Question

In $△ABC$, given below, $AB=8\text{cm}$, $BC=10\text{cm}$ and $AC=6\text{cm}$. If a point $P$ lies on $AB$ and $Q$ on $AC$ such that $PQ\parallel BC$ and $PQ=5\text{cm}$, then find $AP$.

• A. $4\text{cm}$
• B. $5\text{cm}$
• C. $3\text{cm}$
• D. $2\text{cm}$

Answer 1

The correct option is A $4\text{cm}$

In $△APQ$ and $△ABC$,

line $PQ\parallel$ side $BC$ ...given

$\therefore \angle APQ\cong \angle PBC$ ...corresponding angles

and $\angle AQP\cong \angle QCB$ ...corresponding angles

Also, $\angle PAQ\cong \angle BAC$ ...common angle

$\therefore △APQ\sim △ABC$ .... by AAA test of similarity

$\therefore \frac{AP}{AB}=\frac{AQ}{AC}=\frac{PQ}{BC}$ ( In a pair of similar triangles, the corresponding sides are proportional )

$\therefore \frac{AP}{AB}=\frac{AQ}{AC}=\frac{5}{10}=\frac{1}{2}$

$\therefore AP=\frac{1}{2}AB$ and $AQ=\frac{1}{2}AC$

.

$\therefore AP=\frac{1}{2}\times 8$ and $AQ=\frac{1}{2}\times 6$

$\therefore AP=4\text{cm}$ and $AQ=3\text{cm}$

So, option A is correct.