of geometry than
he could possibly have achieved. In many textbooks on the history of mathematics
Thales is credited with five theorems of elementary geometry:
(i) A circle is bisected by any
diameter.(ii) The base angles
of an isosceles triangle are equal.(iii) The angles between
two intersecting straight lines are equal.(iv) Two triangles
are congruent if they have two angles and one side equal.(v)
An angle in a semicircle is a right angle.
What is the basis for these claims? Proclus,
writing around 450 AD, is the basis for the first four of these claims,
in the third and fourth cases quoting the work History of Geometry
by Eudemus of Rhodes, who was a pupil of Aristotle, as his source. The
History of Geometry by Eudemus is now lost but there is no reason
to doubt Proclus. The firth theorem is believed to be due to Thales because
of a passage from Diogenes Laertius book Lives of eminent philosophers
written in the second century AD:
Pamphile says that Thales, who learnt
geometry from the Egyptians, was the first to describe on a circle a
triangle which shall be right-angled, and that he sacrificed an ox (on
the strength of the discovery). Others, however, including Apollodorus
the calculator, say that it was Pythagoras.
A deeper examination of the sources, however,
shows that, even if they are accurate, we may be crediting Thales with
too much. For example Proclus uses a word meaning something closer to
'similar' rather than 'equal- in describing (ii). It is quite likely that
Thales did not even have a way of measuring angles so 'equal- angles would
have not been a concept he would have understood precisely. He may have
claimed no more than "The base angles of an isosceles triangle look
Question: According to the text, what did the Egyptians primarily know about geometry? Answer: They knew mainly rules of thumb, not deductive proofs.
Question: What is one of the methods Thales used to demonstrate a property of pyramids? Answer: He used the ratio of the pyramid's shadow to a stick's shadow, which is the same as the ratio of the pyramid to the stick.
Question: Which of Thales' contributions is believed to be due to him because of a passage from Diogenes Laertius? Answer: The theorem that an angle in a semicircle is a right angle.
Question: Who is the source for the first four of these claims about Thales' contributions? Answer: Proclus, writing around 450 AD. | 677.169 | 1 |
DIHEDRAL ANGLES are the angles between triangles. They are useful if you plan to bevel the skin panels or use bevelled struts.
2v and 3v breakdowns with original Icosa face dotted
The diagram above shows the statistics for a Three Frequency Icosahedron Dome. To make a 5/8 sphere you will need:
Choose your method of model assembly (following pages) and assemble as in the diagram on page 8. This model will sit ?at on the 3/8 and 5/8 levels. It is simpler to join all the pentagon and hexagon spokes together ?rst. Make six pentagons from "A" struts, ten hexagons from "C" struts and ?ve "halt" hexagons for the base.
A simpler model can be made using only three strut lengths by substituting "D" struts with "B" struts and using the new chord factors of:
A = .34862 need 30 B = .40355 need 55 C = .41241 need 80
This model will not sit flat however but will rest on the five half-hex hubs under the pentagons. For a very similar truncation (fraction of a dome), the Kruschke method is used (in this case a 5/9).
Three-frequency icosahedron spheroid
This e-booklet has been compiled to assist potential owner-builders of Geodesic Domes to prefabricate and erect a variety of geometric configurations utilising a number of possible construction techniques.
Much of the data is gleaned from long-out-of-date publications and from the author's practical experience and experimentation in building domes since 1973.
Most of the ideas within have been tested and proven successful but are certainly open to further refinement and development. Rather than promoting any one system as being the 'best', I would prefer to offer ideas and possibilities to the great Australian tradition of "Do-it -yourself" and share my own experiences of what can work.
With this in mind I welcome any feedback about your own experiments, learning experiences and successes and invite any correspondence to share such knowledge. This manual is intended for educational purposes only.
It presents information on principles and techniques that the author has not necessarily employed. It is not suggested that these methods must be followed or that if followed will result in a safe or satisfactory building.
Due to variations in materials, quality of workmanship, tools and equipment, materials and components, local building codes etc., the author assumes no liability for any structure designed or constructed from information in this manual unless it is built by The Dome Company or our nominated sub-contractor.
Geodesic Geometry
GEODESIC = The shortest line between two points on a sphere.
There are only five different structures that we can build where all the sides, faces and angles are equal. These regular solids are called PLATONIC solids.
Question: What is the main goal of the author in compiling this e-booklet? Answer: The main goal is to assist potential owner-builders of Geodesic Domes to prefabricate and erect a variety of geometric configurations.
Question: How many pentagons are needed to make the model? Answer: Six pentagons are needed from "A" struts.
Question: What is the purpose of knowing Dihhedral Angles? Answer: They are useful for beveling skin panels or using beveled struts.
Question: What does the term "Geodesic" mean? Answer: Geodesic means the shortest line between two points on a sphere. | 677.169 | 1 |
Of these five, we see that three are made of triangles. As we might expect, the tetrahedron,octahedron and icosahedron are rigid; while the cube and dodecahedron are not. The cube is the basis for most types of buildings. The icosahedron is the basis for most Geodesic Domes.
You can make small structures with icosahedrons, but if you begin to make bigger structures, the triangles get large and heavy and you begin to need big timbers for the members.
Dome geometry is all about the subdivision of large triangles into smaller ones.
An Icosahedron has twenty equilateral triangular 'faces'. When the face is subdivided. it is not
done equally, but done so that the faces begin to curve outward;this gives you more strength. Each face can be divided by a line (arcs on a sphere) parrallel to the edge.
The number of divisions is called "FREQUENCY".
As the frequency increases, the number of members or "STRUTS" and the numbel of triangles increases also, and the closer you get to a sphere.
The following diagram below is of a THREE FREQUENCY (3v) ICOSA 5/8 SPHERE.
Each of the original Icosa faces is heavily outlined. Looking at a geodesic dome on can determine the frequency by counting the number of struts between the apexes 0f the pentagons.
Note: divisions are different for different frequencies. See chart below for correct nomenclature
Question: What is the term used to describe the number of divisions in a geodesic dome? Answer: Frequency
Question: What happens to the number of members or'struts' and the number of triangles as the frequency increases? Answer: They both increase | 677.169 | 1 |
greatest intercept is ^ c' = f/^' = 30 ft. The pole distance
H — 2,400 lb. Hence, the bending moment = 2,400 X 30 =
72,000 ft. -lb. = 72,000 X 12 = 864,000 in. -lb.
(616) ^^ in Fig. 55
is the given diagonal 3.5 in.
3. 5' = 1 2.25 -f- 2 = 6.125 in.
/6.125 = 2.475 in. = side of
the required square. From
A and B as centers with
radii equal to 2.475 in., de-
scribe arcs intersecting at
C and D. Connect the ex-
tremities A and B with the
points C and D by straight
lines. The figure A C B D is the required square.
(617) Let A B, Fig. 56, be the given shorter side of the
rectangle, 1.5 in. in length. At A erect an indefinite per-
pendicular A C to the line A B. Then, from B as a. center
with a radius of 3 in. describe an arc intersecting the per-
pendicular A C in the point D. This will give us two
234
SURVEYING.
Fig. 56.
adjacent sides of the required rectangle. At B erect an
indefinite perpendicular B E to A B, and at D erect an in-
definite perpendicular DF to AD. These perpendiculars
(618) (See Fig. 57.) With
the two given points as cen-
ters, and a radius equal to
3.5'^ 2 = 1.75'= If', describe
short arcs intersecting each
other. With the same radius
and with the point of intersec-
tion as a center, describe a
circle; it will pass through
the two given points.
Fig. 57
(619) A B \n Fig.
C
58
Fig. 68.
is the given line, A the given
point, A C and C B = A B.
The angles A, B, and C are
each equal to 60°. From B
and C as centers with equal
radii, describe arcs intersect-
ing at D. The line A D bi-
sects the angle A ; hence,
angle B A D = 30°.
(620) Let A B in Fig. 59 be one of the given lines,
whose length is 2 in., and let A C, the other line, meet A B
j^t A^ forming an angle of 30°. From A and B as centers,
SURVEYING.
235
Question: What is the length of the given line A B in Figure 58? Answer: A B | 677.169 | 1 |
An ellipse is the locus of a point which moves so that its distance from e fixed point (called the focus) bears a constant ratio, always less than 1, to its perpendicular distance from a straight line (called the directrix). An ellipee has two foci and two directrices.
Hg. 11/4 shows how to draw an aHips« given tha relative positions of the focus and tha directrix, and the eccentricity In this case the focus and the directrix are 20 mm apart and the eccentricity is J.
The first point to plot is the one thet lies between the focus and the directrix. This is done by dividing OF in the same ratio as the eccentricity. 4: 3. The other end of the ellipee. point P. is found by working out the simple algebraic sum shown on Fig. 11/4.
The condition for the locus is that it ia always 4 as far from the focus es it is from the directrix. It is therefore j as fsr from the directrix as it is from the focus Thus, if the point is 30 mm from F, it is If mm from the directrix; if the point is 20mm from F it is fx20mm from the directrix; if the point is 30 mm from F. It is J x 30 mm from the directrix. This is continued for as many points as may be necessary to draw an accurate curve. The intersections of radii drawn from F and lines drawn parallel to the directrix, their distance from the directrix being proportional to the radii, give the outline of the ellipse These points ere joined together with a neat freehand curve.
STAGE 3
Fig 11/5
STAGE 3
To construct an ellipse by concentric circles We now come to the first of three simple methods of constructing an ellipse. All three methods need only two pieces of information for the construction—the lengths of the major and minor axes.
Stage 1. Draw two concentnc circles, radii equal to \
major and } minor axes. Stege 2. Divide the cirde into a number of sectors. If the ellipee is not too large, twelve will suffice as shown in Fig. 11/5.
To construct an elllpee in a rectangle
Stage 1. Draw a rectangle, length and breadth equal to the mejor and minor exe» Stage 2. Divide the two »horter »idea of the rectangle into the »ame tvtn number of equal part*. Divide the major axis into the »ame number of equal parts Stage 3. From the point» where the minor axis crosses the edge of the rectangle, draw intersecting lines as shown in Fig. 11 /6. Stage 4. Draw a neat curve through the intersections.
Question: How many foci and directrices does an ellipse have? Answer: Two foci and two directrices
Question: What is the eccentricity given in the text? Answer: 1/4 | 677.169 | 1 |
Angles/346961: The supplement of an angle is 36 degrees less than twice the supplement of the complement of the angle. Find the measure of the supplement. Can you also explain it too? Thank you so much 1 solutions Answer 248109 by mananth(12270) on 2010-09-22 21:24:13 (Show Source):
You can put this solution on YOUR website! let the angle be x degrees
the supplement of this angle will be 180-x
..
The complement of the angle will be 90-x
supplement of this complement angle will be 180 - (90-x)
Twice this = 2[180-(90-x)]
Less 36
=2[180-(90-x)]-36
..
180-x = 2[180-(90-x)]-36
...
