Activator Solve the proportion: 4/ (x+2) = 16/(x + 5) Simplify: 5weeks/30days; 85cm/.5m

UNIT E.Q How can I use all what I am learning on polygon similarity to solve real life problems. Which careers use similarity most? Is similarity shown and used only in geometry?

Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others.   4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning.

8.1 Ratio and Proportion

Essential Question: 1. How do we use proportions to solve problems? 2.How do we use properties of proportions to solve real- life problems, such as using the scale of a map?

Computing Ratios Ratio of a to b : Ex: Simplify the ratio of 6 to 8.

Simplifying Ratios Simplify the ratios.

The perimeter of rectangle ABCD is 60 cm. The ratio of AB:BC is 3:2 The perimeter of rectangle ABCD is 60 cm. The ratio of AB:BC is 3:2. Find the length and width of the rectangle. Hint: draw a rectangle ABCD.

Using Extended Ratios The measure of the angles in ∆JKL are in the extended ratio of 1:2:3. Find the measures of the angles. 2x 3x x

Using Ratios The ratios of the side lengths of ∆DEF to the corresponding side lengths of ∆ABC are 2:1. Find the unknown lengths. A D 3 L B C L F E 8

Properties of proportions Cross product property then ad = bc. Reciprocal property Introduce means and extremes

Solve the proportions.

Given that the track team won 8 meets and lost 2, find the ratios. What is the ratio of wins to loses? What is the ratio of losses to wins? What is the ratio of wins to the total number of track meets?

Simplify the ratio. 3 ft to 12 in 60 cm to 1 m 350g to 1 kg 6 meters to 9 meters

Solve the proportion

Tell whether the statement is true.

8.2 Problem Solving in Geometry with Proportions

Additional Properties of Proportions If , then If , then

Using Properties of Proportions Tell whether the statement is true.

In the diagram . Find the length of BD. A 30 16 B C x 10 D E

In the diagram Solve for DE. A 5 2 D B 3 E C

Geometric mean The geometric mean of two positive numbers a and b is the positive number x such that Find the geometric mean of 8 and 18.

Geometric Mean Find the geometric mean of 5 and 20. The geometric mean of x and 5 is 15. Find the value of x.

Different perspective of Geometric mean The geometric mean of ‘a’ and ‘b’ is √ab Therefore geometric mean of 4 and 9 is 6, since √(4)(9) = √36 = 6.

Geometric mean Find the geometric mean of the two numbers. 3 and 27 √(3)(27) = √81 = 9 4 and 16 √(4)(16) = √64 = 8 5 and 15 √(5)(15) = √75 = 5√3

Ex. 3: Using a geometric mean PAPER SIZES. International standard paper sizes are commonly used all over the world. The various sizes all have the same width-to-length ratios. Two sizes of paper are shown, called A4 and A3. The distance labeled x is the geometric mean of 210 mm and 420 mm. Find the value of x. Slide #26

Write proportion 210 x = x 420 X2 = 210 ∙ 420 X = √210 ∙ 420 Solution: The geometric mean of 210 and 420 is 210√2, or about 297mm. 210 x Write proportion = x 420 X2 = 210 ∙ 420 X = √210 ∙ 420 X = √210 ∙ 210 ∙ 2 X = 210√2 Cross product property Simplify Factor Simplify Slide #27

Using proportions in real life In general when solving word problems that involve proportions, there is more than one correct way to set up the proportion. Slide #28

Ex. 4: Solving a proportion MODEL BUILDING. A scale model of the Titanic is 107.5 inches long and 11.25 inches wide. The Titanic itself was 882.75 feet long. How wide was it? Width of Titanic Length of Titanic = Width of model Length of model LABELS: Width of Titanic = x Width of model ship = 11.25 in Length of Titanic = 882.75 feet Length of model ship = 107.5 in. Slide #29

Reasoning: Write the proportion. Substitute. Multiply each side by 11.25. Use a calculator. Width of Titanic Length of Titanic = Width of model Length of model x feet 882.75 feet = 11.25 in. 107.5 in. 11.25(882.75) x = 107.5 in. x ≈ 92.4 feet So, the Titanic was about 92.4 feet wide. Slide #30

Note: Notice that the proportion in this Example contains measurements that are not in the same units. When writing a proportion in unlike units, the numerators should have the same units and the denominators should have the same units. The inches (units) cross out when you cross multiply. Slide #31

A model truck is 13. 5 inches long and 7. 5 inches wide A model truck is 13.5 inches long and 7.5 inches wide. The original truck was 12 feet long. How wide was it?

