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CAT 2022 Slot 3 — QA Question 2

2 CirclesEasy

In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through B and C. Then, the area in sq. cm, of the overlapping region between the two circles is

Answer & solution

  • A

    16(π - 1)

  • 32(π - 1)

  • C

    16π

  • D

    32π

Solution

Easy

AA lies on the circle with diameter BCBC, so ∠BAC=90∘\angle BAC=90^\circ (angle in a semicircle). That fixes both radii. The overlap is the half of the small circle on AA's side plus a circular segment of the large circle — add them up.

A B C
1

Find the two radii. BCBC is the diameter of the small circle and ∠BAC=90∘\angle BAC=90^\circ, so by Pythagoras:

BC=AB2+AC2=82+82=82rsmall=12⋅82=42,rlarge=AB=8\begin{aligned} &BC=\sqrt{AB^2+AC^2}=\sqrt{8^2+8^2}=8\sqrt2\\ &r_{\text{small}}=\tfrac12\cdot8\sqrt2=4\sqrt2,\qquad r_{\text{large}}=AB=8 \end{aligned}
2

Small-circle part. Chord BCBC is a diameter, so the part of the small circle on AA's side is a semicircle:

12πrsmall2=12π(42)2=16π\tfrac12\pi r_{\text{small}}^2=\tfrac12\pi(4\sqrt2)^2=16\pi
3

Large-circle part. With ∠BAC=90∘\angle BAC=90^\circ, the overlap from the big circle is the quarter-sector CABCAB minus triangle ABCABC:

14π(8)2−12(8)(8)=16π−32\tfrac14\pi(8)^2-\tfrac12(8)(8)=16\pi-32
4

Add the two pieces:

16π+(16π−32)=32π−32=32(π−1)16\pi+(16\pi-32)=32\pi-32=32(\pi-1)
Overlap area=32(π−1) sq. cm\text{Overlap area}=\mathbf{32(\pi-1)}\text{ sq. cm}
CAT 2022 Slot 3 QA Q2: In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn w — Solution | TheCATExam