Exercise 19 Page 566

Let's review the theorems that can help us prove that two triangles are similar.

1. Angle-Angle Similarity Theorem: If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.
2. Side-Side-Side Similarity Theorem: If the corresponding side lengths of two triangles are proportional, then the triangles are similar.
3. Side-Angle-Side Similarity Theorem: If the lengths of two sides of one triangle are proportional to the lengths of two corresponding sides of another triangle and the included angles are congruent, then the triangles are similar.

Now we will identify the similar triangles and find the measures, one at a time.

Similar Triangles

We can see that ∠ J is congruent to ∠ Q, and that ∠ H is congruent to ∠ N. This means that two angles of △ JHK are congruent to two angles of △ QNP. Therefore, by the Angle-Angle Similarity Theorem, △ JHK and △ QNP are similar. △ JHK ~ △ QNP

Finding the Measures

Using our similarity statement from above, we can identify three pairs of corresponding sides that will help us find the requested lengths. HJ corresponds with NQ HK corresponds with NP JK corresponds with QP Recall that corresponding segments of similar figures will have proportional lengths. We are given expressions for the lengths of these sides which we can use to write a proportion. HJ/NQ = HK/NP = JK/QP ⇕ 4x+7/12 = 6x-2/8 = 25/20 Let's solve the equation 6x-28 = 2520 to find x.

6x-2/8 = 25/20

▼

Solve for x

CrossMult

Cross multiply

(6x-2)20=8 * 25

Multiply

Multiply

(6x-2)20=200

Distr

Distribute 20

120x-40=200

AddEqn

LHS+40=RHS+40

120x=240

DivEqn

.LHS /120.=.RHS /120.

x=2

Now that we know the value of x, we can find HJ and HK. We will substitute x=2 in the expressions for the lengths.

Measure	Expression	x=2	Simplified
HJ	4x+7	4(2)+7	15
HK	6x-2	6(2)-2	10

We found that HJ=15 and HK=10.