Ratios and Proportions

Transcription

1 Ratios and Proportions ( 7-1/8-1) 1 Proportions & Similarity 1.1 Ratios A ratio is a comparison of two numbers by way of division. Given two numbers a and b, such that b 0, the ratio of a to b is written as: a : b or a b. NB: a ratio should always be reduced to lowest terms. A proportion is an equality or equation between two ratios. Means: b and c a b = c d Extremes: a and d Example: The ratio of the diagonals in a rhombus are 24 to 32. Write the ratio as a simplified fraction (i.e. reduce to lowest terms). We begin by writing the ratio and reducing = 8/ 3 = 3 8/ 4 4 1

2 1.1.1 Means-Extremes Property To solve a proportion, we use this property from algebra. The product of the means is equal to the product of the extremes. Given: a, b, c, and d, such that b 0 and d 0, a b = c d = ad = bc. Example: Solve the proportion 3 5 = x 75. We write the proportion, then use means-extremes property. 3 5 = x 75 = (5)(x) = (3)(75) = x = 45. Example: Solve the proportion 3 x = 6 x+8. Keep in mind that we must distribute when multiplying. 3 x = 6 x + 8 = (6)(x) = (3)(x + 8) = 6x = 3x Thus, working out the algebra we have 3x = 24 = x = 8. Example: Solve the proportion 4 x = x 16. Begin with the original proportion again. 4 x = x 16 = (x)(x) = (4)(16) = x2 = 64. We can use the square root to find x as follows: x2 = 64 = x = 8. 2

3 1.1.2 Extended Ratios An extended ratio can be used to compare three or more numbers. See example below. Example: The ratio of the angles in a triangle are 2 : 3 : 4. Find the measure of each angle. The extended ratio 2 : 3 : 4, tells us that the ratio between the angles are as follows: 2 3, 3 4, and 2 4. Since the angles in a add up to 180, we let the angles have measures 2x, 3x, and 4x. This gives us the same proportions 2x : 3x : 4x. Now, 2x + 3x + 4x = 180; thus, x = 20. the angle measures are 40, 60, and Geometric Mean The means-extreme property can be used to find the geometric mean between any two numbers. Given two numbers a and b, the geometric mean is the positive number g given by: a g = g b. This means that g 2 = ab, thus g = ab. 3

4 Example: Find the geometric mean between 24 and 48. We can set up a proportion 24 g = g 48, or just write g = (24)(48). Thus, g = (24)(24 2) = (24 2 )(2) = Example: The geometric mean between 12 and what number is 6 6? Let x be the other number. We set up a proportion as before = 6 6 x. Now, we use the means-extremes property to write 12x = (6 6)(6 6) = 12x = 216 = x = 18. Thus, 6 6 is the geometric mean between 12 and Proportion Rules Exchange of Means & Extremes a b = c d Adding the Denominator a b = c d a + b b a c = b d = c + d d 4

5 Similarity ( 7-2/3) 1.2 Simliar Polygons Definition. Two shapes are similar ( ) iff all corresponding s are =, and all corresponding sides are proportional ( ). NB: This means the shapes look the same but they have different sizes. In the diagram below, we are given that LMNOP QRST U. By the definition we conclude the corresponding angles are congruent, i.e L = Q, M = R,... and so on. Q L M R P U O N T S But the sides are proportional. So they have the same ratio when compared. This ratio of the sides is known as the scale factor. LM QR = MN RS = NO ST =.... 5

6 1.3 Similar Triangles AA Postulate. If two angles in one triangle are congruent to two angles in another triangle, then the triangles are similar. Example. Decide why the triangles below are similar. Then find x and y. T x 14 I y 5 R 9 E 3 G Since T R IE, the corresponding angles ( T and EIG) are congruent. Also, G = G. Now, we have two congruent angles, so EIG RT G by the AA similarity postulate. To find x and y, we use the fact that all corresponding sides are proportional. EG RG = IG T G = EI RT = 3 12 = x = y 14 Notice that the scale factor is 1:4 from the first ratio EG : RG. Using the first proportion, we have x = 60 = x = 15. 6