180-x= 2[180-90+x]-36
180-x=2[90+x]-36
180-x=180+2x-36
3x=180-180+36
3x=36
x= 12
supplement of 12 = 180-12 = 168
Travel_Word_Problems/346953: A bullet is fired horizontally at a target, and the sound of its impact is heard 1.5 seconds later. If the speed of the bullet is 3300 ft/sec and the speed of sound is 1100 ft/sec how far away is the target? 1 solutions Answer 248100 by mananth(12270) on 2010-09-22 21:15:05 (Show Source):
Triangles/346942: 1.) The perimeter of a triangle is 76 centimeters. The second side is twice as long as the first side. The third side is four centimeters shorter than the second side. How long is each side? 1 solutions Answer 248080 by mananth(12270) on 2010-09-22 20:56:03 (Show Source):
You can put this solution on YOUR website! let first side be x
second side= = 2x
third side= 2x-4
...
x+2x+2x-4=76
5x-4=76
5x=76+4
5x=80
x=16 cm first side
second side = 2*16= 32 cm
third side = 2x-4 = 2*16 -4 =28 cm
Question: On which date and time was the solution to the first problem posted? Answer: 2010-09-22 21:24:13
Question: What is the measure of the supplement of an angle that is 36 degrees less than twice the supplement of its complement? Answer: 168 degrees
Question: What is the speed of the bullet in the second problem? Answer: 3300 ft/sec | 677.169 | 1 |
However, a connected pair of octahedra share two vertices; no worry, we just have to find the second set of vertices which are be connected, and we can represent this second connection with a dashed line like so:
Thus, the map of the compound of five octahedra will look like this (the colors correspond to the one pictured above; also note that when two dashed lines cross they do not represent a connection, any one dashed green line is only connecting two vertices of two octahedra):
What may, initially, look to be a jumbled mess, now, having its relations mapped out, shows a beautiful symmetry; but what can we make of this compound mirificum now that we have this map? The specific way the octahedra connect to one another (with any one octahedron not being connected to all the others directly, only to two specific others) is shown clearly here. An investigation of these interconnections leads to the question of "topology."
Question: How are the octahedra connected to each other? Answer: Each octahedron is connected to only two others, not all of them directly. | 677.169 | 1 |
Trig: Radians, Arc Length, Solve 1st Deg Equations This 83 minute trigonometry lesson introduces radian measure, changing from degrees to radians and radians to degrees. This lesson will show you how to solve first degree trig equations in radians: - algebraically using the calculator in the radian mode - using graphing technology - using exact values for special angles in radians You will also learn how to: - change from degrees to radians and radians to degrees - find the length of an arc of a circle subtended by an angle in radians Thislesson contains explanations of the concepts and 22 example questions with step by step solutions plus 7 interactive review questions with solutions. Lessons that will help you with the fundamentals of this lesson include: - 300 Basic Trigonometry Part I ( - 315 Basic Trigonometry Part II ( - 320 Trigonometry on the Coordinate Plane (
Twin Radian My second flight with my twin Radian or as I like to call it my Radii. video recorded at FLYRC field in southwestern CT.
HD Video From The Parkzone Radian RC Plane at over 1000 Feet! This is video shot with a HD camera mounted on my Radian RC plane. You get to see some high quality video from over 1000 feet in the air and also take in a great sunset. I will be demonstrating the cam mount in a later video. As always remember to..... Rate, Comment and Subscribe to support my channel Thanks, Eric Jordon
Radian Pro by Parkzone.wmv Check out this first glimpse of the Radian Pro from Parkzone! The Radian Pro comes equipped with ailerons and functioning flaps. This makes it possible for experienced sailplane pilots with advanced programmable radio systems to take advantage of mixing functions like reflex, camber and crow.
Parkzone Radian Chases Birds and Shoots Rocket A peaceful morning was disrupted by the startled cries of many a bird, as a low flying RC plane woke them from their slumber.
Radian Pro with GoPro HD A video from my 3rd of 6 flights yesterday. Lots of thermals during my last 3 flights which really took this Radian up! I couldn't get the Radian with the Gopro to rise with the thermals...probably to heavy. This is my second Radian Pro. My first RPro crashed due to a bad elevator servo, so on this PNP version I put a Hitec 85MG in...no screwing around this time! So far the flights have been good. I've been getting about 30min from a 3S 2200mah battery...which should be good for about an hour I've been pending on how strong the thermals are.
A-level Maths: what is a radian? - ExamSolutions In this tutorial you are shown what a radian is and some common angles in radians. See this and other tutorials at along with worked solutions to past papers
Question: What is the total duration of the trigonometry lesson? Answer: 83 minutes
Question: Which of the following is NOT a topic covered in the lesson? A) Radian measure B) Solving first degree trig equations C) Calculus D) Arc length Answer: C) Calculus | 677.169 | 1 |
Geometric Inequalities
Nicholas Kazarinoff
Kazarinoff's Geometric Inequalities will appeal to those who are already inclined toward mathematics. It proves a number of interesting inequalities; for example, of all triangles with the same perimeter, the equilateral triangle has the greatest area; of all quadrilaterals with a given area, the square has least perimeter; and the famous Steiner theorem, the circle has more area than any other plane figure with the same perimeter. The writing is honest. The author labels difficult what is difficult and does not pretend that to the master mind (who is usually the author) all things are simple. The text suggests guessing, conjecturing and then proving. The author does not hesitate to offer a proof of his own which; he points out, he later found to be incorrect. The device of putting a proof of a general theorem in one column and a concrete case alongside could be more widely employed by others. — MAA Reviewer
Anybody who liked their first geometry course (and some who did not) will enjoy the simply stated geometric problems about maximum and minimum lengths and areas in this book. Many of these already fascinated the Greeks, for example, the problem of enclosing the largest possible area by a fence of given length, and some were solved long ago; but others remain unsolved even today. Some of the solutions of the problems posed in this book, for example the problem of inscribing a triangle of smallest perimeter into a given triangle, were supplied by world famous mathematicians, other by high school students.
Question: Who has supplied some of the solutions to the problems in this book? Answer: World famous mathematicians and high school students.
Question: What is one of the devices used by the author in the book? Answer: Putting a proof of a general theorem in one column and a concrete case alongside. | 677.169 | 1 |
Learn more
NCDigIns Help students become proficient in mathematics with this formative assessment diagnostic tool.
Related pages
Noodles away: This lesson will assist students to see angle relationships and the relationship of parallel lines and transversals. This exercise is good for visual and tactile learners since it is of a concrete nature. Students of all academic levels can enjoy this.
Origami geometry: Students apply their knowledge of geometric terms to follow directions while folding an origami sculpture. Students then solve math problems which relate to the project with follow-up discussions about the project relating to geometric terms such as symmetry, faces, edges, square, triangle, plane, etc.
Teacher planning
Time required for lesson
Materials/resources
Pre-activities
Copy diagrams 1 and 2 on the board and give students the standard definition of corresponding, alternate interior, and same-side interior angles and point them out.
Corresponding angles are angles in the same place relative to the transversal and lines being intersected (examples: angles 1 and 5, 2 and 6, 3 and 7, 4 and 8).
Interior angles are the angles between the two coplanar lines being intersected by the transversal. (examples: angles 3, 4, 5, and 6).
Alternate interior angles are non-adjacent interior angles on different sides of the transversal. (examples: angles 3 and 6, 4 and 5).
Same-side interior angles are interior angles on the same side of the transversal. (examples: angles 3 and 5, 4 and 6).
Activities
Have students copy diagrams 1 and 2 on their paper by using a ruler. Trace on top and bottom of the ruler to get the coplanar lines a and b. Next use the ruler to help draw the transversal t that intersects both a and b. Hint: When tracing out the following interior angles, only trace the part of the transversal that is between the two lines being cut.
Have the students trace out the alternate interior angles 3 and 6 found on diagram 1. Ask them: "Do these angles form a particular shape?" Response should be they look they a Z shape.
Now have them trace out angles 4 and 5 on the same diagram. Now ask: "Do they form the same type of shape?" These two will form a backwards Z shape.
Repeat the above using diagram 2. Have them trace out angles 3 and 6. Ask: "Do these form a particular shape?" Should resemble a N shape. Repeat with angles 4 and 5. Ask: "Do these form the same shape?" These two angles will give a N shape that have been flipped.
Next have them trace out the same-side interior angles 4 and 6 on diagram 1. Ask: "What type of shape do they form?" Should resemble a C shape. Repeat with angles 3 and 5. Ask: "Do they form a similar shape?" These two form a backwards C shape.
Question: What is the relationship between corresponding angles and parallel lines?
Answer: Corresponding angles are equal when two lines are cut by a transversal and the lines are parallel.
Question: Which angles in diagram 2 form a 'N' shape?
Answer: Angles 3 and 6. | 677.169 | 1 |
Transformations on theCoordinatePlane (Pages 506–511) ... Transformations are movements of geometric figures, such as translations, rotations, and reflections. •Atranslation is a slide where the figure is moved horizontally or vertically or both.
Transformations on theCoordinatePlane (Pages 197–203) NAME _____ DATE _____ PERIOD _____ Example Practice 4-2 Which type of transformation does this picture show? The figure has been rotated around the point ...
TransformationsintheCoordinatePlane You identified translations, reflections, and rotations. You'll graph transformationsin a coordinateplane. So you can describe movement in a cartoon, as in Ex. 17. Describe the translation using coordinate notation. 1.
Investigate and draw reflections on a coordinateplane. Worksheet (Cabri): Reflections on a CoordinatePlane Answers: 2. a. Answers will vary. ... Summarize transformations using theworksheetTransformations Answers: 1. Reflection with a pre-image, image, and line of reflection.
Unit: Transformations Vertical and Horizontal Shifting Worksheet #3 Name ... coordinateplane. a. b. c. d. e. f. 2. If the point (-7 , 3) lies on the graph of an elementary function y = g(x), find a point that lies on the graph on the ...
A plane can be defined by a point and two vectors or by three points P(a,b)=R+au+bv P(a,b)=R+a(Q-R)+b(P-R) u v R P R Q. Triangles convex sum of P and Q ... In a fixed coordinate system, transformations modify an object's shape and location (coordinates)
transformationsin a coordinateplane and describe the results. Indicator 8 Derive coordinate rules for translations, reflections and rotations of geometric ... worksheet Activity Worksheet, Attachment A. For those who do not have time to complete
coordinateplane. Transformations (also known as mappings) are geometric figures that have been changed by reflection, rotation, or translation (also known ... Transformation WorksheetCoordinate Grid Symmetry Project and rubric
each ordered pair on the back of theworksheet. Lesson 3 Preassessment • Distribute SRS #8 to students as they enter the classroom. Tell students that you want to move the ... On your coordinateplane, identify and label the new coordinates. 4.
