Triangle $A B C$ is isosceles with $A B=A C$ and $B C=65 \mathrm{~cm} . P$ is a point on $B C$ such that the perpendicular distances from $P$ to $A B$ and $A C$ are $24 \mathrm{~cm}$ and $36 \mathrm{~cm}$, respectively. The area of $\triangle A B C$, in $\mathrm{cm}^{2}$, is
(A) 1254
(B) 1640
(C) 1950
(D) 2535
(E) 2942

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-119.jpg?height=309&width=314&top_left_y=908&top_left_x=1426)
Since triangle $A B C$ is isosceles, $\angle A B C=\angle A C B$, and so triangle $B R P$ is similar to triangle $C S P$ (equal angle, right angle).

Therefore, $\frac{B P}{R P}=\frac{C P}{S P}$ or $\frac{B P}{24}=\frac{C P}{36}$ or $\frac{B P}{C P}=\frac{2}{3}$.

But since $B C$ has length $65 \mathrm{~cm}$, then $B P+C P=65 \mathrm{~cm}$, and so $B P=26 \mathrm{~cm}$ and $B P=39 \mathrm{~cm}$. Since $B P=26 \mathrm{~cm}$ and $R P=24 \mathrm{~cm}$, then by the Pythagorean Theorem, $B R=10 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-233.jpg?height=374&width=385&top_left_y=1683&top_left_x=1447)
Next, we drop a perpendicular from $A$ to $F$ on $B C$. Since triangle $A B C$ is isosceles, $B F=\frac{1}{2}(B C)=\frac{65}{2} \mathrm{~cm}$.

Also, triangle $B F A$ is similar to triangle $B R P$ (common angle, right angle).

Thus, $\frac{B R}{R P}=\frac{B F}{F A}$, so $F A=\frac{\left(\frac{65}{2}\right)(24)}{10}=78 \mathrm{~cm}$.

The area of triangle $A B C$ is then $\frac{1}{2}(B C)(F A)=\frac{1}{2}(65)(78)=2535 \mathrm{~cm}^{2}$.