Question

In a $△ABC$, points P and Q are on sides AB and AC respectively. If AP = 3 cm, PB = 6 cm. AQ = 5 cm and QC = 10 cm, then BC = ____.

• A. 4PQ
• B. $\frac{PQ}{2}$
• C. 3PQ
• D. $P{Q}^2$

Answer 1

The correct option is C 3PQ

Given: AP = 3 cm, PB = 6 cm,
AQ = 5 cm and QC = 10 cm


We have, AB = AP + PB = (3 + 6) cm = 9 cm
and, AC = AQ + QC = (5 + 10) cm = 15 cm

$\therefore \frac{AP}{AB}=\frac{3}{9}=\frac{1}{3}$

and $\frac{AQ}{AC}=\frac{5}{15}=\frac{1}{3}$

$\Rightarrow \frac{AP}{AB}=\frac{AQ}{AC}$

Thus, in $△PAQ$ and $△BAC$,

$\frac{AP}{AB}=\frac{AQ}{AC}$

$\angle PAQ=\angle BAC$
(Common in both triangles)

$\therefore$ $△APQ\sim △ABC$ (by SAS similarity criterion)

$\Rightarrow \frac{AP}{AB}=\frac{PQ}{BC}=\frac{AQ}{AC}$

$\Rightarrow \frac{PQ}{BC}=\frac{AQ}{AC}=\frac{1}{3}$

$\Rightarrow BC=3PQ$