6-8: Dilations intheCoordinatePlane Definition: A dilation of k is a transformation of theplane such that: 1. The image of point O, the center of dilation, is O. 2. ... transformations: followed by . 2 D
Investigation 1: Transformations on a CoordinatePlaneIn this investigation you will explore four ordered pair rules. Follow Steps 1 and 2 in your book. Draw your original polygon in Quadrant I, II, or IV. (Here are some examples
Question: What is the name of the worksheet that involves finding a point on the graph of an elementary function after a translation?
Answer: Vertical and Horizontal Shifting Worksheet #3.
Question: In which quadrants can you draw your original polygon for the Investigation 1: Transformations on a Coordinate Plane?
Answer: Quadrant I, II, or IV.
Question: Which transformation involves moving a figure horizontally, vertically, or both?
Answer: Translation. | 677.169 | 1 |
DIHEDRAL ANGLES are the angles between triangles. They are useful if you plan to bevel the skin panels or use bevelled struts.
2v and 3v breakdowns with original Icosa face dotted
The diagram above shows the statistics for a Three Frequency Icosahedron Dome. To make a 5/8 sphere you will need:
Choose your method of model assembly (following pages) and assemble as in the diagram on page 8. This model will sit ?at on the 3/8 and 5/8 levels. It is simpler to join all the pentagon and hexagon spokes together ?rst. Make six pentagons from "A" struts, ten hexagons from "C" struts and ?ve "halt" hexagons for the base.
A simpler model can be made using only three strut lengths by substituting "D" struts with "B" struts and using the new chord factors of:
A = .34862 need 30 B = .40355 need 55 C = .41241 need 80
This model will not sit flat however but will rest on the five half-hex hubs under the pentagons. For a very similar truncation (fraction of a dome), the Kruschke method is used (in this case a 5/9).
Three-frequency icosahedron spheroid
This e-booklet has been compiled to assist potential owner-builders of Geodesic Domes to prefabricate and erect a variety of geometric configurations utilising a number of possible construction techniques.
Much of the data is gleaned from long-out-of-date publications and from the author's practical experience and experimentation in building domes since 1973.
Most of the ideas within have been tested and proven successful but are certainly open to further refinement and development. Rather than promoting any one system as being the 'best', I would prefer to offer ideas and possibilities to the great Australian tradition of "Do-it -yourself" and share my own experiences of what can work.
With this in mind I welcome any feedback about your own experiments, learning experiences and successes and invite any correspondence to share such knowledge. This manual is intended for educational purposes only.
It presents information on principles and techniques that the author has not necessarily employed. It is not suggested that these methods must be followed or that if followed will result in a safe or satisfactory building.
Due to variations in materials, quality of workmanship, tools and equipment, materials and components, local building codes etc., the author assumes no liability for any structure designed or constructed from information in this manual unless it is built by The Dome Company or our nominated sub-contractor.
Geodesic Geometry
GEODESIC = The shortest line between two points on a sphere.
There are only five different structures that we can build where all the sides, faces and angles are equal. These regular solids are called PLATONIC solids.
Question: What is the purpose of knowing Dihhedral Angles? Answer: They are useful for beveling skin panels or using beveled struts.
Question: What are the five regular solids called that have all sides, faces, and angles equal? Answer: They are called Platonic solids.
Question: How many pentagons are needed to make the model? Answer: Six pentagons are needed from "A" struts.
Question: What does the term "Geodesic" mean? Answer: Geodesic means the shortest line between two points on a sphere. | 677.169 | 1 |
Of these five, we see that three are made of triangles. As we might expect, the tetrahedron,octahedron and icosahedron are rigid; while the cube and dodecahedron are not. The cube is the basis for most types of buildings. The icosahedron is the basis for most Geodesic Domes.
You can make small structures with icosahedrons, but if you begin to make bigger structures, the triangles get large and heavy and you begin to need big timbers for the members.
Dome geometry is all about the subdivision of large triangles into smaller ones.
An Icosahedron has twenty equilateral triangular 'faces'. When the face is subdivided. it is not
done equally, but done so that the faces begin to curve outward;this gives you more strength. Each face can be divided by a line (arcs on a sphere) parrallel to the edge.
The number of divisions is called "FREQUENCY".
As the frequency increases, the number of members or "STRUTS" and the numbel of triangles increases also, and the closer you get to a sphere.
The following diagram below is of a THREE FREQUENCY (3v) ICOSA 5/8 SPHERE.
Each of the original Icosa faces is heavily outlined. Looking at a geodesic dome on can determine the frequency by counting the number of struts between the apexes 0f the pentagons.
Note: divisions are different for different frequencies. See chart below for correct nomenclature
Question: Which of the following shapes are made of triangles? (a) Cube (b) Tetrahedron (c) Octahedron (d) Icosahedron Answer: (b) Tetrahedron, (c) Octahedron, (d) Icosahedron | 677.169 | 1 |
greatest intercept is ^ c' = f/^' = 30 ft. The pole distance
H — 2,400 lb. Hence, the bending moment = 2,400 X 30 =
72,000 ft. -lb. = 72,000 X 12 = 864,000 in. -lb.
(616) ^^ in Fig. 55
is the given diagonal 3.5 in.
3. 5' = 1 2.25 -f- 2 = 6.125 in.
/6.125 = 2.475 in. = side of
the required square. From
A and B as centers with
radii equal to 2.475 in., de-
scribe arcs intersecting at
C and D. Connect the ex-
tremities A and B with the
points C and D by straight
lines. The figure A C B D is the required square.
(617) Let A B, Fig. 56, be the given shorter side of the
rectangle, 1.5 in. in length. At A erect an indefinite per-
pendicular A C to the line A B. Then, from B as a. center
with a radius of 3 in. describe an arc intersecting the per-
pendicular A C in the point D. This will give us two
234
SURVEYING.
Fig. 56.
adjacent sides of the required rectangle. At B erect an
indefinite perpendicular B E to A B, and at D erect an in-
definite perpendicular DF to AD. These perpendiculars
(618) (See Fig. 57.) With
the two given points as cen-
ters, and a radius equal to
3.5'^ 2 = 1.75'= If', describe
short arcs intersecting each
other. With the same radius
and with the point of intersec-
tion as a center, describe a
circle; it will pass through
the two given points.
Fig. 57
(619) A B \n Fig.
C
58
Fig. 68.
is the given line, A the given
point, A C and C B = A B.
The angles A, B, and C are
each equal to 60°. From B
and C as centers with equal
radii, describe arcs intersect-
ing at D. The line A D bi-
sects the angle A ; hence,
angle B A D = 30°.
(620) Let A B in Fig. 59 be one of the given lines,
whose length is 2 in., and let A C, the other line, meet A B
j^t A^ forming an angle of 30°. From A and B as centers,
SURVEYING.
235
Question: What is the radius used to describe arcs in Figure 56? Answer: 3 in.
Question: What is the measure of angle B A D in Figure 58? Answer: 30° | 677.169 | 1 |
what is a vertex angle
Answers
I"m not sure what this has to do with quadratic equations, but a vertex is at the intersection of the sides of a polygon. If you stand on the intersection point, looking into the polygon, the two rays left and right form the vertex angle.
In quadratic equations, the vertex is the turning point of the parabola with the coordinates of (-b/2a, y), but I don't know of any angle associated with that term.
Question: What are the two rays that form the vertex angle in a polygon? Answer: The rays left and right from the vertex. | 677.169 | 1 |
Loci: Convergence
Van Schooten's Ruler Constructions
by C. Edward Sandifer
Solution to Problem IV
Problem IV: Above a given indefinitely long straight line, to construct a perpendicular.
Construction: Conceive the given straight line as going through points A and B, and a perpendicular is to be constructed above it; make BC equal to AB [along the same line] and from B draw BD, making with AB any angle whatsoever, and locate D on that line so that it equals BA or BC, and draw the line from point D through point C. If in that line CF is made equal to CA, and in the line ABC, CE is made equal to CD, I say joining EF makes it be perpendicular to AB.
This may be van Schooten's trickiest construction. The key to the proof of correctness is to note that the points A, C and D are all the same distance from B, so they lie on a circle centered at B and with AC as a diameter. This makes angle ADC a right angle. Now you only have to figure out why angle CEF is also a right angle.
Question: What is the given problem in the text? Answer: The problem is to construct a perpendicular above a given indefinitely long straight line. | 677.169 | 1 |
Re: Geometry Question
the thing that needs to be kept in mind is:
area of the traingle = 1/2 (base)*(height)
in the question we have a right angled triangle.
area = 1/2 * (HJ) * (JK) ---- > (1)
also since JL is perpendicular to HK. we can write the area in terms of HK and JL as
Area = 1/2 * (HK) * (JL) -----> (2)
from (1) and (2),
HJ * JK = HK *JL
back to the question and its options.
1) says Hk = 12. ok fine. but we still dont know what JL is. so, data given is not sufficient.
2) says HJ * JK = 48. so, from what we derived earlier. HK*JL should also be equal to 48. Hence, given data is sufficient.
Re: Geometry Question
Careful, I think that's exactly the trap you're supposed to fall into - almost went for it myself...
There is nothing in the question that indicates that angle JHL is 45. It looks that way in the diagram, but DS diagrams are not drawn to scale unless that's specifically stated.
In particular, you can't assume that the line JL bisects the angle HJK - the only thing you can be sure of is that HJK is a right angle. For example, you can easily draw a 30-60-90 triangle with the right angle at HJK, and drop a vertical line from J to a point L on the hypotenuse. The length of JL in that case won't be the same as the length of JL in a 45-90-45 triangle.
Question: In the given problem, what type of triangle is mentioned? Answer: A right-angled triangle is mentioned.
Question: What is the relationship between HJ JK and HK JL, according to the text? Answer: HJ JK = HK JL | 677.169 | 1 |
A pyramid (from ) is a structure whose shape is roughly that of a pyramid in the geometric sense; that is, its outer surfaces are triangular and converge to a single point at the top. The base of a pyramid can be trilateral, quadrilateral, or any polygon shape, meaning that a pyramid has at least three outer triangular surfaces (at least four faces including the base). The square pyramid, with square base and four triangular outer surfaces, is a common version.
Question: Can a pyramid have more than three outer triangular surfaces? Answer: Yes | 677.169 | 1 |
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Angles
This unit uses one of the digital learning objects, Angles, to support students as they investigate measuring and drawing angles using other angles as units of measurement. It is suitable for students working at level 2 of the curriculum because they estimate and measure the size of other angles using other angles and not with compasses and protractors. This unit includes background information on teaching about angles, a sequence that can be used by the teacher when working with a group of students on the learning object, and ideas for independent student work.
Angle can be seen as and thought of in at least three ways. These are as:
the spread between two rays
the corner of a 2-dimensional figure
an amount of turning
The final one of these underpins the others and leads on naturally to the definition of degree and the ability to measure angles with a standard unit. This leads students on to being able to apply their knowledge of angle in a variety of situations.