7 Then, using the first and last ratios for our proportion, we have 1 4 = y 14 = 4y = 14 = y = 7 2 Similarity Theorems. or 3.5. SSS If three sides of one are proportional with the corresponding sides of another, then the two s are similar. Example. In the diagram below SUN RY Z since all corresponding sides are proportional. This is because, SU RY = UN Y Z = NS ZR = 1 : 4. This is the scale factor, which is also known as the ratio of similarity. U Y S 5 N R 20 Z SAS If two sides of one are proportional with the corresponding sides of another, and the included angles of those two sides are congruent, then the two s are similar. 7

8 Triangle Proportionality ( 7-4) 1.4 Parallel Lines & Proportionality In the example above for the AA similarity, recall that the parallel line to the third side created two similar triangles. Using this fact and the rule of adding denominator for proportions, we have the following theorem. Triangle Proportionality Theorem. If a segment connecting two sides of a triangle is parallel to the third side, then it divides the two sides proportionally. Using the diagram below: ES RM T E ER = T S SM. T E S R M The converse is also true. This means if the segment connecting two sides of a triangle divides the two sides proportionally, then the segment is parallel to the third side. This can be called the Converse Theorem. 8

9 1.4.1 Midsegment Theorem A midsegment is a segment inside a triangle that connects the midpoints of two of its sides. Since the midsegment divides both of the sides into equal parts, the sides are proportional. Then, by the converse of the theorem, we conclude it is parallel to the third side. T E S R M Also, because the two triangles are similar ( T ES T RM), the scale factor is T E : T R = 1 : 2. This means that the midsegment is half of the length of the third side. Midsegment Theorem. The midsegment of a triangle is parallel to the third side and is half the length of the third side. 9

10 1.4.2 Corollaries Since parallel lines can create more similar triangles, the proportionality theorems can be extended as follows. If three or more parallel lines intersect two transversals, then the transversals will be divided into segments of proportional lengths. In the diagram below, ST EA CR. S E C T A R Thus, the ratio of the segments T A : T R = SE : SC. Also, T A : AR = SE : EC. Example. Suppose T A = 5, AR = 3, and SC = 6. Find the length of EC. Solution. We can use the ratio AR T R = EC SC. We know by addition that T R = 8. Let EC = x and solve. 3 8 = x 6 = 8x = 18 = x = 9 or

11 Triangle Parts ( 7-5) 1.5 Special Segments and Perimeter If two triangles are similar, then their perimeters have the same ratio as the scale factor. If two triangles are similar, then all the corresponding special segments such as medians, altitudes, and angle bisectors have the same ratio as the scale factor. 1.6 Angle Bisector Theorem M 4 P A E L If P E bisects AP L then AP P L = AE EL. 11

12 Right Triangles ( 8-1) 1.7 Geometric Mean in Right Triangles If we draw the altitude to the hypotenuse of a right triangle, we have two new triangles that are similar to the original triangle and to each other. B a c 1 D c h c 2 C b A ABC ACD CBD This can be shown to be the result of the AA Postulate since they each share an angle and are all right triangles. Now, with the corresponding sides being proportional, we obtain some very interesting results. ABC ACD c 2 b = b c = h a ABC CBD c 1 a = a c = h b 12

13 1.8 Pythagorean Theorem We can now prove the Pythagorean Theorem using the previous proportions. First, we will summarize those results as follows. In a right triangle with sides a, b, and c, when the altitude (t) is drawn to the hypotenuse (c with segments c 1 and c 2 ), then the following two statements hold. Each leg is the geometric mean between the hypotnuse and the segment of the hypotenuse adjacent (connected) to (or shadowing) that leg. c 1 a = a c and c 2 b = b c The altitude (height to the hypotenuse) is the geometric mean between the the segments of the hypotenuse. (1) c 1 t = t c 2 (2) Keep in mind the other very important but obvious fact that: c = c 1 + c 2. (3) Proof. From equation (1) above, we have two key equations: Thus, a 2 + b 2 Therefore, a 2 + b 2 = c 2. a 2 = c 1 c and b 2 = c 2 c. = c 1 c + c 2 c by substitution = c(c 1 + c 2 ) by factoring = c(c) by substitution from (3) 13

14 Special Right Triangle ( 8-2) 2 Solving Right Triangles To solve a triangle in geometry means to find the length of all its sides and the measures of all its angles. This involves a specialized topic in geometry known as Trigonometry. Definition. The roots of Trigonometry are trigon meaning triangle and metr meaning measurement. Together, it means to measure triangles or measure using triangles. 2.1 Special Right Triangles There are two special right triangles. They are special because we can use a simple pattern in the relation of their sides to their angles to solve them Isosceles Right or We consider the diagonal of the square ABCD below. Each side has length r. Use the Pythagorean Theorem to find the length of the diagonal. d 2 = r 2 + r 2 = d 2 = 2r 2 = d = 2r 2. Since r 2 = r, the length of the diagonal is d = r 2. 14

