Question

Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC is

• A. \(9\pi-18\)
• B. 18
• C. \(9\pi\)
• D. 9

Answer 1

The Correct Option is B

Step 1: Known values

Let: \[ AB = a, \quad a = 6 \]

Step 2: Radii of the arcs

• Arc \( BQC \) is a semicircle with radius: \[ \frac{a}{\sqrt{2}} \]
• Arc \( BPC \) is a quarter circle (quadrant) with radius: \[ a \]

Step 3: Areas of the arcs

Area of semicircle \( BQC \): \[ \text{Area} = \frac{\pi \left( \frac{a}{\sqrt{2}} \right)^2}{2} = \frac{\pi a^2}{4} \]

Area of quadrant \( BPC \): \[ \text{Area} = \frac{\pi a^2}{4} \]

Step 4: Enclosed region

The region enclosed by curves \( BPC \) and \( BQC \) corresponds exactly to the area of triangle \( ABC \).

Step 5: Area of triangle ABC

Given: \[ \text{Area}_{\triangle ABC} = 18 \] Thus, the enclosed region area is also: \[ \boxed{18} \]