In an isosceles right - angled triangle , the perimeter is 30 m . Find its area (Approximate) .

Options

A. `38.63 m^(2)`
B. `40 m^(2)`
C. `39.60m^(2)`
D. `37.86 m^(2)`

Answer

To solve the problem step-by-step, we will follow the approach outlined in the video transcript.

### Step 1: Understand the Triangle
In an isosceles right-angled triangle, the two sides that form the right angle are equal. Let's denote the lengths of these two equal sides as \( a \). The length of the hypotenuse will then be \( a\sqrt{2} \).

### Step 2: Set Up the Perimeter Equation
The perimeter \( P \) of the triangle is the sum of all its sides:
\[ P = a + a + a\sqrt{2} = 2a + a\sqrt{2} \]
According to the problem, the perimeter is given as 30 m:
\[ 2a + a\sqrt{2} = 30 \]

### Step 3: Factor Out \( a \)
We can factor out \( a \) from the left-hand side of the equation:
\[ a(2 + \sqrt{2}) = 30 \]

### Step 4: Solve for \( a \)
To find \( a \), divide both sides by \( (2 + \sqrt{2}) \):
\[ a = \frac{30}{2 + \sqrt{2}} \]

### Step 5: Rationalize the Denominator
To simplify \( a \), we will rationalize the denominator:
\[ a = \frac{30(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} \]
Calculating the denominator:
\[ (2 + \sqrt{2})(2 - \sqrt{2}) = 4 - 2 = 2 \]
Thus, we have:
\[ a = \frac{30(2 - \sqrt{2})}{2} = 15(2 - \sqrt{2}) \]

### Step 6: Calculate the Area
The area \( A \) of the triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In our case, both the base and height are \( a \):
\[ A = \frac{1}{2} \times a \times a = \frac{1}{2} a^2 \]

Substituting \( a = 15(2 - \sqrt{2}) \):
\[ A = \frac{1}{2} \times (15(2 - \sqrt{2}))^2 \]

### Step 7: Expand and Simplify
Calculating \( (15(2 - \sqrt{2}))^2 \):
\[ = 225(2 - \sqrt{2})^2 = 225(4 - 4\sqrt{2} + 2) = 225(6 - 4\sqrt{2}) \]
Thus,
\[ A = \frac{1}{2} \times 225(6 - 4\sqrt{2}) = 112.5(6 - 4\sqrt{2}) \]

### Step 8: Approximate the Area
Now, we can approximate \( \sqrt{2} \approx 1.414 \):
\[ A \approx 112.5(6 - 4 \times 1.414) = 112.5(6 - 5.656) = 112.5(0.344) \approx 38.7 \text{ m}^2 \]

### Final Result
Thus, the approximate area of the isosceles right-angled triangle is:
\[ \text{Area} \approx 38.7 \text{ m}^2 \]