15 We consider an equilateral triangle DEF. Draw the altitude ET. Now suppose each side of the triangle is 2t. Then, since the altitude divides DEF into two congruent s, DT = F T = t. We have two sides of a right triangle, so we use the Pythagorean Theorem again to find the length of the altitude. t 2 +a 2 = (2t) 2 = a 2 = 4t 2 t 2 = a 2 = 3t 2 = a = 3t 2. Since t 2 = t, the length of the altitude is a = t 3. 15

16 Trigonometric Ratios ( 8-3) 2.2 SOH-CAH-TOA There are three trigonometric ratios relating the sides of the right triangle: sine, cosine, and tangent. The side opposite to an angle is the side that does not contain either of the rays that make up that angle. Angle Opposite Side Length B A BC a B AC b c a C AB c A b C sin A = opposite hypotenuse = a adjacent, cos A = c hypotenuse = b c, opposite tan A = adjacent = a b Finding Side Length Set up the trigonometric ratio by looking at the angle and notice whether the given side and the unknown side are opposite, or adjacent to the angle, or the hypotenuse of the right Finding Angle Measure 16

17 Trigonometric Application Problems ( 8-4) 2.3 Application Problems Angles of Elevation. Angles of Depression. More Examples. 17

18 Law of Sines ( 8-5) 2.4 Non-Right Triangles Law of Sines When we don t have a right triangle, we can sometimes use the law of sines to solve a triangle. B c sin A a = sin B b = sin C c A b C a Law of Sines Special Cases Finding Area in Non-Right s 18

19 Law of Cosines ( 8-6) Law of Cosines Heron s Formula 19

20 Diameter and Circumference ( 9-1) 3 Circles A circle is not a polygon. It is not made of lines, but of arcs. 3.1 Basic Definitions A circle is defined as the set of all points equidistant from a fixed point in space. The fixed point is called the center of the circle. The distance from the fixed point to any point on the circle is called the radius. Such a segment is also called the radius. A chord is a segment with endpoints on the circle. The diameter is a special chord that contains the center and its length is twice the radius. We name circles based on their center. Below is C. radius C diameter 20

21 3.2 Circumference The distance around the circle is called the circumference. This is equivalent to the perimeter of a polygon. The ratio of the circumference to the diameter, is a special number called pi (pronounce pie ). C d = π So we can write a formula for circumference of a circle as follows. C = πd Since d = 2r, we could also write C = 2rπ or C = 2πr, however, the formula above with the diameter has two advantages: (1) it shows clearly how π is related to the parts of a circle, and (2) it avoids the confusion with the formula for area (A = πr 2 ). Example. Find the exact circumference for a circle with radius 8 cm (NB: the word exact means pi is not to be approximated). Solution. Since r = 8, the diameter is d = 16. Thus, C = 16π. Example. Approximate the radius of a circle with C = 25 m. Solution. Since C = 25, we can write πd = 25 Thus, r = d/ meters. d = 25/π d

22 Arcs and Angles ( 9-2) 3.3 Arcs and Central Angles Recall that circles are made of arcs. between arcs and angles. We now study the relation Definition. A central angle is an angle with its vertex at the center of the circle and its sides are radii (plural for radius). In C below, BCD is a central angle. A E C D B What is AE called? What is AD called? What is BC called? Theorem. The sum of the measures of all adjacent central angles in a circle is

23 Definition. A semicircle is an arc that covers half of the circle. Arc Type Naming Measurement Minor Arc smaller than semicircle Major Arc larger than semicircle Semicircle half of the circle Arc Addition Postulate. Two letters: ȦB equals central angle Three BDA letters: Two or Three letters: AD or ABD equals 360 minus central angle equals 180 If ĖA and ȦB are adjacent, then m EAB = mėa+mȧb. Arc Length is the distance around an arc and depends on the circumference of the circle. Example. Suppose ACB = 120 and the radius of C is 9. Find the length of ȦB. Solution. The measure of the minor arc equals the central angle, so mȧb = 120. First we find how much of the circle is covered by the arc: = 1 3. Next, we find the circumference: d = 2 9 = 18, thus C = 18π. Therefore, the length of ȦB is 1 18π = 6π or approximately