We see angle as developing over the following progression:
Level 1: quarter and half turns as angles
Level 2: quarter and half turns in either a clockwise or anti-clockwise direction and angle as an amount of turning
Level 3: sharp (acute) angles, blunt (obtuse) angles and right angles degrees applied to simple angles – 90°, 180°, 360°, 45°, 30°, 60°
Level 4: degrees applied to all acute angles, degrees applied to all angles, angles applied in simple practical situations
Level 5: angles applied in more complex practical situations
The concept of angle is something that we see students developing gradually over several years. As their concept matures, they will be able to apply it in a range of situations including giving instructions for directions and finding heights.
Outside school and university, angle is something that is used regularly by surveyors and engineers both as an immediate practical tool and as a means to solve mathematics that arises from practical situations. So angle is important in many applications in the 'real' world as well as an 'abstract' tool. This all means that angles have a fundamental role to play in mathematics and its application.
Prior to using the Angles Learning Objects
The unit Clockwise helps students to explore the idea that turns can be clockwise and anticlockwise in their direction. This would also be a useful unit to do prior to this Angles unit.
Activity:
Working with the learning object with students (Measure)
Show students the learning object and explain that it provides a tool for measuring angles.
Choose the Measure button from the front page of the learning object.
Choose Measuring Angle.
Discuss the first screen with students and clarify with them what the problem is asking.
Discuss ways to solve the question. For example, start by estimating how many red angles would be in half the circle. Ask for a volunteer to discuss a way to solve the question and enter an answer.
After a correct answer has been entered wait to click the Make an Angle to Measure button.
Discuss with the students ways to solve the question. Ask for a volunteer to enter a number.
Question: What is the primary tool used for measuring angles in this unit? Answer: The Measure button in the Angles learning object.
Question: What is the first step in the 'Measure' activity with the learning object? Answer: Choosing the 'Measure' button from the front page of the learning object. | 677.169 | 1 |
This blog is still alive, just in semi-hibernation. When I want to write something longer than a tweet about something other than math or sci-fi, here is where I'll write it.
Sunday, February 27, 2011
Sunday Numbers 2.0, Vol. 2: Perpendicular
90 degrees. Rectangular or square corners. Perpendicular. Vertical meets horizontal. All these words and phrases mean the same thing, and just to add to the confusion slightly, mathematicians will refer to it as orthogonal. But really, there is little confusion in the general public on this mathematical concept because we see it so often.
In some ways, the 90 degree angle is more a part of human nature than it is of "nature" nature. There are a few crystals that naturally place a flat surface perpendicular to another flat surface, and in studies of electrical fields, an electric current creates a second current that flows orthogonally to the original. But in nature, other forces interact with perpendicular surfaces, so even a tree that wants to grow straight up and down from the level surface of a prairie will be bent by the wind or rain or erosion.
In contrast, humans make any of a number of things with nice right angles, including the computer screen you are reading this post on and most the corners of the room where you are reading this. Rectangular two dimensional shapes and rectangular solids show up everywhere in human designs.
One of the reasons for the ubiquity of the 90 degree angle is that we know so much about it. The Pythagorean Theorem is ancient wisdom, sussed out by every civilization worth being called a civilization. There are proofs from the Chinese, the Indians, pre-Colombian American tribes like the Mayans, the ancient Egyptians, the Persians and, of course, the Greeks. There is a nice useful example of a² + b² = c², 3² + 4² = 5². Archeologists has found sticks of length 3 units, 4 units and 5 units in ancient Egyptian ruins, and the best guess is that these were used to make sure walls were perpendicular to the floor and to each other.
Of course, vertical meets horizontal is just one way to make two lines perpendicular. In the picture above, all the lines of length c are perpendicular to the ones they touch and parallel to the opposite side of the square.
Question: What is the angle that defines perpendicular lines? Answer: 90 degrees
Question: Which mathematical term is equivalent to perpendicular? Answer: Orthogonal | 677.169 | 1 |
Triangles/92988: If a base angle of an isoceles triangle is 16 degrees more than half of the measure of its vertex angle, than the vertex angle is ? degrees?
Please show solution using equation.
Thank You! 1 solutions Answer 67780 by scott8148(6628) on 2007-08-17 16:55:27 (Show Source):
the angles of a triangle (ANY TRIANGLE) sum to 180 degrees ... the 2 base angles of an isoceles triangle are equal
so, ((x/2)+16)+((x/2)+16)+x=180 ... 2x+32=180 ... 2x=148 ... x=74
Circles/93066: This question is from textbook
you 1 solutions Answer 67778 by scott8148(6628) on 2007-08-17 15:46:15 (Show Source):
Expressions-with-variables/92918: Hey this is Debbie and I am having trouble understanding this.
For these linear functions complete the table of values by substitution.
b=a+3
a 0 1 2 3 4 5 10
b
I am having trouble working how to do substitution because I forgot how to do it.Must help appreciated. 1 solutions Answer 67632 by scott8148(6628) on 2007-08-16 11:30:51 (Show Source):
then put the (b) value in the table next to the (a) value that you used to get it
Linear-equations/92938: In the x,y plane, what is perpendicular to the graph of the linear equation 3x-4y=0
I've done this over and over but i can't figure it out. The answer is supposed to be y=-(4/3)x+12 but i keep getting y=(3/4)x+0. Thank you for your help! 1 solutions Answer 67631 by scott8148(6628) on 2007-08-16 11:23:19 (Show Source):
Equations/92844: I really need help on this problem. I haven't taken algebra in years so im a little rusty.
Q: What is the remainder when x^3+5x^2-6x+10 is divided by x+3?
The book says that the answer is 46 but i'm not sure how to get that.
Thanks alot.
Quadratic_Equations/92892: I am having major problems with this word problem and need some help
Question: What is the relationship between the two base angles of an isosceles triangle? Answer: They are equal
Question: Which of the following is NOT a linear equation? A) y = 3x + 2 B) y = x^2 + 3x + 2 C) y = 3x - 2 Answer: B) y = x^2 + 3x + 2
Question: What is the equation of the line that is perpendicular to the graph of the linear equation 3x - 4y = 0? Answer: y = -(4/3)x + 12
Question: In the expression b = a + 3, what is the value of b when a is 5? Answer: 8 | 677.169 | 1 |
An example of an azimuth is the measurement of the position of a star in the sky. The star is the point of interest, the reference plane is the horizon or the surface of the sea, and the reference vector points to the north. The azimuth is the angle between the north vector and the perpendicular projection of the star down onto the horizon.[2]
In land navigation, azimuth is usually denoted as alpha, , and defined as a horizontal angle measured clockwise from a north base line or meridian.[3][4]Azimuth has also been more generally defined as a horizontal angle measured clockwise from any fixed reference plane or easily established base direction line.[5][6][7]
Today, the reference plane for an azimuth in a general navigational context is typically true north, measured as a 0° azimuth, though other angular units (grad, mil) can also be employed.
For example, moving clockwise on a 360 degree circle, a point due east would have an azimuth of 90°, south 180°, and west 270°. However, there are exceptions: some navigation systems use geographic south as the reference plane. Any direction can potentially serve as the plane of reference, as long as it is clearly defined for everyone using that system.
Quite commonly, azimuths or compass bearings are stated in a system in which either north or south can be the zero, and the angle may be measured clockwise or anticlockwise from the zero. For example, a bearing might be "south, thirty degrees east", abbreviated "S30°E", which is the bearing 30 degrees in the eastward direction from south, i.e. a bearing of 150 degrees clockwise from north. The reference direction, stated first, is always north or south, and the other direction, stated last, is east or west. The directions are chosen so that the angle, stated between them, is positive, between zero and 90 degrees. If the bearing happens to be exactly in the direction of one of the cardinal points, a different notation, e.g. "due east", is used instead.
We are standing at latitude , longitude zero; we want to find the azimuth from our viewpoint to Point 2 at latitude , longitude L (positive eastward). We can get a fair approximation by assuming the Earth is a sphere, in which case the azimuth is given by
A better approximation assumes the Earth is a slightly-squashed sphere (a spheroid); "azimuth" then has at least two very slightly different meanings. "Normal-section azimuth" is the angle measured at our viewpoint by a theodolite whose axis is perpendicular to the surface of the spheroid; "geodetic azimuth" is the angle between north and the geodesic – that is, the shortest path on the surface of the spheroid from our viewpoint to Point 2. The conceptual difference is that a plumb-line measures a gradient in the gravitational potential associated with the Newtonian forces, whereas the shortest path to the surface is a purely geometric entity. The difference is usually unmeasurably small; if Point 2 is not more than 100 km away the difference will not exceed 0.03 arc second.
Question: Can any direction serve as the reference plane for measuring an azimuth? Answer: Yes, as long as it is clearly defined for everyone using that system.
Question: Which direction has an azimuth of 0°? Answer: North
Question: What is the difference between normal-section azimuth and geodetic azimuth? Answer: Normal-section azimuth is measured at the viewpoint using a theodolite, while geodetic azimuth is the angle between north and the geodesic, the shortest path on the Earth's surface. | 677.169 | 1 |
However, this does not contain a sign (i.e. it doesn't distinguish between a clockwise or counterclockwise rotation).
I need something that can tell me the minimum angle to rotate from a to b. A positive sign indicates a rotation from +x-axis towards +y-axis. Conversely, a negative sign indicates a rotation from +x-axis towards -y-axis.
What about a = (-1,1) and b = (-1,-1), where the answer should be pi/2? You should check if the absolute value is bigger than pi, and then add or subtract 2*pi if it is. – Derek LedbetterJan 27 '10 at 21:52
@Derek
Good catch. I actually discovered this myself while implementing the solution. – CerinJan 28 '10 at 13:06
Question: What points does Derek Ledbetter use as an example (a and b)?
Answer: Derek Ledbetter uses a = (-1,1) and b = (-1,-1) as an example. | 677.169 | 1 |
Desargues' theorem
In a projective space, two triangles are in perspective axially if and only if they are in perspective centrally.
To understand this, denote the three vertices of one triangle by (lower-case) a, b, and c, and those of the other by (captial) A, B, and C. Axial perspectivity is the condition satisfied iff the point of intersection of ab with AB, and that of intersection of ac with AC, and that of intersection of bc with BC, are collinear, on a line called the axis of perspectivity. Central perspectivity is the condition satisfied iff the three lines Aa, Bb, and \Cc are concurrent, at a point called the center of perspectivity.
In an affine space nothing similar is true unless one lists various exceptions involving accidentally parallel lines. Desargues' theorem is therefore one of the most basic of simple and intuitive geometric theorems whose natural home is in projective rather than affine space.
The truth of Desargues' theorem in the plane is most readily deduced by getting it as a corollary to its truth in 3-space. Two triangles cannot be in perspective unless they fit into a space of dimension 3 or less; thus in higher dimensions the span of the two triangles is always a subspace of dimension no higher than 3.
