Module 6

Module 6 - SIMILARITY

Regional Mass Training of Grade 9 Mathematics Teachers

May 17 – 21, 2014

1. Solve problems involving similar polygons.2. Prove certain triangles are similar by using AA,

SSS, and SAS.3. Solve problems involving Basic Triangle

Proportionality Theorem and Triangle Angle Bisector Theorem.

4. Solve problems involving Right Triangle Similarities and Special Right Triangles.

DEPARTMENT OF EDUCATION

Objectives

Warm UpSolve each proportion.

1. 2. 3.

4. If ∆QRS ~ ∆XYZ, identify the pairs of congruent angles and

write 3 proportions using pairs of corresponding sides.

Review: Proportion


z = ±10 x = 8

Q X; R Y; S Z;

Similar Polygons


Two polygons are said to be similar if and only if:i. corresponding angles are congruent.ii. corresponding sides are proportional.

A

B

C

D

EM

NO

P

Q A↔M

E↔Q

C↔OD↔P

B↔N↔ ↔ ↔ ↔ ↔

AM

EQCO

DPBN

ABMN=

BCNO=

CDOP=

DEPQ=

AEMQ

ABCDEMNOPQ

Similar Polygons


Example: If ABCDEMNOPQ, determine the value of x and y. Figure is drawn not to scale.

A

B

C

D

E

Answer:y=3x=2

9

M

NO

P

Q6

y+3

4x+2

5

2x+2 6

7.5

4y-3

Triangle Similarities


Example 1: Using the AA Similarity Postulate

Explain why the triangles are similar and write a similarity statement.

Since , B E by the Alternate Interior Angles Theorem. Also, A D by the Right Angle Congruence Theorem. Therefore ∆ABC ~ ∆DEC by AA~.

Similar Triangles


Similar Triangles


In ∆ABC and ∆DEF, Prove that ∆ABC∆DEF.

B

AC

D F

E

Similar Triangles


Proof:Construct in ∆DEF such that G and H are in and respectively, , and .

B

AC

D F

E

G H

Similar Triangles


B

AC

D F

E

G H

By AA Sim. Postulate, ∆EGH∆EDF which implies .

Since , , then and by substitution.

Similar Triangles


B

AC

D F

E

G H

And since from the given and , then by transitivity PE, which implies AC=GH by multiplication PE.

Similar Triangles


B

AC

D F

E

G H

Meaning ∆ABC ∆GEH by SSS congruence postulate. Since ∆EGH∆EGF, then ∆ABC∆DEF by substitution.

Similar Triangles


Given ∆ABC and ∆DEF such that and

B

AC

D F

E

Similar Triangles


B

AC

D F

E Locate P on so that PE=AB. Draw so that ║ By the AA Sim. Theo., we have ∆PEQ∆DEF, and thus P Q

Similar Triangles


B

AC

D F

E

P Q

Since PE=AB, we can substitute this in the given proportion and find EQ=BC and QP=CA. By SSS congruence Theo., it follows that ∆PEQ∆ABC.

Similar Triangles


B

AC

D F

E

P Q

And since ∆PEQ∆DEF and ∆PEQ∆ABC, by substitution, then ∆ABC∆DEF

Example : Verifying Triangle Similarity

Verify that the triangles are similar.

∆PQR and ∆STU

Therefore ∆PQR ~ ∆STU by SSS ~.

Similar Triangles


Example : Verifying Triangle Similarity

∆DEF and ∆HJK

Verify that the triangles are similar.

D H by the Definition of Congruent Angles.

Therefore ∆DEF ~ ∆HJK by SAS ~.

Similar Triangles


A A by Reflexive Property of , and B C since they are both right angles.

Example : Finding Lengths in Similar Triangles

Explain why ∆ABE ~ ∆ACD, and then find CD.

Step 1 Prove triangles are similar.

Therefore ∆ABE ~ ∆ACD by AA ~.

Similar Triangles


Example Continued

Step 2 Find CD.Corr. sides are proportional. Seg. Add. Postulate.

Substitute x for CD, 5 for BE, 3 for CB, and 9 for BA. Multiplication PE x(9) = 5(3 + 9)Simplify. 9x = 60

Similar Triangles


Example : Writing Proofs with Similar Triangles

Given: ∆BAC with medians and .

Prove: ∆BAC ~ ∆BDE

Similar Triangles


A

B

C

D E

Example ContinuedSimilar Triangles


Statement Reason1. and are medians of ∆BAC

1. Given

2. 2BD=BA ; 2BE = BC 2. Definition of Medians

3. ; 3. Properties of Ratio

4. 4. Transitive PE

5. 5. Reflective PE

6. ∆BAC ~ ∆BDE 6. SAS ~

A

B

C

D E

Check It Out!

Given: M is the midpoint of JK. N is the midpoint of KL, and P is the midpoint of JL.

Similar Triangles


Statements Reasons

Check It Out! Example 4 Continued

1. Given1. M is the mdpt. of , N is the mdpt. of ,

and P is the mdpt. of .

