The perimeter of a right angled isosceles triangle (T) is `(24+12sqrt2)` cm if one of T's smaller sides is equal to the side of a square(R) then what is R's area

Options

A. 72
B. 288
C. 144
D. 196

Answer

To solve the problem, we need to find the area of square R given the perimeter of a right-angled isosceles triangle T.

### Step-by-Step Solution:

1. **Understanding the Triangle**:
- Let the lengths of the two equal sides (the smaller sides) of the right-angled isosceles triangle T be \( x \) cm.
- The hypotenuse (the third side) can be calculated using the Pythagorean theorem:
\[ AC = \sqrt{AB^2 + BC^2} = \sqrt{x^2 + x^2} = \sqrt{2x^2} = x\sqrt{2} \]

2. **Calculating the Perimeter**:
- The perimeter \( P \) of triangle T is the sum of all its sides:
\[ P = AB + BC + AC = x + x + x\sqrt{2} = 2x + x\sqrt{2} \]
- We know from the problem that the perimeter is given as \( 24 + 12\sqrt{2} \) cm. Therefore, we can set up the equation:
\[ 2x + x\sqrt{2} = 24 + 12\sqrt{2} \]

3. **Factoring Out \( x \)**:
- Rearranging the equation gives:
\[ x(2 + \sqrt{2}) = 24 + 12\sqrt{2} \]

4. **Solving for \( x \)**:
- To find \( x \), divide both sides by \( (2 + \sqrt{2}) \):
\[ x = \frac{24 + 12\sqrt{2}}{2 + \sqrt{2}} \]

5. **Rationalizing the Denominator**:
- Multiply the numerator and denominator by the conjugate of the denominator:
\[ x = \frac{(24 + 12\sqrt{2})(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} \]
- The denominator simplifies to:
\[ 2^2 - (\sqrt{2})^2 = 4 - 2 = 2 \]
- The numerator expands to:
\[ 24 \cdot 2 - 24\sqrt{2} + 12\sqrt{2} \cdot 2 - 12 \cdot 2 = 48 - 24\sqrt{2} + 24\sqrt{2} - 24 = 24 \]
- Thus:
\[ x = \frac{24}{2} = 12 \text{ cm} \]

6. **Finding the Area of Square R**:
- Since the side of square R is equal to the smaller side \( x \) of triangle T, the side of square R is 12 cm.
- The area \( A \) of square R is given by:
\[ A = \text{side}^2 = 12^2 = 144 \text{ cm}^2 \]

### Final Answer:
The area of square R is \( 144 \text{ cm}^2 \).