ABC is a triangle right angled at A AB = 6 cm and AC = 8 cm. Semi-circle are drawn outside the triangle having area `x, y, z` units, respectively on the sides AB, AC andBC.What is `x + y` - zequal to?

Options

A. `48 cm^(2)`
B. `32 cm^(2)`
C. 0
D. None of these

Answer

To solve the problem, we need to calculate the areas of the semicircles drawn on the sides of the right-angled triangle ABC, where AB = 6 cm, AC = 8 cm, and we need to find the relationship between the areas of these semicircles.

### Step-by-Step Solution:

1. **Identify the sides of the triangle:**
- Given triangle ABC is right-angled at A.
- AB = 6 cm (one leg)
- AC = 8 cm (the other leg)

2. **Calculate the length of side BC using the Pythagorean theorem:**
- According to the Pythagorean theorem, \( BC^2 = AB^2 + AC^2 \)
- Thus, \( BC^2 = 6^2 + 8^2 = 36 + 64 = 100 \)
- Therefore, \( BC = \sqrt{100} = 10 \) cm.

3. **Calculate the areas of the semicircles:**
- The radius of the semicircle on AB (x) is \( \frac{AB}{2} = \frac{6}{2} = 3 \) cm.
- The area of the semicircle on AB is:
\[ x = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (3^2) = \frac{1}{2} \pi (9) = \frac{9\pi}{2} \text{ square cm} \]

- The radius of the semicircle on AC (y) is \( \frac{AC}{2} = \frac{8}{2} = 4 \) cm.
- The area of the semicircle on AC is:
\[ y = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (4^2) = \frac{1}{2} \pi (16) = \frac{16\pi}{2} = 8\pi \text{ square cm} \]

- The radius of the semicircle on BC (z) is \( \frac{BC}{2} = \frac{10}{2} = 5 \) cm.
- The area of the semicircle on BC is:
\[ z = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (5^2) = \frac{1}{2} \pi (25) = \frac{25\pi}{2} \text{ square cm} \]

4. **Set up the equation based on the relationship of the areas:**
- From the problem, we know that \( x + y = z \).
- Substituting the values we found:
\[ \frac{9\pi}{2} + 8\pi = \frac{25\pi}{2} \]

5. **Combine the areas:**
- Convert \( 8\pi \) to a fraction with a common denominator:
\[ 8\pi = \frac{16\pi}{2} \]
- Now, add \( x \) and \( y \):
\[ x + y = \frac{9\pi}{2} + \frac{16\pi}{2} = \frac{25\pi}{2} \]

6. **Conclusion:**
- Since \( x + y = z \), we find that \( x + y - z = 0 \).

### Final Answer:
Thus, \( x + y - z = 0 \).