8.3 Similar Polygons

Activator

Identifying Similar Polygons When there is a correspondence between two polygons such that their corresponding angles are congruent and the lengths of corresponding sides are proportional the two polygons are called similar polygons.

Similar polygons If ABCD ~ EFGH, then G H C D B F A E

Similar polygons Given ABCD ~ EFGH, solve for x. 2x = 24 x = 12 6 x 2

Is ABC ~ DEF? Explain. B D 6 E 13 12 5 7 A C F 10 ABC is not similar to DEF since corresponding sides are not proportional. ? ? yes no

Similar polygons Given ABCD ~ EFGH, solve for the variables. 5 x 10 2 6 A E

Ex: Scale factor of this triangle is 1:2 If two polygons are similar, then the ratio of the lengths of two corresponding sides is called the scale factor. Ex: Scale factor of this triangle is 1:2 4.5 9 3 6

Quadrilateral JKLM is similar to PQRS. Find the value of z. 15 Q z 6 P J M 10 15z = 60 z = 4

Theorem If two polygons are similar, then the ratio of their perimeters is equal to the ratios of their corresponding side lengths. If KLMN ~ PQRS, then

Given ABC ~ DEF, find the scale factor of ABC to DEF and find the perimeter of each polygon. = 40 B P = 4 + 6 + 10 = 20 12 20 6 10 A C D F 4 8 CORRESPONDING SIDES 4 : 8 1 : 2

8.4 Similar Triangles

In the diagram, ∆BTW ~ ∆ETC. Write the statement of proportionality. Find m<TEC. Find ET and BE. T 34° E C 3 20 79° B W 12

Postulate 25 Angle-Angle (AA) Similarity Postulate If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar.

Similar Triangles Given the triangles are similar. Find the value of the variable. )) m )) ) 6 8 11m = 48 ) 11

Similar Triangles Given the triangles are similar. Find the value of the variable. Left side of sm Δ Base of sm Δ Left side of lg Δ Base of lg Δ = 6 5 > 2 6h = 40 > h

∆ABC ≈ ∆DBE. A 5 D y 9 x B C 8 E 4

Determine whether the triangles are similar. 6 32° 33° 9 18 No, because two angles of one triangle are not congruent to two angles of another triangle.

Determine whether the triangles are similar. 60° 60° 60° 60° Yes, because two angles of one triangle are congruent to two angles of another triangle.

Given two triangles are similar, solve for the variables. 14 16 15 ) ) 10 15(a+3) = 10(16) 15a + 45 = 160 15a = 115

Decide whether two triangles are similar, not similar, or cannot be determined. 92° 31° 47° S 41° 92° 57° S + 92 + 41 = 180 S + 133 = 180 S = 47 A + 92 + 57 = 180 A + 149 = 180 A = 31 Not similar

8.5 Proving Triangles are Similar

Objectives: Use similarity theorems to prove that two triangles are similar Use similar triangles to solve real-life problems such as finding the height of a climbing wall.

Using Similarity Theorems In this lesson, you will study 2 alternate ways of proving that two triangles are similar: Side-Side-Side Similarity Theorem and the Side-Angle-Side Similarity Theorem. The first theorem is proved in Example 1 and you are asked to prove the second in Exercise 31.

Side Side Side(SSS) Similarity Theorem If the corresponding sides of two triangles are proportional, then the triangles are similar. THEN ∆ABC ~ ∆PQR AB BC CA = = PQ QR RP

Side Angle Side Similarity Thm. If an angle of one triangle is congruent to an angle of a second triangle and the lengths of the sides including these angles are proportional, then the triangles are similar. ZX XY If X  M and = PM MN THEN ∆XYZ ~ ∆MNP

Ex. 1: Proof of Theorem 8.2 Given: Prove RS ST TR ∆RST ~ ∆LMN = = LM MN NL Locate P on RS so that PS = LM. Draw PQ so that PQ ║ RT. Then ∆RST ~ ∆PSQ, by the AA Similarity Postulate, and RS ST TR = = LM MN NL Because PS = LM, you can substitute in the given proportion and find that SQ = MN and QP = NL. By the SSS Congruence Theorem, it follows that ∆PSQ  ∆LMN Finally, use the definition of congruent triangles and the AA Similarity Postulate to conclude that ∆RST ~ ∆LMN.

Ex. 2: Using the SSS Similarity Thm. Which of the three triangles are similar? To decide which, if any, of the triangles are similar, you need to consider the ratios of the lengths of corresponding sides. Ratios of Side Lengths of ∆ABC and ∆DEF. AB 6 3 CA 12 3 BC 9 3 = = = = = = DE 4 2 FD 8 2 EF 6 2 Because all of the ratios are equal, ∆ABC ~ ∆DEF.