24 Arcs and Chords ( 9-3) 3.4 More Definitions & Theorems Concentric circles are circles that have the same center. Congruent circles are circles with equal radii. In the same circle or in two congruent circles, two arcs with equal measures are congruent arcs. A minor arc that shares the same endpoints as a chord is called the arc of the chord. Congruent Arcs Theorem. Two minor arcs are congruent if and only if their chords are congruent. E N K R O Example. In K, OR = 8 units. Find NE. 24

25 Perpendicular Diameter Theorem. If the diameter of a circle is perpendicular to a chord, then it bisects the chord and its arc. D E C A F B Proof. Draw the radii CE and CB. Notice that two right triangles are formed with equal legs (CA = CA) and hypotenuse (CE = CB). Thus, CAE = CAB by HL Theorem. EA = BA and ACE = ACB by definition of = s. As a result, ĖF = ḂF, showing that ĖB is also bisected. Congruent Chords Theorem. Two chords in a circle (or in two = s) are congruent if and only if they are equidistant from the center. 25

26 Inscribed Angles ( 9-4) 3.5 Inscribed Angles An inscribed angle is an angle with its vertex on the circle. In C, since A is a point on the circle, EAD is an inscribed angle. A E C D Theorem. The measure of an inscribed angle is half the measure of its intercepted arc: m EAD = 1 2 (mėd). Suppose mėd = 70. What is m DCE (a central angle)? Recall that ACE is isosceles, so CEA = EAC. Since DCE is an exterior angle to ACE, then m DCE = m CEA + m EAC = 2(m EAC). This means m EAC is half of m DCE or equivalently half of mėd, which is 1 2 (70) =

27 Example. Let m 1 = 4x 7 and m 2 = 2x Find x. Solution. Since 1 and 2 both intercept ḊC, then m 1 = 1 (mḋc) = m 2. 2 Substituting, we find 4x 7 = 2x + 11 = x = 9. Theorem. Any angle inscribed in a semicircle is a right angle. M I S E Proof. Since ĔI is a semicircle in S, the mĕi = 180. We notice that EMI is an inscribed angle intercepting ĔI. Thus, m EMI = 1 2 (mĕi). Therefore, m EMI =

28 Inscribed Polygons ( 9-4) Inscribed Polygons A polygon with its vertices on the circle with its sides as chords is an inscribed polygon. Alternately, we can say that the circle circumscribes the polygon. Theorem. If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Corollary. Any square inscribed in a circle is unique. Corollary. If the side of a square inscribed in a circle is s units, then the radius of the circle is r = s 2. 28

29 Tangents of a Circle ( 9-5) 3.6 Tangents A tangent of a circle is a line in the exterior of a circle that intersects it at a single point, called the point of tangency. S T Theorem. A line is tangent to a circle if and only if it is perpendicular to the radius at the point of tangency. Notice that any radii extended outside the circle will be longer that ST. Since ST is the shortest distance from the center to the tangent line, it must be perpendicular. 29

30 Theorem. If two segments from one exterior point to a circle are both tangents to the circle, then the segments are congruent. S E A M Proof. The radii SA and SE are perpendicular to the tangents at the point of tangency. Since SAM = SEM (why?), we conclude that MA = ME.. Example. in S, let SA = 6 and SM = 14. Find the exact length of ME. Solution. Since SA AM, we use Pythagorean Theorem (SA) 2 + (AM) 2 = (SM) 2 = AM = 160. Since the tangents are from the one point, AM = EM =

31 Secants and Angles ( 9-6) 3.7 Secants A secant to a circle is a line that intersects the circle at two points. Therefore, it contains a chord of a circle. Below is a graphic summary of the important relations between angles and arcs of circles. a a a x x x b x = 1 2 a x = 1 2 a x = 1 (a + b) 2 a x b x b a a b x x = 1 2 (a b) x = 1 2 (a b) x = 1 (a b) 2 31

32 Secant Segments ( 9-7) Segments of Secants The segments that are formed by intersecting secants are called secant segments. We already know that the segments of tangents to a circle from the same point are congruent. Now we consider the relations between the segments formed by secants to a circle. a c a d b a b c d c d a b = c d a (a + b) = c (c + d) a 2 = c (c + d) 32

