Question

Triangle $ABC$ is isosceles with $AB=AC$ and $BC=65$ cm. $P$ is a point on $BC$ such that the perpendicular distance from $P$ to $AB$ and $AC$ are $24$ cm and $36$ cm respectively.

The area of triangle $ABC$ in sq-cm is?

• A. $1254$
• B. $1950$
• C. $2535$
• D. $5070$

Answer 1

The correct option is C $2535$

From the fig:

$AB=AC$; $B=C$

$BC=BP+PC=65$ ...(1)

and $PD=24$; $PE=36$

In $△PBD$

$BP=\frac{PD}{\sin B}=\frac{24}{\sin B}$ ....(2)

In $△PCE$

$PC=\frac{PE}{\sin C}=\frac{36}{\sin B}$ ....(3)

From (2) & (3), we get $BP=\frac{2PC}{3}$ ....(4)

from (1) & (4),

$BP=39$ and $PC=26$

From (2), $\sin B=\frac{24}{39}=\frac{12}{13}$

$\Rightarrow \cos B=\frac{5}{13}$

$\because A=\pi -B-C=\pi -2B$

$\Rightarrow \sin A=\sin 2B=2\sin B\cos B=\frac{120}{169}$

using sine rule in $\Delta ABC$:

$AB=AC=\frac{BC\sin B}{\sin A}=\frac{65\sin B}{\sin 2B}=\frac{65}{2\cos B}=\frac{169}{2}$

Therefore, $△=\frac{\left(AB\right)\left(AC\right)\sin A}{2}=2535$