Exercise 36 Page 579

Let's analyze the given figure. Since we are given a triangle with a line that is parallel to one of its sides, we can use the Triangle Proportionality Theorem.

The lengths of the segments intercepted by the parallel line are proportional. Let's write a proportion using the expressions for the lengths of the segments. WV/VZ=WX/XY ⇕ 6/a-9=7/XY Since we are given the lengths of WX and WY, we can find the length of XY.

WX+XY=WY

SubstituteII

WX= 7, WY= a

7+XY= a

SubEqn

LHS-7=RHS-7

XY=a-7

Now we can substitute the missing length into our proportion. 6/a-9=7/XY ⇕ 6/a-9=7/a-7 Finally, we will solve this equation for a.

6/a-9=7/a-7

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Solve for a

MultEqn

LHS * (a-9)=RHS* (a-9)

6=7/a-7(a-9)

MultEqn

LHS * (a-7)=RHS* (a-7)

6(a-7)=7(a-9)

Distr

Distribute 6

6a-42=7(a-9)

Distr

Distribute 7

6a-42=7a-63

AddEqn

LHS+63=RHS+63

6a+21=7a

SubEqn

LHS-6a=RHS-6a

21=a

RearrangeEqn

Rearrange equation

a=21

Since WY=a, we found that WY=21.