The ten lines involved. (It is an amusing exercise to show that those incidence conditions can also be satisfied by a configuration of ten points and ten lines that is not incidence-isomorphic to the Desargues configuration.) The statement of the theorem above may misleadingly connote that the Desargues configuration has less symmtery than it really has: Any of the ten points may be chosen to be the center of perspectivity, and that choice determines which six points will be the vertices of triangles and which line will be the axis of perspecitivity
Question: Is Desargues' theorem applicable in affine space? Answer: No, unless there are specific exceptions involving parallel lines. | 677.169 | 1 |
Last week, we started our triginometry unit in geometry. We have been using the Sin, Cos, and Tan formulas in class and on the homework for the last few classes. The trig formulas are really interesting, because hey allow you to find a side length of a right triangle, when you only know the measures of a side and two angles. Its also cool that for eah angle measure, the ratio is always the same, no matter how big or small the triangle is. This makes finding the ratios so much easier!
Last week in geometry, we talked about, and did a lot of work with circles. We learned the formulas to find the area and circunfrance of a circle. We also learned a lot about the radius and diameter of a circle , and their special properties, and their relationships with pi. As well as learning about circles, we learned how to find the area of regular polygons, trapezoids, kites, and rhombuses. We also had a test at the end of the week. I thought it was going to be a lot harder than it actually was.
There are a few tricks you can use in certain right triangles to find out the lengths of all the sides if you only know the length of one side. The triangle either has to have two 45 degree angles, and a right angle; or it can have 30, 60, and 90 degree angles. In the 45, 45, 90 triangle; the hypotenuse is the length of one of the legs times the square root of two. in the 30, 60, 90; the long leg is the short leg times the square root of three, and the hypotenuse is twice the length of the short leg.
Last week in geometry we learned about radicals. Sometimes the square roots of numbers are never ending, unrepeting decimals; so we use simplified radicals to write them instead. Radicals can sometimes take a while to simplify, but i think that once i memorize a lot of the perfect squares, it will be much easier.
One week in geometry we learned about reflections. Reflections are when a shape is reflected across a line. This makes the line of reflection very similar to a line of symmetry. Reflections can sometimes be difficult, because ou have to reflect each point seperately, and if you mess one up, the whole thingis wrong; although it usually easy to notice.
That week in geometry we learned about translations, dialations and rotatins, which are all forms of transformation. Translations are when the hape is moved, dialations are when the shape changes size, and are when the shape is rotated around one point.
Last week in geometry, we learned a lot about quadrilaterals and the different properties of the different types. We also learned a lot of new theorems that have to do with the different quadrilaterals. It was interesting how a lot of the properties and characteristics of the quadrilaterals, were also theorems that we could prove.
Question: What is unique about the ratios of trigonometric functions? Answer: They are always the same, regardless of the size of the triangle
Question: What is the relationship between the hypotenuse and the short leg in a 30-60-90 triangle? Answer: The hypotenuse is twice the length of the short leg
Question: What other topics did the speaker learn about in geometry, besides trigonometry? Answer: Circles, regular polygons, trapezoids, kites, rhombuses, and transformations (reflections, translations, dilations, and rotations)
Question: Which trigonometric ratios does the speaker mention using in class? Answer: Sin, Cos, and Tan | 677.169 | 1 |
My favorite theorem/postulate/property is the Substitution Property of Equality. I think that is the one that I use the most, and that is why it is my favorite. I wish that you could use it for things that are congruent, because that would remove two or three steps in a proof.
I think i will look over my tests and quiz's for all my class's to study for my midterms.
Last week in geometry, we talked a lot about triangles, and the theorems that have to do with triangles. Like the triangle sum theorem, which says that all the measures of all the angles in any triangle must add up to 180 degrees. We also got the proof quiz back, and I wish I had done better. Maybe I should have studied harder.
This week in geometry, we took a quiz. It was on proofs. It was three questions and each questions was a proof. I thought the third proof was the hardest. But that might have been because I started running out of time towards the end of the block. I hope I did okay on the quiz.
Last week we had a two day school week, because of thanksgiving. In geometry, we had a binder quiz. I got a D, maybe I should stay more organized. We are also still doing proofs. I wish we didn't get any homework for the vacation, but we did in a few classes.
this week in geometry, we are doing more with proofs. We are also doing a lot with parallel lines and proofs having to do with them. We are proving that lines a parallel using the converse's of the postulates and theorems our class learned last week.
This week in geometry, he have been learning about proofs. I don't really like proofs, because they are a little tedious. And sometimes, all the steps you have to do just seem pointless. We have also been doing some more work with angle pairs, and parallel lines.
A construction is recreating line segments, angles, and other shapes. It is also bisecting lines and angles, and making perpendicular lines. All using a compass, straight edge, and a pencil.
My favorite construction is bisecting angles, because I like the steps.
To bisect an angle; you must first have an angle, then put the point end of the compass on the vertex of the angle and draw a curve that touches both sides of the angle. Then put the point end on the point that the curve touches the ray, and draw another curve above the first curve. Do the same for the second ray, and where the second and third curves meet, draw a point. Connect the vertex with this point, and you have bisected an angle.
Question: What did the student find difficult about the proofs in their geometry class? Answer: They found them tedious and sometimes pointless.
Question: Which theorem is mentioned as being studied in the student's geometry class? Answer: The Triangle Sum Theorem. | 677.169 | 1 |
You can put this solution on YOUR website! given:
angle
find: the supplementary angle
Supplementary angles are pairs of angles that add up to degrees. Thus the supplement of an angle of degrees is an angle of degrees.
...plug in degrees
a diagonal, width, and a height form right triangle; so, you need to use Pythagorean theorem
let's , then you have
.....solve for ...plug in values
....use only positive rood (height could be only positive number)
Once you've learned to translate phrases into expressions and sentences into equations, you are ready to dive into word problems.
Translate each phrase into an algebraic expression means:
"the of and " translates to ""
" multiplied by " translates to ""
"the of than to " translates to ""
You can put this solution on YOUR website! .........->.......}
if varies directly as , we have:
...first find ...->....->......
Now that we have the value of , we can plug in the new value, and solve for the new value of
if
Question: What should you do after learning to translate phrases into algebraic expressions and sentences into equations? Answer: Dive into word problems | 677.169 | 1 |
Tangents Teacher Resources
Title
Resource Type
Views
Grade
Rating
In this circles worksheet, students construct circles and determine the distance of its diameter. They explore the methods to construct inscribed and circumscribed figures. This one-page worksheet contains 5 multiple-choice problems.
Learners solve problems with circles and their properties. In this geometry lesson plan, students calcualte the diameter, radius, circumference and area of a circle. They identify the secant and tangent lines. They find the measurements of the chords.
Students use Geometer's Sketchpad or Patty Paper Geometry to explore and write conjectures about chords, tangents, arcs and angles. In this geometric conjecture lesson, students examine what a conjecture is as it relates to geometric properties. Students explore central angles and inscribed angles while writing conjectures about the relationship between the measure of these angles.
In this circles worksheet, 10th graders solve and complete 13 various types of problems. First, they solve each set of triangles for the given variable. Then, students name a radius, a diameter, a tangent, a secant, a point of tangency, and a chord in the circles shown.
In this tangent lines worksheet, 10th graders solve and complete 13 various types of problems. First, they solve the right triangles for the variables shown. Then, students name a radius, a diameter, a tangent, a secant, a point of tangency and a chord in the circle illustrated.
Students play a game based on the unit circle used in trigonometry. In this instructional activity on the unit circle, students draw a card a perform a given task including measuring degrees, radians, locating coordinates and finding the value of various functions.
Students match points on a unit circle to corresponding angles. For this geometry lesson, students evaluate the six trigonometric values and their angles. Students convert between degrees and radians and find the Cartesian equivalent coordinates.
Students apply the properties of trigonometric ratios to solve problems. In this calculus instructional activity, students apply the ratios of sine, cosine and tangent as they solve problems with vectors.
Question: What is the focus of the instructional activity on the unit circle? Answer: Students draw a card and perform a given task, including measuring degrees, radians, locating coordinates, and finding the value of various functions.
Question: What is the total number of problems in the first circles worksheet? Answer: 13 | 677.169 | 1 |
Apolyhedron is a three-dimensional solid
figure in which each side is a flat surface. These flat surfaces
are polygons and are joined at their edges. The word
polyhedron is derived from the Greek
poly (meaning many) and the
Indo-European hedron (meaning seat
or face).
A polyhedron has no curved surfaces.
The common polyhedron are pyramids and
prisms.
pyramid
prism
A polyhedron is called
regular if the faces are congruent, regular polygons and the
same number of faces meet at each vertex. There are a total of
five such convex regular polyhedra called the
Platonic solids.
tetrahedron
octahedron
icosahedron
hexahedron
dodecahedron
Euler's Polyhedron Theorem: Euler discovered that
the number of faces (flat surfaces) plus the number of vertices (corner
points) of a polyhedron equals the number of edges of the polyhedron
plus 2.
F + V = E + 2
Non-Polyhedra The following solids are not polyhedra since a part or all
of the figure is curved.
Cylinder
Cone
Sphere
Torus
A torus is a "tube
shape". Examples include an inner tube, a doughnut, a tire and a
bagel. Small r is the radius of the tube and capital R
is the distance from the centre of the torus to the center of the tube.
Question: Is a polyhedron a two-dimensional or three-dimensional figure? Answer: Three-dimensional
Question: Which of the following is NOT a polyhedron? A) Cube B) Sphere C) Cylinder D) Tetrahedron Answer: B) Sphere | 677.169 | 1 |
This activity initially helps students to recognise that the circumference is approximately six times the length of the radius. However, the objective of the activity is to give them practice in using the compass to make geometric designs beginning with a circle. This becomes a satisfying artistic activity, but you could also encourage them to use their design or logo to explore symmetry.
For example, the logo shown in the book has rotational symmetry of order 3 and three lines of reflective symmetry (m1, m2, m3). (Point A on the diagram below is the centre of rotation.)
Another design made from the original inscribed hexagon could be:
This design has rotational symmetry of order 6 and six lines of reflective symmetry. Examining designs for reflective or rotational symmetry follows on from the activity on page 21 of Geometry, Figure It Out, Level 3.
Answers to Activity
1. a. Six times
b. Your diagram should look like this:
2. Make six equally spaced marks on the circumference using the radius as the length of each space. Join every second point to draw an equilateral triangle.
3. Practical activity
4. Practical activity
Question: What is the order of rotational symmetry for the second example design? Answer: The order of rotational symmetry for the second example design is 6. | 677.169 | 1 |
Calculate, draw and learn the geometric shapes, with calculators which can draw parabolas, circles in the coordinate system, draw triangles with compass, ruler and protractor with step by step instructions and a lot more within geometry.
* Solid geometry
Cylinder, cone, cuboid, prism, pyramid, sphere and truncated cone
Calculate the solids and see the formulas for volume, surface area.
* Conversions
Area, degrees and radians, length, volume, currency
Convert between different units, learn how to convert between different currencies and convert between degrees and radians, directly in the app.