33 Equation of Circle ( 9-8) 3.8 Graphing Circles The equation of a circle describes the relation between the center (h, k) and any point on the circle (x, y). Using the distance formula, we find the radius to be (x h)2 + (y k) 2 = r. This is true for any point (x, y) on the circle because of the definition of the circle (cf: 3.1 in notes). Squaring both sides we have the equation below. (x h) 2 + (y k) 2 = r 2 Example. Find the equation of a circle with center (3, 7) and radius 5. Solution. Just substitute the values into the equation above. Let h = 3, k = 7, r = 5 = r 2 = 25 : (x 3) 2 + (y + 7) 2 =

34 Area of Triangle and Quads ( 10-3/4) 4 Area Area is the measure of the quantity that can be placed in the interior of a closed shape. It is measured using square units; that is, we find how many unit squares (1 1 squares) can be placed inside the shape. It is different from perimeter which only measures length. The altitude of a shape is important in finding area because we use squares as our basic unit. Recall that an altitude is the segment from a vertex which is perpendicular to the opposite side. The length of the altitude is called the height and the opposite side the base of the shape. Theorem. If two shapes are congruent, then they have equal areas. 4.1 Areas of Basic Shapes Shape Rectangle Triangle Parallelogram Area A = bh A = 1 2 bh A = bh Rhombus A = 1 2 d 1d 2 Trapezoid A = 1 2 (b 1 + b 2 )h 34

35 Polygons and Regular Polygon Angles ( 10-1) 4.2 Interior Angles of Polygons Recall that a polygon is (1) a closed two dimensional shape, having (2) its sides made of line segments connected by their endpoints known as vertices, and (3) the sides are not curved, do not overlap nor intersect each other. Polygons with more than 4 sides are named using Greek prefixes added on to the suffix -gon, meaning sides. So, polygon means many sides. penta = 5, hexa = 6, hepta = 7, octa = 8, nona = 9, deca = 10. We often refer to polygons as n-gons indicating that n stands for the number of sides. To find the sum of interior angles for any n-gon, we use the fact that in a, the sum is 180. Consider the pentagon below. Drawing all the diagonals from a single vertex results in 3 triangles. Since each has a total sum of 180 and no angles overlap, we conclude the sum of the interior angles in a pentagon is =

36 We can generalize this as follows for n 3. Theorem. The sum of the interior angles for any n-gon is: (n 2) Regular Polygons A regular polygon is any polygon which is equilateral and equiangular. Two examples are equilateral triangles and a square. Since all angles are congruent, we can find each interior angle by dividing the formula above by the number of sides (n). each interior for regular n-gon = (n 2) 180 n Recall that exterior angles are supplementary to interior angles. Let x be the exterior angle measure. Then, (n 2) 180 x + = 180 n 180n 360 x + = 180 n x n = 180 x = 360 n. 36

37 Area of Regular Polygons ( 10-5) 4.3 Area of Regular Polygons We inscribe a regular polygon in a circle as shown below. The sides of the polygon are the chords of the circle and they intercept equal arcs. The perpendicular segment from the center to the sides of the polygon (the circle s chord) is called the apothem (ap e θ im).! radius apothem By drawing the radii (dotted lines) to each vertex of the polygon, we form n congruent triangles (in the above case 8 s). As a result, the measure of γ (gamma) is half the measure of each interior angle of the regular polygon. ((1/2n)(n 2)(180)) To find the area, we define the following variables: 37

38 a = apothem n = number of sides l = length of polygon side γ = 1/2 of interior r = radius P = perimeter A = area of polygon A = area of central The following steps lead us to a formula and can be used to find the area of a regular polygon. (1) To find the area of one central triangle, we must know l, one side of the polygon and a, the apothem, which are its base and height. (2) The following relations will help find these values as long as one of them is known. tan γ = a 1 2 l, sin γ = a r, cos γ = 1 2 l r. (3) Thus, after finding l and a, find the area of: A = 1 2 la. (4) We have n of these triangles, so the area of the polygon is : A = n Ñ 1 2 la é = A = 1 2 nla ; And because nl = P, we conclude that the area is A = 1 2 P a. Summary. Find the apothem. Find the length of one side and multiply by n to get the perimeter. Use the formula above to get the area of a regular polygon. 38

