LMN is a triangle right angled at L. Semi-circles have been drawn on its sides. If MN = x, then sum of the areas of these semi-circles is:

$\frac{πx^2}{4}$

The correct answer is Option (3) → $\frac{πx^2}{4}$

Given, Hypotenuse MN, H = x

Assume, Perpendicular NL, P = y

Thus, the base B = (\( {x}^{2} \) - \( {y}^{2} \))^1/2

Area of Semi Circle at H with dia x, A1 = $\frac{πx^2}{8}$ 

Area of Semi Circle at P with dia y, A2 = $\frac{πy^2}{8}$

Area of Semi Circle at B with dia x, A3 = $\frac{πB^2}{8}$

Total Area = A1 + A2 + A3 = $\frac{πx^2}{4}$