Question

1. if \\( \triangle gfe \sim \triangle cbe \\), find the value of \\( x \\).

Explanation:

Step1: Use the property of similar triangles

Since \(\triangle GFE\sim\triangle CBE\), the ratios of corresponding sides are equal. That is \(\frac{GE}{CE}=\frac{FE}{BE}\).
We know that \(GE = 60\), \(CE=60 + 24=84\), \(FE = 4x - 1\), and \(BE=x + 5\).
So, \(\frac{60}{84}=\frac{4x-1}{x + 5}\).

Step2: Cross - multiply

Cross - multiplying the proportion \(\frac{60}{84}=\frac{4x-1}{x + 5}\) gives us \(60(x + 5)=84(4x-1)\).
Expand both sides: \(60x+300 = 336x-84\).

Step3: Solve for \(x\)

First, move the \(x\) terms to one side and the constants to the other side.
\(300 + 84=336x-60x\).
\(384 = 276x\).
Then \(x=\frac{384}{276}=\frac{32}{23}\approx1.39\).

Answer:

\(x = \frac{32}{23}\)