See the rules for exponentiations and parentheses or learn how to divide two numbers on paper. The calculator sets up the division with step by step explanations. Learn how to solve equations or see if a number is a prime or a composite number.
* Games and training
Learn how to multiply the numbers from 1 to 10 with a fun little game. Easy at first, but very soon challenging.
Question: Can the app determine if a number is prime or composite? Answer: Yes, the app can check if a number is prime or composite. | 677.169 | 1 |
Geometric Vectors
In this lesson our instructor talks about geometric vectors. He discusses magnitude and direction. He talks about describing quantities, William Rowan Hamilton, and James Maxwell. He talks about representing vector. He talks about algebraically and geometrically representing vectors. He also discusses adding and subtracting vectors and multiplying vectors. Lastly, he talks about unit vectors and the standard vectors along x and y axis. Four extra example videos round up this lesson.
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Geometric Vectors
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
Question: In what two ways can vectors be represented? Answer: Algebraically and geometrically. | 677.169 | 1 |
how can they be different if triangle is same dimensions?
zjmna
Quite a cool little puzzle, I can solve it but I don't think I could ever create it.
zjmna
It's easy to see if you try to line a piece of paper up along the "hypotenuse" of the shape.
z7q2
More specifically, the slope of the red triangle's hypotenuse is gentler than that of the hunter green triangle. So the top composite triangle's 'hypotenuse' is indented at the join, and the bottom triangle is pushed out at the join.
Love this puzzle, it's a classic. And nice to have an excuse to say 'hypotenuse' on BoingBoing.
Teresa Nielsen Hayden / Community Manager
We don't see it because the slight irregularity of the figures is below our normal noise level. We automatically correct for it.
retchdog
I think if it weren't for the thick black outlines it'd be easier to see… It's more of a trick than a puzzle.
trr
also notice that the point (8,3) is ON the top triangle but it is INSIDE the bottom triangle,(0,0) being the leftmost point of each triangle, so they are clearly not the same figures.
felixjawesome
Everyone is wrong!
2 + 2 = 5
Anonymous
I remember this from grade school.
lrn2geometry
JoshuaZ
Yes, this is a classic geometric puzzler. It was invented about 50 years ago by Paul Curry who was an amatuer magician and amateur mathematician. Very talented guy. Similar puzzles have been around for about a hundred fifty years but Curry's is noteworthy for being especially simple to see that there's a problem and being especially trick to solve.
Jason Rizos
They are both 13 squares long, 5 squares high. Fibonacci sequence. The extra square of area is redistributed by switching the triangles. Nice!
Now do it in three dimensions!
Thad E Ginataom
And it should be from whence comes this hole?
Aloisius
The slope of the hypotenuse changes. You can visually see it by comparing the square three up from the missing piece.
Basically, one of those is not a triangle.
Anonymous
@ #9 No. It would be whence comes this hole.
Brainspore
Ow! My Hippocampus!
richlb
This illusion is most effective because the grid lines throw your eyes off. With the gridlines removed, the different angles on the two different triangles becomes more apparent.
Simple: the two triangles are drawn to look similar (in the geometry sense of that word) but they cannot be, unless arcsin(0.4) = arcsin(0.375)
n
Ceci n'est pas une triangle
SamSam
An old one, but a classic.
Question: What is the mathematical reason why the two triangles cannot be the same? Answer: arcsin(0.4) ≠ arcsin(0.375)
Question: Who created this geometric puzzle? Answer: Paul Curry | 677.169 | 1 |
Theresa wrote @45:
We don't see it because the slight irregularity of the figures is below our normal noise level. We automatically correct for it.
I spotted something was off straight away, but had to count squares to be sure. The hypotenuse didn't look straight in either one.
I do a lot of on-paper designing things, though, and have a pretty well developed eye for lines that should line up and don't ("The hull plates and bottom aren't faired together quite right, that's not really straight! D'oh!")
So it's just all a trick. And here I thought we had invented a new math that would allow us to travel across the universe through measured triangular wormholes in space and time, from whence we could go forward.
misshallelujah
It's not that tough IMHO– we learnt it in I guess what would be the equivalent of grade school. The cyan and red triangles are not similar triangles and so don't form a straight line, so the combined shape is not an actual triangle for real.
Question: What did Theresa initially think the irregularity meant? Answer: She thought it might allow them to travel across the universe through measured triangular wormholes in space and time. | 677.169 | 1 |
In the first section of the report, we present a text from Piero della Francesca (1416-1492), proposition I.25 of De Prospettiva Pingendi. Piero gives instructions to construct the perspective image (in the painter's canvas) of a square given in the horizontal plane. The solution of Piero is to define a map from a square (interior and border), representing the horizontal plane, onto a trapezium (interior and border) representing the perspective plane. As usual, the drawing of Piero that illustrates his explanations depicts a 2D situation, and we will find in the following text some animations made with GSP that will try to explain the 3D ideas behind the 2D illustration. But perhaps the main contribution of the dynamic geometry, in this case, is to show that the map defined by Piero, when extended to the whole plane, may be identified with a plane projective transformation. This is not surprising, because the origin of this transformation is a central projection, but the software enables us to "confirm" this conjecture and to easily draw the respective vanishing line.
In section III, the subject is the tracing of tangents to the cycloid and the solutions proposed are those by Gilles Persone de Roberval (1602-1675 ) and by René Descartes (1596-1650). One of the most interesting aspects is the fact that the construction of the cycloid proposed by Roberval, that is, to consider the cycloid as generated by a point subjected to two uniform motions, one circular and the other straight, is the most appropriate one to be used with dynamic software to trace the cycloid as a locus of a point that moves around a circular path that simultaneously moves in a straight path. Also, the consideration and construction of prolate and curlate cycloids is very straightforward with the help of GSP.
The subject of section IV is based on the work of Gaspard Monge (1746-1818) (extracted from his Descriptive Geometry of 1799), describing one method of finding the tangent planes to a given sphere containing a given line. As with the previous tasks, the work in descriptive geometry is greatly simplified by the software for dynamic geometry. But for anyone who is a newcomer to descriptive geometry – and this is the case for almost all prospective and inservice teachers in Portugal – it is not an easy task to follow Monge's drawings in double projection. But with a tool that can represent each step of a construction in cavalier perspective – that is not difficult to build in GSP – we may follow step by step those drawings and visualize the situation very easily. The main ideas of Monge's method to find the two tangent planes are the following:
• To construct two conic surfaces with vertices in two points of the given line e and touching the sphere (in two circles c and c') ;
• To obtain the two points of tangency of the planes and the sphere, i.e. R and S;
• Finally defining the two planes: R and e and S and e.
Question: Who was the author of the text presented in the first section? Answer: Piero della Francesca
Question: What is the title of the work from which the first section's text is taken? Answer: De Prospettiva Pingendi
Question: Which scientist proposed the construction of the cycloid by considering it as generated by a point subjected to two uniform motions? Answer: Gilles Persone de Roberval
Question: What shape is given in the horizontal plane and needs to be represented in the perspective image? Answer: A square | 677.169 | 1 |
Parts of a Circle - Vocabulary and Application Lesson PPTX this 31 slide presentation on learning and applying the vocabulary of parts of a circle, including radius, center, diameter, and chord.
This presentation includes highly- contrasting slides for students to learn and apply the different parts of a circle. Vocabulary includes center, radius, diameter, and chord. Students will learn, review, and apply knowledge to find the lengths of radii and diameters.
Please be aware that this presentation is designed to be introductory in nature and does not include tangents, points outside the circle, or circumference.
PowerPoint is in 2010 format.
As always, we would like you to be happy with your purchases from us. Please let us know if we may be of service to you.
Be sure that you have an application to open this file type before downloading and/or purchasing.
3737.01
Question: What should you ensure before downloading or purchasing the presentation? Answer: That you have an application to open this file type | 677.169 | 1 |
Now, let's
consider what would happen if we were to let Barney start outside of the
room.Would he still return to his
starting point after 5 bounces?To
explore this idea with GSP, click HERE.
It
appears that Barney will return to his starting point after 5 bounces.
If we
assume there is an X on the line BC beyond B that is not P, a similar proof to
the one given above shows that triangles PHB and CGF are congruent and PH =
CG.Therefore, CF = PB and that
would hold for any point P, so X = P.
Question: What is the purpose of the proof mentioned in the text? Answer: The purpose of the proof is to show that CF equals PB for any point P on the line BC. | 677.169 | 1 |
Computes the topological relationship (Location) of a single point to a Geometry. It handles both single-element and multi-element Geometries. The algorithm for multi-part Geometries takes into account the boundaryDetermination rule.
Returns:
the Location of the point relative to the input Geometry
The documentation for this class was generated from the following file:
Question: If the input is a single point, what will the function do? Answer: It will determine the point's location relative to the input geometry | 677.169 | 1 |
As can be seen, the left projection works the same as the regular projection with the exception that the projection is perpendicular to the destination vector(vector B). there are several things that we can conclude from this. First, if the dot product is positive, the projection is toward the left of vector 'B'
if the dot product is negative the projections is toward the right of vector 'B" and if its 0 its on the vector 'B' and third the absolute value of the dot product is the distance the end of vector 'a' is from vector b. This is going to become very useful to understand in future explanations. One useful way is if there is a point moving toward a line(vector) or away from it, one can determine how far the dot is from the line.
I hope this is becoming clearer as all of this examples are just illustrations of what has been discussed in earlier posts.
A very important and useful part of using vectors is being able to project them. Projection is mainly used to find distances. Lets start defining it: To project a vector, two vectors are need and always either vector can be projected into the other. Lets assume two vectors are connected by a common starting point(don't have to be but that's going to be our representation and is going to be easier to understand) and both vectors can be moving in any direction. lets use this example for ease of use:
first example:
vector C projected onto vector A. The projection is a vector from the start of vector A to a line perpendicular to vector A extending from the end point of Vector C
Second example:
Now lets Assume vector B is a line perpendicular to Vector A this creates an x,y Graph. For general purpose, one angle have to be at any degree while the second one has to be at 90 degrees from the first vector or visa versa to make the graph. While keeping that in mind (and it's quite important to note), note that vector B is in the direction of Vector A left normal. Now lets project vector C onto Vector B. The projection is a vector from the start of vector B to a line perpendicular to vector B extending from the end point of vector C.
