Question

In triangle ABC, \( AB=26 \) cm, \( AC=24 \) cm and \( \angle C=90^{\circ} \). If \( r_1, r_2, r_3 \) are the areas of the semi-circles with AB, AC, BC as diameters respectively, then \( \frac{1}{\pi}(r_1 + 2r_2 + 3r_3) \) (in \( cm^2 \)) is

• A. 246
• B. 266
• C. 194
• D. 169

Hint:

The relationship between areas of semi-circles on the sides of a right triangle is a specific application of the Pythagorean theorem.

Answer 1

The Correct Option is B

Concept: The area of a semi-circle with diameter \( d \) is given by \( \frac{\pi d^2}{8} \). We use the Pythagorean theorem for the right-angled triangle ABC: \( AB^2 = AC^2 + BC^2 \).

Step 1: Find the length of side BC.
\[ BC = \sqrt{AB^2 - AC^2} = \sqrt{26^2 - 24^2} = \sqrt{676 - 576} = 10 \text{ cm} \]

Step 2: Calculate the areas of the semi-circles.
\[ r_1 = \frac{\pi(26^2)}{8} = 84.5\pi, \quad r_2 = \frac{\pi(24^2)}{8} = 72\pi, \quad r_3 = \frac{\pi(10^2)}{8} = 12.5\pi \]

Step 3: Evaluate the expression \( \frac{1}{\pi}(r_1 + 2r_2 + 3r_3) \).
\[ \frac{1}{\pi}(84.5\pi + 2(72\pi) + 3(12.5\pi)) = 84.5 + 144 + 37.5 = 266 \] \centerline{{266}}