Question

1. △bcd ~ △fge; find fe

Explanation:

Step1: Set up proportion

Since $\triangle BCD\sim\triangle FGE$, the ratios of corresponding sides are equal. Let's assume the side corresponding to $BD = 42$ is $FE=5x - 2$ and the side corresponding to $BC = 39$ is $FG = 4x+2$. Then $\frac{BC}{FG}=\frac{BD}{FE}$, so $\frac{39}{4x + 2}=\frac{42}{5x - 2}$.

Step2: Cross - multiply

Cross - multiplying gives us $39(5x - 2)=42(4x + 2)$.
Expanding 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 the expression for $FE$ which is $5x-2$.
$FE=5\times6-2=30 - 2=28$.

Answer:

28