Question

1. if $\triangle cdg \sim \triangle edf$, find the value of $x$.

Explanation:

Step1: Use the property of similar triangles

Since \(\triangle CDG\sim\triangle EDF\), the ratios of corresponding sides are equal. That is \(\frac{CD}{ED}=\frac{GD}{FD}\).
Given \(CD = 48\), \(GD=40\), \(ED = 5x - 2\), \(FD=2x + 7\). So we have the equation \(\frac{48}{5x-2}=\frac{40}{2x + 7}\).

Step2: Cross - multiply

Cross - multiplying gives us \(48(2x + 7)=40(5x - 2)\).
Expand both sides: \(96x+336 = 200x-80\).

Step3: Solve for \(x\)

First, move the \(x\) terms to one side and the constants to the other side.
\(200x-96x=336 + 80\).
\(104x=416\).
Divide both sides by \(104\): \(x=\frac{416}{104}\).

Answer:

\(x = 4\)