The equation to find the projection(or the new Projected Vector). from vector C into Vector A is done with a couple of equations. First we find the distance from the start of Vector A to the perpendicular line extended from the end of Vector C to the intersection point on Vector A. The equation is the dot Product between vector C and the normal of Vector A:
Remember that the normal of a vector is calculated like this; vector A components are A.vx and A.vy to find its norma first we find its length:
length = Sqrt(A.vx*A.vx+A.vy*A.vy)
then the normal(or unit vector):
A.dx = A.vx/length
A.dy = A.vy/length
Then to find the distance (or dotProduct in this special case) from the start of Vector A to its perpendicular line extending from the endpoint of vector C is done this way:
DotProduct = C.vx * A.dx + C.vy * A.dy
Question: Which of the following is NOT a use of vector projection? A) Finding distances B) Determining angles C) Measuring speed D) Calculating area Answer: D) Calculating area
Question: In the provided text, which vectors are used in the first example of vector projection? Answer: Vector C and Vector A | 677.169 | 1 |
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Hi Guys, Geometry, Algebra and Number system form the major chunk of our QA section for CAT. Proficiency in these three sections would definitely boost our Quants scores. Contents of Geometry 1. Plane Geometry - Basics and Triangles 2. Polygons and Quadrilaterals 3. Circle 4....
Three balls touching each other can placed like :-(:-( :-( consider these connected to each other, And a ball is above in the center, now the ball center from earth above earth is radius of any ball. now the centers of 3 balls makes the equilateral triangle. try to imagine a...Three balls touching each other can placed like :-(:-( :-( consider these connected to each other, And a ball is above in the center,
now the ball center from earth above earth is radius of any ball.
now the centers of 3 balls makes the equilateral triangle. try to imagine a ball above it... if you keep a ball above it
it will take a shape like pyramid. base is Equi. triangle and top is in middle if we draw a perpendicular on the triangle it will meet centroid. so centroid is 2/3 of perpendicular inside the triangle.
Now you need to again apply one time for hight. than hight+ radius is your answer. I hope I am pretty clear on that.Diagonals in a Cyclic Quadrilateral In Triangles PAD and PB...Thanks in advance, Sandeep
Diagonals in a Cyclic Quadrilateral
In
Triangles PAD and PBC are similar, so that PA/PB = AD/BC = PD/PC, which can be also written as (2) ABAD/PA = ABBC/PB, and BCCD/PC = ADCD/PD, In the same manner, the similarity of triangles PAB and PDC implies (3) ABAD/PA = ADCD/PD. which shows that four expressions (4) ABAD/PA, ABBC/PB, BCCD/PC, and ADCD/PD are all equal. (1) follows by combining the first and the thrid terms and also the second and the fourth.puys i have a question in which i am stuck :help C T...
C The distance of P from its center is: (a) (sqrt2)/3 (b) 2/(sqrt3) (c) 3/(sqrt2) (d) 1
From top city,3 routes are possible Since there are 3 edges leading from each city and also the figure is perfectly symmetrical, these 3 routes are possible from each edge. So,total number of routes = 4 * 3 = 12. ps : one more ceo from DelhiFrom top city,3 routes are possible
Since there are 3 edges leading from each city and also the figure is perfectly symmetrical, these 3 routes are possible from each edge.
Question: How many edges lead from each city in the given figure? Answer: 3
Question: Which of the following is NOT a topic covered in the 'Geometry' section of the QA section for CAT? (Multi-choice)
A: Plane Geometry
B: Algebra
C: Number System
D: Calculus
Answer: D: Calculus
Question: What is the distance of point P from its center in the given figure? Answer: (a) (sqrt2)/3
Question: What are the three main sections of the QA (Quantitative Ability) section for CAT? Answer: Geometry, Algebra, and Number System. | 677.169 | 1 |
Hi puys, Please solve the below qstion: Two circles C(O,r) and C(O',r') intersect at two points A and B. A tangent CD is drawn to the circle C(O',r') at A. then 1) /_OAC=/_OAB 2) /_OAB=/_AO'O 3) /_AO'B=/_AOB 3) /_OAC=/_OAB the question is from LOD2(Geometry) Arun Sharma Qno. 21. 22,23,24,25 4 values
A triangle has sides of lengths 10, 24 and n, where n is a positive integers. The number of values of n for which this triangle has three acute angles is A. 1 B. 2 C. 3 D. 4 E. 5If a circle is provided with a measure of 19 degrees at the centre. Is it possible to divide circle into 360 equal part.
1) never. 2) possible when one more measure of 20 degrees is given 3) always 4) noneHi Puys, I got a problem in Geometry . If a circle is provided with a measure of 19 degrees at the centre. Is it possible to divide circle into 360 equal part. 1) never. 2) possible when one more measure of 20 degrees is given 3) always 4) none Hi I got the answer as option D.)4 let me know if I am ri...
option 1) is the answer let each side of hexagon be 1 unit the triangle containing the larger circle is an equilateral tr. with side = root3 now use area = inradius * semiperimeter 1/2*root3*3/2 = r * 3root3/2 r = 1/2 so area of bigger tr. = 1/4 now , for smaller triangle...
SET - 30 Guys, This is a very good problem !! In the regular hexagon shown below, what is the ratio of the area of the smaller circle to that of the bigger circle? a. 3 : 7 + 43 b. 3 : 7 + 163 c. 3 : 7 + 3 d. 3 : 7 + 23
a. 120 b. 160 c. 200 d. 240B'C' cut by D'E' Now to form a parallelogram, trapezium B'C'D'E' will be turned upside down and will be joined to BCDE. Now the sides of the parallelogram are.. E'BDC'.. and the perimeter = BC + (DE + DB) + (B'C' + C'E') + E'D' = 40 + 40 + 40 + 40
Question: What is the area of the larger triangle in the hexagon problem, given that each side of the hexagon is 1 unit? Answer: 1/4
Question: Can a circle with a central angle of 19 degrees be divided into 360 equal parts? Answer: Never (Option 1) | 677.169 | 1 |
Each face has three numbers: they are arranged such that the upright number (which counts) is the same on all three visible faces. Four-sided dice are often used in Role-playing games such as Dungeons & Dragons, to get small numbers for things such as damage or character statistic increasesA tetrahedron (plural tetrahedra) is a Polyhedron composed of four triangular faces three of which meet at each vertex. Alternatively, all of the sides have the same number in the lowest edge and no number on the top. This die does not roll well and thus it is usually thrown into the air instead.
Each face is triangular; looks something like two Egyptian pyramids attached at the base. An octahedron (plural octahedra is a Polyhedron with eight facesA pyramid is a Building where the upper surfaces are triangular and converge on one point Usually, the sum of the opposite faces is 9.
Each face is kite-shaped; five of them meet at the same sharp corner (as at the top of the diagram in this row), and five at another equally sharp one; about halfway between them, a different group of three faces converges at each of ten blunter corners. As a die Some Role-playing games and Miniature wargames use ten-sided dice typically pentagonal trapezohedra to get random decimal numbers such as percentagesAs a die Some Role-playing games and Miniature wargames use ten-sided dice typically pentagonal trapezohedra to get random decimal numbers such as percentagesIn Geometry a kite, or deltoid, is a Quadrilateral with two disjoint pairs of Congruent Adjacent sides in contrast The ten faces usually bear numbers from zero to nine, rather than one to ten (zero being read as "ten" in many applications), and often all odd numbered faces converge at the same sharp corner, and the even ones at the other. In Mathematics, the parity of an object states whether it is even or oddIn Mathematics, the parity of an object states whether it is even or odd
Faces are equilateral triangles. In Geometry, an icosahedron ( Greek: eikosaedron, from eikosi twenty + hedron seat /ˌaɪProperties The area of an equilateral triangle with sides of length a\\! Typically, opposite faces add to twenty-one. A 2nd century CE Roman icosahedron die is in the collection of the British Museum, though the game it was used for is not known. The British Museum is a Museum of human history and culture in London.[6]
Rarer variations
Most commonly a joke die, this is just a sphere with a 1 marked on it. "Globose" redirects here See also Globose nucleus. A sphere (from Greek σφαίρα - sphaira, "globe There exist spherical dice constructed of a spherical weight glued between two hemispherical pieces, which have been hollowed to create a certain number places where the weight can rest. The weight effectively pins one side of the sphere down, revealing the number directly opposite.
Question: What is the typical opposite face sum of an icosahedron die? Answer: Twenty-one | 677.169 | 1 |
I cannot paste the figure so I will describe it. It is a triangle with A on the top and B, C, D, E at the bottom. In the middle of the triangle is C and D is 1 above C and 2 above D.
A
1 2
B C D E 1 solutions Answer 375149 by solver91311(16877) on 2012-03-22 16:11:46 (Show Source):
Since segment AB and segment AE are congruent (given), the triangle ABE must be isoscles by definition of an isosceles triangle.
From that it follows that angle ABC must be congruent to angle AED, again by definition of an isosceles triangle.
Then because you are given that segment BC and segment DE are congruent, triangle ABC must be congruent to triangle AED by SAS.
Now you aren't clear whether angle 1 is angle ACB or ACD. Assume it is ACB, then ACB is congruent to ADE by CPCT. Therefore angle 1 equals angle 2. QED.
If angle 1 is ACD and angle 2 is ADC, then since angle ACB is supplementary to angle ACD, angle ADE is supplementary to angle ADC, and from the step above angle ACB is congruent to angle ADE, then angle ACD is congruent to angle ADC by transitive equality. Therefore angle 1 equals angle 2. QED.
a quadrilateral, a parallelogram, a rhombus, a rectangle, and a square.
John
My calculator said it, I believe it, that settles it
Triangles/590383: triangle ABC has vertices A(–2, 3), B(2, 0), and C(–1, –4). Is triangleABC a right triangle?
triangleABC is a right triangle because a.
triangleABC is a right triangle because a.
triangleABC is a right triangle because s.
triangleABC is not a right triangle. 1 solutions Answer 375145 by solver91311(16877) on 2012-03-22 15:54:55 (Show Source):
Use the slope formula to calculate the slopes of the three lines that contain the three line segments AB, AC, and BC that form the sides of the triangle. If any pair of slopes are negative reciprocals of each other, then those two segments are perpendicular, meaning that the angle between them is a right angle and yea verily the triangle is a right triangle. On the other hand, if no pair of slopes are negative reciprocals, then the triangle is not a right triangle.
Question: Which type of triangle is triangle ABE in the text? Answer: Isosceles
Question: Who provided the solution for triangle ABC? Answer: solver91311 (16877)
Question: Is triangle ABC in the text a right triangle? Answer: Yes. | 677.169 | 1 |
Irregularities in the form of the Bent Pyramid, which Dorner attributes to settlements in the core-masonry, are found to reflect the complexities encountered by the builders in the fulfilment of an exceptionally ambitious project.
Soon after the construction of the Bent Pyramid, the measure of 280 cubits was used for the height of the Great Pyramid, which was divided into parts of 82 and 198 or 2 x 99 cubits at the level of the 'King's Chamber'.
It is generally assumed that the unique double-sloping profile of the Bent Pyramid was brought about during the construction when the builders, noticing a settlement in the masonry, decided to reduce the pyramid's eventual volume by lessening the external casing-angle.
This squaring of the circle works with a right triangle that represents the apothem(ZY) (a line drawn from the base of the center of one of the sides to top of the pyramid), down to the center of the base (ZE), and out to the point where the apothem touches the Earth (EY).
