A right angled isosceles triangle is inscribed in a semi-circle of radius 7 cm. The area enclosed by the semi-circle but exterior to the triangle is 

Options

A. `14 cm^(2)`
B. `28 cm^(2)`
C. `44 cm^(2)`
D. `68 cm^(2)`

Answer

To find the area enclosed by the semi-circle but exterior to the right-angled isosceles triangle inscribed in it, we will follow these steps:

### Step 1: Calculate the area of the semi-circle.
The formula for the area of a semi-circle is given by:
\[ \text{Area of semi-circle} = \frac{1}{2} \pi r^2 \]
where \( r \) is the radius of the semi-circle.

Given that the radius \( r = 7 \, \text{cm} \):
\[ \text{Area of semi-circle} = \frac{1}{2} \pi (7)^2 = \frac{1}{2} \pi (49) = \frac{49\pi}{2} \, \text{cm}^2 \]

### Step 2: Determine the dimensions of the inscribed triangle.
In a right-angled isosceles triangle inscribed in a semi-circle, the hypotenuse is equal to the diameter of the semi-circle. The diameter \( d \) is:
\[ d = 2r = 2 \times 7 = 14 \, \text{cm} \]

Let the lengths of the two equal sides of the triangle be \( a \). By the Pythagorean theorem:
\[ a^2 + a^2 = d^2 \quad \Rightarrow \quad 2a^2 = (14)^2 \quad \Rightarrow \quad 2a^2 = 196 \quad \Rightarrow \quad a^2 = 98 \quad \Rightarrow \quad a = \sqrt{98} = 7\sqrt{2} \, \text{cm} \]

### Step 3: Calculate the area of the triangle.
The area \( A \) of a right-angled triangle is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In this case, both the base and height are equal to \( a \):
\[ A = \frac{1}{2} \times a \times a = \frac{1}{2} \times (7\sqrt{2}) \times (7\sqrt{2}) = \frac{1}{2} \times 98 = 49 \, \text{cm}^2 \]

### Step 4: Calculate the area exterior to the triangle.
To find the area enclosed by the semi-circle but exterior to the triangle, we subtract the area of the triangle from the area of the semi-circle:
\[ \text{Area exterior to triangle} = \text{Area of semi-circle} - \text{Area of triangle} \]
\[ = \frac{49\pi}{2} - 49 \]

### Step 5: Approximate the value.
Using \( \pi \approx 3.14 \):
\[ \text{Area of semi-circle} \approx \frac{49 \times 3.14}{2} = \frac{153.86}{2} \approx 76.93 \, \text{cm}^2 \]
Now, substituting this value:
\[ \text{Area exterior to triangle} \approx 76.93 - 49 \approx 27.93 \, \text{cm}^2 \]

### Conclusion:
The area enclosed by the semi-circle but exterior to the triangle is approximately \( 28 \, \text{cm}^2 \).