39 Areas of Circles & Sectors ( 10-5) 4.4 Circle Area If we consider a regular polygon inscribed in a circle and gradually increase the number of sides, the polygon begins to take the shape of a circle. Using the formulas used to find area of a regular polygon, we notice: a = r sin γ, l = 2r cos γ, P = 2rn cos γ. Thus, the area formula can be written as A = (1/2)(2rn cos γ)(r sin γ), which simplifies to: A = r 2 n cos γ sin γ. As we increase n, the value of n cos γ sin γ approaches or gets closer and closer to π. Therefore, the area of the polygon approaches: A = πr 2. A sector is a slice of the circle made by two radii. To find its area, we need to know the measure of the central angle (or the intercepted arc). The process is similar to finding arc length, so we have A sector = central 360 πr 2. 39

40 Geometric Probability ( 10-6) 4.5 Probability The probability of an event is a ratio of the expected number of outcomes to the toal possible outcomes: Geometric: Line # expected outcomes # total possible outcomes. The probability of selecting points on a segment (MN) contained in a larger segment (AB), given that we randomly select any point from the larger segment is the ratio of their lengths: Geometric: Area P = MN AB. The probability of selecting points in a particular shape (shaded region) contained in a larger shape, given that we randomly select any point from the larger shape is the ratio of their areas: P = A shaded A total. 40

41 Area Ratios (? -?) 4.6 Areas of Similar Shapes Recall that when two shapes are similar, the ratio of the lengths of their sides determines the scale factor of enlargement or reduction. This ratio does not give the correct comparisons for their areas, since area is two dimensions. Theorem. If the scale factor (ratio of their side lengths or perimeters) of two similar shapes is given as the ratio a : b, then the ratio of their areas is a 2 : b 2. Note: If the ratio is given as a decimal r, then the ratio of their areas is simply r 2. 41

42 Transformations in a Plane ( 13-4,6/12-5) 5 Mappings and Transformations A mapping is when two quantities (numbers) are associated (related) by some sort of rule. Here are some examples you have seen before. When you graph a line, you are associating each x coordinate with a unique y coordinate based on the rule for the line. Corresponding parts of congruent triangles are also examples of mappings. because the parts are associated by the rule of congruence. 5.1 Transformations A transformation in a plane is a type of mapping. The rule for the mapping tells us how to transform any given coordinate (known as the pre-image) to a new coordinate (known as the image). The mapping rule can be in words or may be algebraic. A transformation can be of two types. (1) isometric (keeps its shape but changes) (2) non-isometric (smilarity, changes the) {size position orientation 42

43 Isometries are: Translation, Reflection, and Rotation; while Dilation is a similariy transformation. (1) Translation or slide only changes the position of points to obtain their image. (2) Relfection or flip changes the position and orientation by using a line as a mirror to reflect the point to its image point. (3) Rotation or turn also changes position and most possibly the orientation of the pre-image by turning about a fixed point for a given number of degrees. (4) Dilation results in a change in distance between points, thus enlarging or reducing the shape of the pre-image Translation Example Suppose we are given the mapping rule (x, y) (x + 3, y 8) along with the coordinates for a quadrilateral RAIN as follows: R(3, 9), A(6, 7), I(7, 3), and N(1, 4). Problem: Find the coordinates of the image and graph. Solution: The image coordinates are R (6, 1), A (9, 1), I (10, 5), and N (4, 4). Notice that the image was translated or moved to a new position. The vector 3, 8 can also be used to describe the movement of points. We will discuss vectors later. 43 O y N R N R A I A I x

44 Definition. A vector is a mathematical object with two scalar quantities (direction and magnitude) and (in 2 dimensions it has) two components (horizontal: x, and vertical: y) associated to it. We name vectors by their initial and terminal points: AB. Example. Initial point A(4, 1) and terminal point B( 1, 3): AB = 1 4, 3 1 = 5, 2. Which means to get from A to B, we go 5 in the horizontal direction and 2 in the vertical direction. We can use the distance formula to find the vector s magnitude. AB = 5, 2 = ( 5) = We can also add vectors by their components. Example. Let BC = 3, 1. We can find the sum as follows: AB + BC = 5, 2 + 3, 1 = 5 + 3, 2 + ( 1) This new vector is called the resultant: = 2, 1. AC. Theorem. Given two vectors u = a, b and v = c, d, u = v if a = c and b = d (same direction and magnitude) u b v if a = d c (same slope only) 44

