Question 17:
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
Previous QuestionNext 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
- $9\pi-18$
- 18
- $9\pi$
- 9
Option: 2
Explanation:
Explanation:
The image of the figure is as shown.
AB = AC = 6cm. Thus, BC = $\sqrt{6^2 + 6^2}$ = 6?2 cm
The required area = Area of semi-circle BQC - Area of quadrant BPC + Area of triangle ABC
Area of semicircle BQC
Diameter BC = 6?2cm
Radius = 6?2/2 = 3?2 cm
Area = $\pi r^2 $/2 = $ \pi$ × $(3 \sqrt{2})^2 $/2 = 9$\pi$
Area of quadrant BPC
Area = $\pi r^2$/4 = $\pi×(6)^2$/4 = 9$\pi$
Area of triangle ABC
Area = 1/2 × 6 × 6 = 18
The required area = Area of semi-circle BQC - Area of quadrant BPC + Area of triangle ABC
= 9$\pi$ - 9$\pi$ + 18 = 18
Previous QuestionNext Question