Question

1. klmn ~ pqrs; find x
2. △rst ~ △ysz; find yz

Explanation:

Step1: Use property of similar polygons

For similar polygons, the ratios of corresponding sides are equal. For trapezoids \(KLMN\sim PQRS\), we can set up the proportion \(\frac{4x + 4}{48}=\frac{7x-9}{60}\).
Cross - multiply: \(60(4x + 4)=48(7x - 9)\).
Expand both sides: \(240x+240 = 336x-432\).

Step2: Solve for \(x\)

Move \(x\) terms to one side and constants to the other side:
\(336x-240x=240 + 432\).
\(96x=672\).
Divide both sides by 96: \(x = 7\).

Step3: For similar triangles \(\triangle RST\sim\triangle YSZ\)

Set up the proportion \(\frac{2x + 2}{3x-7}=\frac{5}{40}=\frac{1}{8}\).
Cross - multiply: \(8(2x + 2)=3x-7\).
Expand: \(16x+16 = 3x-7\).

Step4: Solve for \(x\)

Move \(x\) terms to one side and constants to the other side:
\(16x-3x=-7 - 16\).
\(13x=-23\).
\(x=-\frac{23}{13}\) (This seems incorrect as we likely made a wrong side - correspondence. Let's use \(\frac{2x + 2}{3x-7}=\frac{40}{45}=\frac{8}{9}\)).
Cross - multiply: \(9(2x + 2)=8(3x-7)\).
Expand: \(18x+18 = 24x-56\).
Move \(x\) terms to one side and constants to the other side: \(24x-18x=18 + 56\).
\(6x=74\).
\(x=\frac{37}{3}\).
Substitute \(x = \frac{37}{3}\) into \(YZ=2x + 2\).
\(YZ=2\times\frac{37}{3}+2=\frac{74}{3}+2=\frac{74 + 6}{3}=\frac{80}{3}\).

Answer:

For problem 5: \(x = 7\)
For problem 6: \(YZ=\frac{80}{3}\)