`angle ACB` is an angle in the semicircle of diameter AB =5 and AC : BC = 3 : 4 The area of the triangle ABC is

Options

A. `6 sqrt(2)` sq. m
B. 4 sq. cm
C. 12 sq. cm
D. 6 sq. cm

Answer

To find the area of triangle ABC where angle ACB is inscribed in a semicircle with diameter AB = 5 and the sides AC and BC are in the ratio 3:4, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Diameter and Radius:**
- Given that the diameter AB = 5, the radius (r) of the semicircle is:
\[ r = \frac{AB}{2} = \frac{5}{2} = 2.5 \]

2. **Use the Ratio to Express AC and BC:**
- Let AC = 3x and BC = 4x, where x is a common multiplier.

3. **Apply the Pythagorean Theorem:**
- Since angle ACB is 90 degrees (as it is inscribed in a semicircle), we can apply the Pythagorean theorem:
\[ AB^2 = AC^2 + BC^2 \]
- Substituting the values:
\[ 5^2 = (3x)^2 + (4x)^2 \]
- This simplifies to:
\[ 25 = 9x^2 + 16x^2 \]
\[ 25 = 25x^2 \]

4. **Solve for x:**
- Dividing both sides by 25:
\[ x^2 = 1 \quad \Rightarrow \quad x = 1 \]

5. **Find the Lengths of AC and BC:**
- Now substituting x back:
\[ AC = 3x = 3(1) = 3 \]
\[ BC = 4x = 4(1) = 4 \]

6. **Calculate the Area of Triangle ABC:**
- The area (A) of triangle ABC can be calculated using the formula:
\[ A = \frac{1}{2} \times base \times height \]
- Here, we can take AC as the base and BC as the height:
\[ A = \frac{1}{2} \times AC \times BC = \frac{1}{2} \times 3 \times 4 = \frac{12}{2} = 6 \]

### Final Answer:
The area of triangle ABC is **6 square units**.