Triangle `ABC` is isosceles with `AB=AC and BC=65cm.` P is a point on `BC` such that the perpendiculardistances from P to `AB and AC` are `24cm and 36cm,` respectively. The area of triangle `ABC` (in sq cm is)

Options

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

Explanation

To find the area of triangle \( ABC \) given the conditions, we can follow these steps:

### Step 1: Understand the Triangle Configuration
We know that triangle \( ABC \) is isosceles with \( AB = AC \) and \( BC = 65 \, \text{cm} \). Let \( AB = AC = x \).

### Step 2: Set Up the Problem
Let \( P \) be a point on \( BC \) such that the perpendicular distances from \( P \) to \( AB \) and \( AC \) are \( 24 \, \text{cm} \) and \( 36 \, \text{cm} \), respectively. We denote the distances from \( P \) to \( AB \) as \( PD = 24 \, \text{cm} \) and from \( P \) to \( AC \) as \( PE = 36 \, \text{cm} \).

### Step 3: Use the Area Formula
The area \( A \) of triangle \( ABC \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is \( BC = 65 \, \text{cm} \) and the height can be found using the distances from point \( P \).

### Step 4: Calculate the Area Using Heights
The area of triangle \( ABP \) can be calculated as:
\[ A_{ABP} = \frac{1}{2} \times BP \times PD \]
The area of triangle \( ACP \) can be calculated as:
\[ A_{ACP} = \frac{1}{2} \times PC \times PE \]

### Step 5: Express \( BP \) and \( PC \)
Let \( BP = y \) and \( PC = 65 - y \). The total area of triangle \( ABC \) can be expressed as:
\[ A = A_{ABP} + A_{ACP} = \frac{1}{2} \times y \times 24 + \frac{1}{2} \times (65 - y) \times 36 \]

### Step 6: Simplify the Expression
Combining the areas:
\[ A = \frac{1}{2} \left( 24y + 36(65 - y) \right) \]
\[ A = \frac{1}{2} \left( 24y + 2340 - 36y \right) \]
\[ A = \frac{1}{2} \left( -12y + 2340 \right) \]
\[ A = -6y + 1170 \]

### Step 7: Find the Value of \( y \)
To find \( y \), we can use the relationship between the heights and the sides. Since \( \frac{PD}{PE} = \frac{BP}{PC} \):
\[ \frac{24}{36} = \frac{y}{65 - y} \]
Cross-multiplying gives:
\[ 24(65 - y) = 36y \]
\[ 1560 - 24y = 36y \]
\[ 1560 = 60y \]
\[ y = 26 \]

### Step 8: Calculate the Area
Substituting \( y = 26 \) back into the area formula:
\[ A = -6(26) + 1170 \]
\[ A = -156 + 1170 = 1014 \, \text{cm}^2 \]

### Final Answer
The area of triangle \( ABC \) is \( 1014 \, \text{cm}^2 \).

---