Multiple choice

Directions: Solve the following question and mark the best possible option. 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

  1. 9 π - 18

  2. 18

  3. 9 π

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Area of semi-circle BQC = 1/2 * pi * r^2. With AB=6, AC=6, BC = 6*sqrt(2), so radius r = 3*sqrt(2). Area = 1/2 * pi * (3*sqrt(2))^2 = 9*pi. Area of sector BPC = 1/4 * pi * 6^2 = 9*pi. The region enclosed is the difference between the sector and the triangle area, but the question asks for the region between BPC and BQC, which is 18.

AI explanation

The area between the arc BPC centered at A and the line segment BC is the difference between the area of the sector BPC and triangle ABC. The angle at A is 90 degrees, so the area of the quadrant BPC is 9 pi and the area of the right triangle ABC is 18, making the enclosed region 9 pi minus 18. The radius of the semicircle BQC is 3 root 2, making its area 9 pi; subtracting the region between the arc and the line (9 pi minus 18) leaves exactly 18 square cm.

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