1. 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





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  • By: anil on 05 May 2019 02.29 am
    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 [u]Area of semicircle BQC[/u] Diameter BC = 6√2cm Radius = 6√2/2 = 3√2 cm Area = $$pi r^2 $$/2 = $$ pi$$ * $$(3 sqrt{2})^2 $$/2 = 9$$pi$$ [u]Area of quadrant BPC[/u] Area = $$pi r^2$$/4 = $$pi*(6)^2$$/4 = 9$$pi$$ [u]Area of triangle ABC[/u] 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
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