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CAT 2017 Slot 1 — QA Question 17

Basics of TrianglesEasy

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:

Answer & solution

  • A

    9π - 18

  • 18

  • C

    9π

  • D

    9

Solution

AB = a (a = 6)
BC = a√2

BCQB is a semicircle of radius = half of BC = a2
Area of semicircle BCQB = ½ × π × (a2)2 = ¼ × π × a2 

ACPBA is a quarter circle (quadrant) of radius a.
Area of BCPB = Area of sector ABPCA - Area of triangle ABC
                     = ¼ × π × a2 - Area of triangle ABC

Now, Area of CPBQC = Area of semicircle CQB - Area of CPBC
                                   =  ¼ × π × a2 - ( ¼ × π × a2 - Area of triangle ABC)
                                   = Area of triangle ABC

∴ Area of region enclosed by CPBQC = Area of ∆ABC = ½ × 6 × 6 = 18.

Hence, option (b).

CAT 2017 Slot 1 QA Q17: Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with d — Solution | TheCATExam