Ratios of Side Lengths of ∆ABC ~ ∆GHJ 6 CA 12 6 BC 9 = = 1 = = = GH 6 JG 14 7 HJ 10 Because the ratios are not equal, ∆ABC and ∆GHJ are not similar. Since ∆ABC is similar to ∆DEF and ∆ABC is not similar to ∆GHJ, ∆DEF is not similar to ∆GHJ.

Ex. 3: Using the SAS Similarity Thm. Use the given lengths to prove that ∆RST ~ ∆PSQ. Given: SP=4, PR = 12, SQ = 5, and QT = 15; Prove: ∆RST ~ ∆PSQ Use the SAS Similarity Theorem. Begin by finding the ratios of the lengths of the corresponding sides. SR SP + PR 4 + 12 16 = 4 = = = SP SP 4 4

ST SQ + QT 5 + 15 20 = 4 = = = SQ SQ 5 5 So, the side lengths SR and ST are proportional to the corresponding side lengths of ∆PSQ. Because S is the included angle in both triangles, use the SAS Similarity Theorem to conclude that ∆RST ~ ∆PSQ.

Using Similar Triangles in Real Life Ex. 6 – Finding Distance indirectly. To measure the width of a river, you use a surveying technique, as shown in the diagram.

Solution By the AA Similarity Postulate, ∆PQR ~ ∆STR. RQ PQ = Write the proportion. RT ST RQ 63 = Substitute. 12 9 RQ = 12 ● 7 Multiply each side by 12. RQ = 84 Solve for TS. So the river is 84 feet wide.

8.6 Proportions & Similar Triangles

Objectives: Use proportionality theorems to calculate segment lengths. To solve real-life problems, such as determining the dimensions of a piece of land.

Use Proportionality Theorems In this lesson, you will study four proportionality theorems. Similar triangles are used to prove each theorem.

Theorems 8.4 Triangle Proportionality Theorem If a line parallel to one side of a triangle intersects the other two sides, then it divides the two side proportionally. If TU ║ QS, then RT RU = TQ US

Theorems 8.5 Converse of the Triangle Proportionality Theorem If a line divides two sides of a triangle proportionally, then it is parallel to the third side. RT RU If , then TU ║ QS. = TQ US

Ex. 1: Finding the length of a segment In the diagram AB ║ ED, BD = 8, DC = 4, and AE = 12. What is the length of EC?

Triangle Proportionality Thm. Step: DC EC BD AE 4 EC 8 12 4(12) 8 6 = EC Reason Triangle Proportionality Thm. Substitute Multiply each side by 12. Simplify. = = EC = So, the length of EC is 6.

Ex. 2: Determining Parallels Given the diagram, determine whether MN ║ GH. LM 56 8 = = MG 21 3 LN 48 3 = = NH 16 1 8 3 ≠ 3 1 MN is not parallel to GH.

Theorem 8.6 If three parallel lines intersect two transversals, then they divide the transversals proportionally. If r ║ s and s║ t and l and m intersect, r, s, and t, then UW VX = WY XZ

Theorem 8.7 If a ray bisects an angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the lengths of the other two sides. If CD bisects ACB, then AD CA = DB CB

Ex. 3: Using Proportionality Theorems In the diagram 1  2  3, and PQ = 9, QR = 15, and ST = 11. What is the length of TU?

9 ● TU = 15 ● 11 Cross Product property SOLUTION: Because corresponding angles are congruent, the lines are parallel and you can use Theorem 8.6 PQ ST Parallel lines divide transversals proportionally. = QR TU 9 11 = Substitute 15 TU 9 ● TU = 15 ● 11 Cross Product property 15(11) 55 TU = = Divide each side by 9 and simplify. 9 3 So, the length of TU is 55/3 or 18 1/3.

Ex. 4: Using the Proportionality Theorem In the diagram, CAD  DAB. Use the given side lengths to find the length of DC.

Solution: Since AD is an angle bisector of CAB, you can apply Theorem 8.7. Let x = DC. Then BD = 14 – x. AB BD = Apply Thm. 8.7 AC DC 9 14-X Substitute. = 15 X

Ex. 4 Continued . . . 9 ● x = 15 (14 – x) Cross product property Distributive Property Add 15x to each side Divide each side by 24. So, the length of DC is 8.75 units.

Use proportionality Theorems in Real Life Example 5: Finding the length of a segment Building Construction: You are insulating your attic, as shown. The vertical 2 x 4 studs are evenly spaced. Explain why the diagonal cuts at the tops of the strips of insulation should have the same length.

