Question

1. \\( \triangle b c d \sim \triangle f g e \\); find \\( f e \\)

Explanation:

Step1: Set up proportion

Since \(\triangle BCD\sim\triangle FGE\), the ratios of corresponding sides are equal. So, \(\frac{BC}{FG}=\frac{BD}{FE}\). Here \(BC = 39\), \(FG=4x + 2\), \(BD = 42\), \(FE=5x-2\). Then \(\frac{39}{4x + 2}=\frac{42}{5x-2}\).

Step2: Cross - multiply

Cross - multiplying gives \(39(5x - 2)=42(4x + 2)\).
Expand both sides: \(195x-78 = 168x+84\).

Step3: Solve for \(x\)

Subtract \(168x\) from both sides: \(195x-168x-78=168x - 168x+84\), \(27x-78 = 84\).
Add \(78\) to both sides: \(27x-78 + 78=84 + 78\), \(27x=162\).
Divide both sides by \(27\): \(x=\frac{162}{27}=6\).

Step4: Find \(FE\)

Substitute \(x = 6\) into \(FE = 5x-2\). Then \(FE=5\times6-2=30 - 2=28\).

Answer:

\(28\)