2. Midline Theorem2.

3. Mult. PE.3.

4. SSS ~ Step 34. ∆JKL ~ ∆NPM

Similar Triangles


Basic Triangle Proportionality


Proof:In ∆ABC, let D and E be points on and respectively, such that ║ . We have to prove that . Since parallel lines form congruent corresponding angles, we have and . By the AA Sim. Theo., we say ∆ABC∆DBE.

Basic Triangle Proportionality


Since corresponding sides of similar triangles are proportional, . By Reciprocal Property, we have . By Subtraction Property of Proportionality, we have and hence or .

Check It Out! Basic Triangle Proportionality


Find the value of x in the figure below. The figure is not drawn to scale.

4

6

3

x

64=

𝑥−33

4(x – 3)=184x – 12 =18

4x =30x

Triangle Angle Bisector Theorem


If a segment bisects an angle of a triangle, then it divides the opposite side into segments proportional to the other two sides.

Proof:

1 2

3

4

Let bisect A of ∆ABC. We must prove that .

B X C

A

YDraw parallel to . Extend so that it intersects at Y. Since , we have . By theorems involving parallel lines cut by a transversal, we also have and .


If a segment bisects an angle of a triangle, then it divides the opposite side into segments proportional to the other two sides.

Proof:

1 2

3

4

B X C

A

YSince bisect A, then and . Thus, by the Isosceles Triangle Theorem, AY = AB. By substitution, we conclude that .


Check It Out!


In the figure below, . Use the lengths to find the value of y.

𝐴𝐵𝐴𝐶=

𝐵𝑋𝑋𝐶

14

y X C

9

A

15

159 =

𝑦14− 𝑦

15(14 – y) = 9yy = 8.75


B

Consider ∆ABC such that is right and is the altitude to the hypotenuse. We have to prove that ∆ADC∆ACB∆CDB.Since , then . Since =, then by AA Sim. Theo., we have ∆ADC∆ACB. Using the same theorem and that =, we have ∆ACB∆CDB. Hence, ∆ADC∆CDB by transitivity.

Right Triangle Similarity


Consider the proportion . In this case, the means of the proportion are the same number, and that number is the geometric mean of the extremes.The geometric mean of two positive numbers is the positive square root of their product. So the geometric mean of a and b is the positive number x suchthat , or x2 = ab.



You can use Theorem on Right Triangle Similarity to write proportions comparing the side lengths of the triangles formed by the altitude to the hypotenuse of a right triangle. All the relationships in red involve geometric means.





Example 3: Finding Side Lengths in Right Triangles

Find x, y, and z.

62 = (9)(x) 6 is the geometric mean of 9 and x.

x = 4 Divide both sides by 9.

y2 = (4)(13) = 52 y is the geometric mean of 4 and 13.

Find the positive square root.

z2 = (9)(13) = 117 z is the geometric mean of 9 and 13.

Find the positive square root.



Once you’ve found the unknown side lengths, you can use the Pythagorean Theorem to check your answers.

Helpful Hint



Special Right Triangles


Example : Craft Application

Jana is cutting a square of material for a tablecloth. The table’s diagonal is 36 inches. She wants the diagonal of the tablecloth to be an extra 10 inches so it will hang over the edges of the table. What size square should Jana cut to make the tablecloth? Round to the nearest inch.

Jana needs a 45°-45°-90° triangle with a hypotenuse of 36 + 10 = 46 inches.



Check It Out! Example

What if...? Tessa’s other dog is wearing a square bandana with a side length of 42 cm. What would you expect the circumference of the other dog’s neck to be? Round to the nearest centimeter.

Tessa needs a 45°-45°-90° triangle with a hypotenuse of 42 cm.



Example: Using the 30º-60º-90º Triangle Theorem

An ornamental pin is in the shape of an equilateral triangle. The length of each side is 6 centimeters. Josh will attach the fastener to the back along AB. Will the fastener fit if it is 4 centimeters long?

Step 1 The equilateral triangle is divided into two 30°-60°-90° triangles.

The height of the triangle is the length of the longer leg.



Example Continued

Step 2 Find the length x of the shorter leg.

Step 3 Find the length h of the longer leg.

The pin is approximately 5.2 centimeters high. So the fastener will fit.

Hypotenuse = 2(shorter leg)6 = 2x3 = x Divide both sides by 2.



Check It Out! Example What if…? A manufacturer wants to make a larger clock with a height of 30 centimeters. What is the length of each side of the frame? Round to the nearest tenth.

Step 1 The equilateral triangle is divided into two 30º-60º-90º triangles.The height of the triangle is the length of the longer leg.



Check It Out! Example Continued

Step 2 Find the length x of the shorter leg.

Each side is approximately 34.6 cm.

Step 3 Find the length y of the longer leg.

Rationalize the denominator.

Hypotenuse = 2(shorter leg)y = 2x

Simplify.



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Module 6

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