Use proportionality Theorems in Real Life Because the studs AD, BE and CF are each vertical, you know they are parallel to each other. Using Theorem 8.6, you can conclude that DE AB = EF BC Because the studs are evenly spaced, you know that DE = EF. So you can conclude that AB = BC, which means that the diagonal cuts at the tops of the strips have the same lengths.

Ex. 6: Finding Segment Lengths In the diagram KL ║ MN. Find the values of the variables.

Solution To find the value of x, you can set up a proportion. 9 37.5 - x = Write the proportion Cross product property Distributive property Add 13.5x to each side. Divide each side by 22.5 13.5 x 13.5(37.5 – x) = 9x 506.25 – 13.5x = 9x 506.25 = 22.5 x 22.5 = x Since KL ║MN, ∆JKL ~ ∆JMN and JK KL = JM MN

Solution To find the value of y, you can set up a proportion. 9 7.5 = Write the proportion Cross product property Divide each side by 9. 13.5 + 9 y 9y = 7.5(22.5) y = 18.75

8.7 Dilations

Objectives Identify dilations Use properties of dilations to create a real-life perspective drawing.

Identifying Dilations In chapter 7, you studied rigid transformations, in which the image and preimage of a figure are congruent. In this lesson, you will study a type of nonrigid transformation called a dilation, in which the image and preimage are similar.

What is it? A dilation with center C and a scale factor k is a transformation that maps every point P in the plane to a point P’ so that the following properties are true. If P is not the center point C, then the image point P’ lies on CP. The scale factor k is a positive number such that k = and k ≠1. 2. If P is the center point C, then P = P’. CP’ CP

Reduction/Enlargement The dilation is a reduction if 0 < k < 1 and it is an enlargement if k > 1. CP’ 3 1 REDUCTION: = = CP 6 2 6

Is equal to the scale factor of the dilation. CP’ 5 = ENLARGEMENT: CP 2 5 Because ∆PQR ~ ∆P’Q’R’ P’Q’ Is equal to the scale factor of the dilation. PQ

Ex. 1: Identifying Dilations Identify the dilation and find its scale factor. CP’ 2 = REDUCTION: CP 3 2 The scale factor is k = This is a reduction. 3

Ex. 1B -- Enlargement Identify the dilation and find its scale factor. CP’ 2 = 2 = ENLARGEMENT: CP 1 2 The scale factor is k = This is an enlargement. = 2 1

Notes: In a coordinate plane, dilations whose centers are the origin have the property that the image of P (x, y) is P’ (kx, ky)

Ex. 2: Dilation in a coordinate plane Draw a dilation of rectangle ABCD with A(2, 2), B(6, 2), C(6, 4), and D(2, 4). Use the origin as the center and use a scale factor of ½. How does the perimeter of the preimage compare to the perimeter of the image?

SOLUTION: Because the center of the dilation is the origin, you can find the image of each vertex by multiplying is coordinates by the scale factor A(2, 2) A’(1, 1) B(6, 2) B’(3, 1) C(6, 4) C’(3, 2) D(2, 4) D’(1, 2)

Solution continued From the graph, you can see that the preimage has a perimeter of 12 and the image has a perimeter of 6. A preimage and its image after a dilation are similar figures. Therefore, the ratio of perimeters of a preimage and its image is equal to the scale factor of the dilation.

Using Dilations in Real Life—p.508 Finding Scale Factor: Shadow puppets have been used in many countries for hundreds of years. A flat figure is held between a light and a screen. The audience on the other side of the screen sees the puppet’s shadow. The shadow is a dilation, or enlargement of the shadow puppet. When looking at a cross sectional view, ∆LCP ~ ∆LSH.

Shadow Puppet continued The shadow puppet shown is 12 inches tall. (CP in the diagram). Find the height of the shadow, SH, for each distance from the screen. In each case, by what percent is the shadow larger than the puppet? A. LC = LP = 59 in.; LS = LH = 74 in. B. LC = LP = 66 in.; LS = LH = 74 in.

Finding Scale Factor So, the shadow is 25% larger than the puppet. 59 12 LC CP = ENLARGEMENT: = 74 SH LS SH 59SH = 75(12) 59SH = 888 SH ≈15 INCHES To find the percent of the size increase, use the scale factor of the dilation. SH Scale factor = CP 15 1.25 = 12 So, the shadow is 25% larger than the puppet.

Finding Scale Factor Notice that as the puppet moves closer to the screen, the shadow height increase. 66 12 LC CP = ENLARGEMENT: = 74 SH LS SH 66SH = 75(12) 66SH = 888 SH ≈13.45 INCHES Use the scale factor again to find the percent of size increase. SH Scale factor = CP 13.45 1.12 = 12 So, the shadow is 12% larger than the puppet.