We are given a triangle $ABC$ with a line segment $PQ$ parallel to $BC$. We have the lengths $AP = 7$, $PB = 35$, and $QC = 40$. We want to find the length $AQ = x$.

1. Problem Description

We are given a triangle $ABC$ with a line segment $PQ$ parallel to $BC$. We have the lengths $AP = 7$, $PB = 35$, and $QC = 40$. We want to find the length $AQ = x$.

2. Solution Steps

Since $PQ \parallel BC$, we can use the property that the sides are proportional.

That is,

$\frac{AP}{PB} = \frac{AQ}{QC}$.

Substituting the given values, we get

$\frac{7}{35} = \frac{x}{40}$.

Simplifying the left side, we have

$\frac{1}{5} = \frac{x}{40}$.

Multiplying both sides by 40, we get

$x = \frac{1}{5} \times 40 = 8$.

3. Final Answer

The value of x is 8.