All of the Sacred Geometry ratios we will be working with, the square roots of two (1.414), three (1.732) and five (2.238), phi (1.618) and pi (3.1416), are all irrational numbers.
The claim is that the smaller circle (in square abcd) is to the larger circle (in square ABCD) as the Moon is to the Earth.
An interesting problem of spherical trigonometry is that of finding the area of a spherical cap of either a cone or pyramid, with its apex at the center of the sphere.
Tacitly, we assume that the geometry is a plane geometry (curvature equal to zero).
For each law, we give the spherical (a, b, c are the sides; A, B, C are the angles), its dual for the polar triangle (A. C are the angles; a, b, c are the sides), and the plane (a, b, c are the norms of the sides; A, B, C are the angles) version.
The earliest known proprietors of sacred geometry were the Egyptians who embedded its secrets in the ground plans of their temples, their frescoes and, most blatantly, in the Gizeh pyramid which single-handedly contains most of the fundamental universal laws that many a tortured schoolchild now attributes to Pythagoras.
Because sacred geometry reflected the universe, its pure forms and dynamic equilibriums shared a higher purpose: the attainment of spiritual wholeness through self-reflection, thereby giving structural insight into the workings of the inner self.
As a mirror of the heavens sacred geometry was liberally applied across the Egyptian landscape for millennia as a way to bestow universal order on Earth, as reflected in their Hermetic maxim 'As Above, So Below'.
Question: What is the topic of the interesting problem of spherical trigonometry mentioned? Answer: Finding the area of a spherical cap of either a cone or pyramid, with its apex at the center of the sphere
Question: What was the initial issue that led to the Bent Pyramid's unique double-sloping profile? Answer: A settlement in the masonry
Question: What was the height of the Great Pyramid in cubits? Answer: 280 cubits | 677.169 | 1 |
bottom of the castle, or top of the hill, the angle of depression was 4" 2' :
required the horizontal distance of the ship, as also the height of the hill,
that of the castle itself being 60 feet.
(14) The height of the mountain called the Peak of Teneriffe is about
2^ miles ; the angle of depression of the remotest visible point of the
surface of the sea is found to be 1 58' : required the diameter of the
earth, and the utmost distance at which an object can be seen from the
top of the mountain that is, the length of the line from the eye to the
remotest visible point.
(15) The angles of elevation of a balloon were taken by two observers
at the same time ; both were in the same vertical plane as the balloon,
and on the same side of it ; the angles were 35" and 64, and the observers
were 880 yards apart : required the height of the balloon.
224. Oblique-angled Triangles. Every triangle has six parts
as they are called three sides and three angles. If any three of the
six be given, except they be the three angles, the remaining three may
be determined by computation. It is plain that the three angles would
not suffice for the determination of the triangle, because there may be an
infinite variety of equiangular triangles all different in size. There are
three cases to be considered.
225. Case I. When two of the three given parts are opposite parts an
angle and a side.
Question: What is the relationship between the angles of elevation and the height of the balloon? Answer: The angles of elevation help determine the height of the balloon | 677.169 | 1 |
Pie chart
In a pie chart, the arc length (and consequently, the central angle and the area) of each segment, is proportional to the quantity it represents. Together, the wedges create a full disk. A chart with one or more wedge separated from the rest of the disk is called an exploded pie chart.
Example
A pie chart for the above data to the right
An exploded pie chart, with the largest party group exploded
The following example chart is based on the results of the election for the European Parliament in 2004. The following table lists the number of seats allocated to each party group, along with the percentage of the total that they each make up. The values in the last column, is the central angle of each segment, found by multiplying the percentage by 360°
Question: What is the name of a pie chart with one or more wedges separated from the rest of the disk? Answer: Exploded pie chart | 677.169 | 1 |
You have already rated this item,
The Office of Square Trading, the government body overseeing the sale of all rectangular shapes, has been investigating the illegal sale of squares.
"There is a massive black-market trade in squares, oblongs and rectangles," said Rex Tangle, chair of the Office of Square Trading. "People are circumventing the traditional markets of squares, and instead buying them under the counter."
This illegal trading is costing the government millions in lost Shape Tax revenue.
"Our sister organisation, the Office of Elliptical Trading, has not seen this same rise," said Tangle. "Where we're seeing people secretly buying rectangular shapes without paying the necessary shape tax, circles, ovals and ellipses are still being sold legitimately."
This tax evasion scam has gone largely unnoticed for so long due to a mislabelling of shapes, according to Tangle.
"We've seen perfect squares being sold is polygons," he said. "All polygons are tax exempt, due to them not being luxury shapes. We even saw one square being sold as two triangles to avoid paying tax. I think this is why circles and ovals have avoided being scammed this way, as it takes an infinite number of triangles to make a circle. And that would be expensive, even if the triangles are sold at a penny per triangle."
New procedures are being put into place to crack down on this illegal trading in squares, forcing the clear labelling of squares as squares and not as multiple triangles or a four sided polygon. Additionally, the Office of Square Trading is to be given new powers to seize and detain anybody falsifying the shape being sold.
Make IainB
Question: Which shape would be expensive to make by combining other shapes? Answer: A circle, as it would require an infinite number of triangles.
Question: What are the new procedures being put into place to stop this illegal trading? Answer: Clear labelling of squares and giving the Office of Square Trading new powers to seize and detain those falsifying the shape being sold. | 677.169 | 1 |
math go to google and type degree for right triangle trigonometry.and click on the second option which is:Trigonometry of rigth triangle. Topics in trigonometry. you will get ur answer. sorry couldnt paste the whole thing here hope i could help u.:)
Tuesday, February 10, 2009 at 4:10pm by vero
Trigonometry Well, i tried to see the beauty in trigonometry but to me, it is just too HARD!!!!!!!! There r so many formulae in trigonometry and how to i know which one to use. After i stay in the desk for 15 mins, i just wanna throw this stupid book away. I cant gain anything even i try ...
Saturday, July 19, 2008 at 1:03am by Tommy
Trigonometry Just expand it ... 8x^3 - 6x^4 - 5x^2 + 10x all done! Why are you calling this trigonometry?
Sunday, December 26, 2010 at 5:55pm by Reiny
Trigonometry question? Thanks a lot, sorry I am not too good at trigonometry
Thursday, March 28, 2013 at 8:25am by Knights
trig this is trigonometry.. its the class i'm taking and my book is called, functions, statistics, and trigonometry. scott, foresman.
Wednesday, November 12, 2008 at 11:12pm by jerson
Trigonometry The question is not about trigonometry.
Sunday, December 26, 2010 at 5:55pm by drwls
Trigonometry By the way, this is not trigonometry. It is algebra.
Wednesday, August 12, 2009 at 8:48pm by drwls
Trigonometry Reiny, I interpret "inverse" being inverse trigonometric function, since he is doing a trigonometry course. However, I stand to be corrected!
Thursday, March 3, 2011 at 1:35am by MathMate
Trigonometry Why did you title this "trigonometry"?
Saturday, June 4, 2011 at 9:17pm by Reiny
Trigonometry I am surprised that you are studying trigonometry and don't know the basic shape of y = sinx Surely your instructor has taught you how to sketch the basic graphs before jumping to technology methods.
Tuesday, April 3, 2012 at 6:25pm by Reiny
history i know an education is important, but i don't think i'll need to know u.s. history, or trigonometry in the future. i plan on doing something in the cooking field as a career, and most people i know who do that didn't really get good grades on their trigonometry ...
Question: What did MathMate interpret "inverse" as? Answer: Inverse trigonometric function
Question: What is the main topic of discussion in this text? Answer: Trigonometry
Question: What is the title of Jerson's book? Answer: Functions, Statistics, and Trigonometry
Question: What did Knights ask for help with? Answer: A trigonometry question | 677.169 | 1 |
Replies(8)
A PVector(100, 100) is not a point at (100,100) it's a vector from the origin (0,0) to point(100, 100). So your middle PVector is a vector from the origin to (350, 350). The angle of this vector is 45 degrees. Your Pos5 PVector is a vector from the origin to (550, 550);. The angle of this vector is also 45 degrees. So the angleBetween the middle and the Pos5 Pvector is 0. For all the other it's the same, the method gives the angle between one vector and another.
To understand better what angleBetween does add the following lines to your sketch:
// add to draw()
showPVector(new PVector(mouseX, mouseY), color(55, 55, 233));
// add to showPVector()
stroke(col);
line(0,0, myPVector.x, myPVector.y);
You will notice as described above, that a PVector on the 45 degree (middle) line will give 0 degrees, while (0, mouseX) or (mouseY, 0) will give 45 degrees. So it's just the angle from one vector compared to another. Also see below.
I also remember looking at the PVector.angleBetween, and wanting a different approach. I wrote some code as a study, and based my angles on the Unit Circle, with 0 degrees being at 3 o'clock. I chose not to use PVectors, but I'm sure it could easily be refactored using them if desired. Also, my output returns the angular separation between two lines rather than the bisector, but this is also easily modified. The code is archived here, and I'm copying it below also. See if there's anything in it you find useful.
Question: What does the method angleBetween do? Answer: The method angleBetween gives the angle between one vector and another.
Question: What is a PVector(100, 100)? Answer: A PVector(100, 100) is a vector that starts from the origin (0,0) and ends at the point (100, 100). | 677.169 | 1 |
The first two figures are congruent to each other. The third is a different size, and so is similar but not congruent to the first two; the fourth is different altogether. Note that congruences alter some properties, such as location and orientation, but leave others unchanged, like distances and angles. The latter sort of properties are called invariants and studying them is the essence of geometry.
Congruence of triangles
Two triangles are congruent if their corresponding sides and angles are equal in measure. Usually it is sufficient to establish the equality of three corresponding parts and use one of the following results to conclude the congruence of the two triangles:
SAS Axiom (Side-Angle-Side): Two triangles are congruent if a pair of corresponding sides and the included angle are equal.
SSS Theorem (Side-Side-Side): Two triangles are congruent if their corresponding sides are equal.
ASA Theorem (Angle-Side-Angle): Two triangles are congruent if a pair of corresponding angles and the included side are equal.
While the AAS (Angle-Angle-Side) condition also guarantees congruence, SSA (Side-Side-Angle) does not, as there are often two dissimilar triangles with a pair of corresponding sides and a non-included angle equal. This is known as the ambiguous case. Of course, AAA (Angle-Angle-Angle) says nothing about the size of the two triangles and hence shows only similarity and not congruence.
However, a special case of the SSA condition is the HL (Hypotenuse-Leg) condition. This is true because all right triangles (which this condition is used with) have a congruent angle (the right angle). If the hypotenuse and a certain leg of a triangle are congruent to the corresponding hypotenuse and leg of a different triangle, the two triangles are congruent
Question: Are the first two figures described in the text congruent to each other? Answer: Yes.
Question: What kind of relationship does the third figure have with the first two? Answer: It is similar but not congruent. | 677.169 | 1 |
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