Triangle $ABC$ is isosceles with $AB + AC$ and $BC = 65$ cm. $P$ is a point on $\overline{BC}$  such that the perpendicular distances from $P$ to $\overline{AB}$ and $\overline{AC}$ are $24$ cm and $36$ cm, respectively. The area of $\triangle ABC$, in cm$^2$, is 

$ \mathrm{(A) \ } 1254 \qquad \mathrm{(B) \ } 1640 \qquad \mathrm {(C) \ } 1950 \qquad \mathrm{(D) \ } 2535 \qquad \mathrm{(E) \ } 2942$
1. **Identify the given information and setup the problem:**
   - Triangle $ABC$ is isosceles with $AB = AC$.
   - $AB + AC = 65$ cm and $BC = 65$ cm.
   - Point $P$ is on $\overline{BC}$ such that the perpendicular distances from $P$ to $\overline{AB}$ and $\overline{AC}$ are 24 cm and 36 cm, respectively.

2. **Determine the segments $BP$ and $CP$:**
   - Let the perpendicular distances from $P$ hit $AB$ at $D$ and $AC$ at $E$.
   - Since $\triangle BDP \sim \triangle CEP$ in the ratio $2:3$, we have:
     \[
     \frac{BP}{CP} = \frac{2}{3}
     \]
   - Let $BP = 2k$ and $CP = 3k$. Since $BP + CP = BC = 65$ cm, we get:
     \[
     2k + 3k = 65 \implies 5k = 65 \implies k = 13
     \]
   - Therefore, $BP = 2k = 26$ cm and $CP = 3k = 39$ cm.

3. **Calculate the lengths $BD$ and $CE$ using the Pythagorean Theorem:**
   - For $\triangle BDP$:
     \[
     BD^2 + DP^2 = BP^2 \implies BD^2 + 24^2 = 26^2 \implies BD^2 + 576 = 676 \implies BD^2 = 100 \implies BD = 10 \text{ cm}
     \]
   - For $\triangle CEP$:
     \[
     CE^2 + EP^2 = CP^2 \implies CE^2 + 36^2 = 39^2 \implies CE^2 + 1296 = 1521 \implies CE^2 = 225 \implies CE = 15 \text{ cm}
     \]

4. **Set up the equation for $AB = AC = x$ using the Pythagorean Theorem:**
   - For $\triangle ADP$ and $\triangle AEP$:
     \[
     AD = x - 10 \quad \text{and} \quad AE = x - 15
     \]
     \[
     (x - 10)^2 + 24^2 = (x - 15)^2 + 36^2
     \]
     \[
     (x - 10)^2 + 576 = (x - 15)^2 + 1296
     \]
     \[
     x^2 - 20x + 100 + 576 = x^2 - 30x + 225 + 1296
     \]
     \[
     x^2 - 20x + 676 = x^2 - 30x + 1521
     \]
     \[
     10x + 676 = 1521
     \]
     \[
     10x = 845 \implies x = 84.5
     \]

5. **Calculate the altitude of $\triangle ABC$ from side $BC$:**
   - Using the Pythagorean Theorem:
     \[
     \text{Altitude} = \sqrt{84.5^2 - 32.5^2}
     \]
     \[
     \text{Altitude} = \sqrt{7140.25 - 1056.25} = \sqrt{6084} = 78 \text{ cm}
     \]

6. **Calculate the area of $\triangle ABC$:**
   - The area is given by:
     \[
     \text{Area} = \frac{1}{2} \times BC \times \text{Altitude} = \frac{1}{2} \times 65 \times 78 = 65 \times 39 = 2535 \text{ cm}^2
     \]

The final answer is $\boxed{2535}$