45 Reflections ( 13-5) Reflection Examples 45

46 Polyhedrons ( 11-1) 6 General Solid Figures A shape that is three dimensional is refered to as a solid in Geometry. The most common type of solids are polyhedrons. Polyhedrons are closed solids with many faces that meet at the edges and vertces. Faces of a polyhedron make up the surface, which are made of polygons. Edges are the sides of polygons that meet to close the shape Vertices are the points where 3 ormore edges meet. Platonic Solids are regular polyhedrons that have a single regular polygon for all of their faces. Prisms and Pyramids are special polyhedrons. Naming can be done either by using the number of faces, or the type of base that a special polyhedrons may have. Theorem. Euler s Polyhedron Formula states that for any polyhedron where F, V, and E represent the numbers of Faces, Vertice, and Edges respectively, then F + V E = 2. 46

47 Visualizing 3-D ( 11-1/2) 6.1 Isometric Views Isometric drawings are also called perspective drawings. We look at a three dimensional object from a single persepctive and the draw the shape in two dimensions. We generally represent the Top, Front, and Right views. TOP LEFT RIGHT FRONT Isometric Drawing Othogonal Views We can also use an orthogonal drawing to represent the two dimensional views of each side separately. For example, we can have Top, Left, Front, and Right views in two dimensional for greater detail. 6.2 Surface Area The surface area of a solid is the sum of the areas for each part of the surface. In polyhedrons, the surface area is the sum of areas for all the polygons that make up the faces of the polyhedron. 47

48 One way to approach the problem is to make a net: a two dimensional representation of the solid when it is cut open and laid flat. Below is an exmple of a pentagonal pyramid and its net. 48

49 Surface Area of Prism and Cylinder ( 11-3) Prism A prism is a polyhedron with two congruent faces called its bases. Its lateral faces are parallelograms. In a right prism, the lateral faces form right angles with the base. In a regular prism, the bases are regular polygons. The lateral area (LA) is the sum of the areas of all the lateral faces. We can do this quickly by using the perimeter (P ) of the base. The base area (B) is the area of one base. It is not necessary to have any additional formulas other than the area formulas we learned before, but it may be more efficient to use the following formulas. Let h stand for the height of the prism which can also be called the length of the lateral faces in a right prism. LA = P h, SA = P h + 2B, i.e. LA + two Base areas Cylinder A cylinder is not a polyhedron but has a circular base. The formulas are similar except that for P we use C = 2πr, and for B we use πr 2. 49

50 Surface Area of Pyramid and Cone ( 11-4) Pyramid A pyramid is a polyhedron with a single base. Its lateral faces are triangles that meet at a single point (the vertex). The slant height of a pyramid is the height of the lateral faces. The height of a pyramid is the perpendicular distance from the vertex to the base In a regular pyramid, the base is a regular polygon. The lateral area (LA) is the sum of the areas of all the lateral faces. We can do this quickly by using the perimeter (P ) of the base and the slant height. The base area (B) is the area of the base. Let l stand for the slant height. LA = 1 2 P l, SA = 1 P l + B, i.e. LA + the Base area Cone A cone is not a polyhedron but has a circular base. As in the cylinder, for P we use C = 2πr, and for B we use πr 2. Thus, LA = πrl and SA = πrl + πr 2. 50

51 Volume ( 11-5/6) 6.3 Volume The volume of a solid is the amount of space inside the solid. We measure volume in cubic units. Cavalieri s Principle. Two solid objects that have the same cross sectional area have equal volumes as long as they have equal heights Right Prisms and Circular Cylinders B stands for area of the base; in a cylinder, B = πr 2. V = B h Right Pyramids and Circular Cones B stands for area of the base; in a cone, B = πr 2. V = 1 3 B h. 51

52 Spheres ( 11-7) 6.4 Spheres Definition. A sphere is a perfectly round 3 dimensional solid, also known as a ball. The radius of a sphere is the distance from the center to any point on the sphere. The Great Circle is the circle which has the same center and radius as the sphere. It divides the sphere into two halves called hemispheres Surface Area and Volume Archimedes showed that we can find the surface area of a sphere by using the lateral area of a cylinder with the same radius and twice the height. SA = 4πr 2. We can find the volume by extracting small pyramids with vertex at the center of the sphere. Since the height of the pyramids will equal the sphere s radius and the the sum of the base areas of the small pyramids will equal the surface area, we can find the volume. V = 4 3 πr3. 52