In a right -angle triangle, the length of the sides containing the right angles are `a and b`. With the mid-point of each sides as centres three semicircles are drawn outiside the triangle. Find the area of `Delta` and the semicircles together

Answer

To find the area of a right-angle triangle along with the areas of three semicircles drawn on its sides, we can follow these steps:

### Step 1: Find the Area of the Triangle
The area \( A_{\Delta} \) of a right-angle triangle can be calculated using the formula:
\[ A_{\Delta} = \frac{1}{2} \times \text{base} \times \text{height} \]
In this case, let the lengths of the sides containing the right angles be \( a \) and \( b \). Therefore, the area of the triangle is:
\[ A_{\Delta} = \frac{1}{2} \times a \times b = \frac{ab}{2} \]

### Step 2: Find the Area of the Semicircles
Three semicircles are drawn on the sides of the triangle. We will calculate the area of each semicircle separately.

#### Semicircle on side \( a \)
The radius of the semicircle on side \( a \) is \( \frac{a}{2} \). The area \( A_{S_a} \) of this semicircle is given by:
\[ A_{S_a} = \frac{1}{2} \pi \left(\frac{a}{2}\right)^2 = \frac{1}{2} \pi \frac{a^2}{4} = \frac{\pi a^2}{8} \]

#### Semicircle on side \( b \)
The radius of the semicircle on side \( b \) is \( \frac{b}{2} \). The area \( A_{S_b} \) of this semicircle is given by:
\[ A_{S_b} = \frac{1}{2} \pi \left(\frac{b}{2}\right)^2 = \frac{1}{2} \pi \frac{b^2}{4} = \frac{\pi b^2}{8} \]

#### Semicircle on the hypotenuse
To find the area of the semicircle on the hypotenuse, we first need to find the length of the hypotenuse \( c \) using the Pythagorean theorem:
\[ c = \sqrt{a^2 + b^2} \]
The radius of the semicircle on the hypotenuse is \( \frac{c}{2} = \frac{\sqrt{a^2 + b^2}}{2} \). The area \( A_{S_c} \) of this semicircle is:
\[ A_{S_c} = \frac{1}{2} \pi \left(\frac{\sqrt{a^2 + b^2}}{2}\right)^2 = \frac{1}{2} \pi \frac{a^2 + b^2}{4} = \frac{\pi (a^2 + b^2)}{8} \]

### Step 3: Total Area Calculation
Now, we can find the total area \( A_{total} \) by adding the area of the triangle and the areas of the three semicircles:
\[ A_{total} = A_{\Delta} + A_{S_a} + A_{S_b} + A_{S_c} \]
Substituting the values we calculated:
\[ A_{total} = \frac{ab}{2} + \frac{\pi a^2}{8} + \frac{\pi b^2}{8} + \frac{\pi (a^2 + b^2)}{8} \]
Combining the semicircle areas:
\[ A_{total} = \frac{ab}{2} + \frac{\pi a^2}{8} + \frac{\pi b^2}{8} + \frac{\pi a^2}{8} + \frac{\pi b^2}{8} = \frac{ab}{2} + \frac{\pi (a^2 + b^2)}{4} \]

### Final Answer
Thus, the total area of the triangle and the semicircles together is:
\[ A_{total} = \frac{ab}{2} + \frac{\pi (a^2 + b^2)}{4} \]

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