Question

1. if \\( \triangle pts \sim \triangle pqr \\), find the value of \\( x \\).

1. if \\( \triangle fgh \sim \triangle kjh \\), find \\( fh \\).

Explanation:

Step1: Use the property of similar triangles (Problem 8)

Since \(\triangle PTS\sim\triangle PQR\), the ratios of corresponding sides are equal. So, \(\frac{PT}{PQ}=\frac{PS}{PR}\). Here, \(PQ = 5x + 13\), \(PT=36\), \(PR = 6x - 2\), \(PS = 30\). Then \(\frac{36}{5x + 13}=\frac{30}{6x - 2}\).
Cross - multiply: \(36(6x - 2)=30(5x + 13)\)
Expand: \(216x-72 = 150x+390\)
Subtract \(150x\) from both sides: \(216x-150x-72=150x - 150x+390\), \(66x-72 = 390\)
Add \(72\) to both sides: \(66x-72 + 72=390 + 72\), \(66x=462\)
Divide both sides by \(66\): \(x=\frac{462}{66}=7\)

Step2: Use the property of similar triangles (Problem 9)

Since \(\triangle FGH\sim\triangle KJH\), \(\frac{FH}{KH}=\frac{GH}{JH}\). Here, \(FH = 4x - 25\), \(KH = 52\), \(GH=x + 8\), \(JH = 32\). Then \(\frac{4x-25}{52}=\frac{x + 8}{32}\)
Cross - multiply: \(32(4x-25)=52(x + 8)\)
Expand: \(128x-800=52x + 416\)
Subtract \(52x\) from both sides: \(128x-52x-800=52x-52x + 416\), \(76x-800 = 416\)
Add \(800\) to both sides: \(76x-800 + 800=416 + 800\), \(76x=1216\)
Divide both sides by \(76\): \(x=\frac{1216}{76}=16\)
Substitute \(x = 16\) into \(FH=4x-25\), \(FH=4\times16-25=64 - 25=39\)

Answer:

For problem 8: \(x = 7\)
For problem 9: \(FH=39\)