Question

In $△ABC$, if $AB=8,BC=7,CA=6$ and side $BC$ is extended as shown in the figure to a point $P$ such that $△PAB$ is similar to $△PCA$, then length of $PC$ is

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

Answer: (C)

$9$

$△PAB\sim △PCA$

$\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}$

$\frac{PC}{PA}=\frac{6}{8}=\frac{3}{4}$

$PA=\frac{4PC}{3}$

$\frac{PA}{PC+7}=\frac{3}{4}\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}$

$\frac{16PC}{3}=3PC+21\Rightarrow \frac{7PC}{3}=21$

$PC=9$