Question

question 3 (1 point) if $\triangle gfe \sim \triangle cbe$, find the value of $x$.
blank 1:
$x = \underline{\quad\quad}$

Explanation:

Step1: Use similarity of triangles

Since \(\triangle GFE \sim \triangle CBE\), the ratios of corresponding sides are equal. So, \(\frac{GE}{CE}=\frac{FE}{BE}\).
Given \(GE = 60\), \(CE = 24\), \(FE = 4x - 1\), \(BE = x + 5\). Substituting these values, we get \(\frac{60}{24}=\frac{4x - 1}{x + 5}\).

Step2: Simplify the ratio

Simplify \(\frac{60}{24}\) to \(\frac{5}{2}\). So the equation becomes \(\frac{5}{2}=\frac{4x - 1}{x + 5}\).

Step3: Cross - multiply

Cross - multiplying gives \(5(x + 5)=2(4x - 1)\).

Step4: Expand both sides

Expanding, we have \(5x+25 = 8x-2\).

Step5: Solve for x

Subtract \(5x\) from both sides: \(25=3x - 2\).
Add 2 to both sides: \(27 = 3x\).
Divide both sides by 3: \(x = 9\).

Answer:

\(9\)