In the given questions two quantities are given one as quantity I and another as quantity II you have to determine relationship between two quantities choose appropriate option
one of equal sides of right angled isosceles triangle ABD measures 12cm, BC is perpendicular bisector
qunatityI:area of semiircle formed by BC as diameter
quantityII:area of triangle whose height and base measures 12cm and 7cm respectively

Options

A. `IltII`
B. `IgtII`
C. `IleII`
D. `IgeII`

Answer

To solve the problem, we need to calculate the areas of the semicircle and the triangle based on the given dimensions.

### Step 1: Calculate the area of the semicircle formed by BC as the diameter.

1. **Identify the length of BC**:
- Since triangle ABD is a right-angled isosceles triangle with equal sides measuring 12 cm, we can find the length of BC, which is the perpendicular bisector of AD.
- The length of AD can be calculated using the Pythagorean theorem:
\[ AD = \sqrt{AB^2 + BD^2} = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2} \text{ cm} \]
- Since BC is the perpendicular bisector, its length is half of AD:
\[ BC = \frac{AD}{2} = \frac{12\sqrt{2}}{2} = 6\sqrt{2} \text{ cm} \]

2. **Calculate the radius of the semicircle**:
- The radius \( r \) of the semicircle is half of BC:
\[ r = \frac{BC}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \text{ cm} \]

3. **Calculate the area of the semicircle**:
- The area \( A \) of a semicircle is given by:
\[ A = \frac{1}{2} \pi r^2 \]
- Substituting the radius:
\[ A = \frac{1}{2} \pi (3\sqrt{2})^2 = \frac{1}{2} \pi (9 \cdot 2) = \frac{1}{2} \pi \cdot 18 = 9\pi \text{ cm}^2 \]
- Approximating \( \pi \) as \( \frac{22}{7} \):
\[ A \approx 9 \cdot \frac{22}{7} = \frac{198}{7} \approx 28.29 \text{ cm}^2 \]

### Step 2: Calculate the area of the triangle with height and base.

1. **Identify the base and height**:
- The base is given as 12 cm and the height as 7 cm.

2. **Calculate the area of the triangle**:
- The area \( A_{triangle} \) of a triangle is given by:
\[ A_{triangle} = \frac{1}{2} \times \text{base} \times \text{height} \]
- Substituting the values:
\[ A_{triangle} = \frac{1}{2} \times 12 \times 7 = \frac{84}{2} = 42 \text{ cm}^2 \]

### Step 3: Compare the two quantities.

- **Quantity I (Area of semicircle)**: \( \approx 28.29 \text{ cm}^2 \)
- **Quantity II (Area of triangle)**: \( 42 \text{ cm}^2 \)

Since \( 28.29 < 42 \), we conclude:

### Final Conclusion:
**Quantity I is less than Quantity II.**