CAT 2022Slot 3QAQuestion & Solution
Question
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 Band C. Then the area, in sq. cm, of the overlapping region between the two circles is
Options
$16\pi$
$16(\pi - 1)$
$32(\pi - 1)$
$32\pi$
Solution
BC is the diameter of circle C2 so we can say that $\angle BAC=90^{\circ\ }$ as angle in the semi circle is $90^{\circ\ }$
Therefore overlapping area = $\frac{1}{2}$(Area of circle C2) + Area of the minor sector made be BC in C1
AB= AC = 8 cm and as $\angle BAC=90^{\circ\ }$, so we can conclude that BC= $8\sqrt{2}$ cm
Radius of C2 = Half of length of BC = $4\sqrt{2}$ cm
Area of C2 = $\pi\left(4\sqrt{2}\right)^2=32\pi$ $cm^2$
A is the centre of C1 and C1 passes through B, so AB is the radius of C1 and is equal to 8 cm
Area of the minor sector made be BC in C1 = $\frac{1}{4}$(Area of circle C1) - Area of triangle ABC = $\frac{1}{4}\pi\left(8\right)^2-\left(\frac{1}{2}\times8\times8\right)=16\pi-32$ $cm^2$
Therefore,
Overlapping area between the two circles= $\frac{1}{2}$(Area of circle C2) + Area of the minor sector made be BC in C1
= $\frac{1}{2}\left(32\pi\right)\ +\left(16\pi-32\right)=32\left(\pi-1\right)$ $cm^2$