CAT 2022 Slot 3 Paper

Question : 60 of 66

Marks: +1, -0

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 over-lapping region between the two circles is

$16π$
$16π$
$32(π − 1)$
$32π$

Validate

Solution:

{\BC}

is the diameter of circle

{\C} 2

so we can say that

∠ \BAC=90^{∘}

as angle in the semi circle is

90°

Therefore overlapping area

={1}/{2}( \Area \of \circle \C2)+ \Area \of \the \minor \sector \made \be \BC

\C 1

{\AB}= {\AC}=8 {\cm}

and as

∠ \BAC=90°

, so we can conclude that

{\BC}=8 {√{2 \cm}}

Radius of

{\C} 2=

Half of length of

{\BC}=4 {√{2}} {\cm}

Area of

{\C} 2=π(4 {√{2}})^2=32 π {\cm}^2

\A

is the Centre of

\C 1

and

\C 1

passes through

\B

, so

\AB

is the radius of

\C 1

and is equal to

8 {\cm}

Area of the minor sector made be

{\BC}

{\C} 1={1}/{4} (\Area \of \circle {\C} 1) - \Area \of \triangle

{\ABC}={1}/{4} π(8)^2- ({1}/{2} × 8 × 8) =16 π-32 {\cm}^2

Therefore,
Overlapping area between the two circles

={1}/{2} (\Area \of \circle \C2) + \Area \of \the \minor \sector \made \be {\BC}

{\C} 1

={1}/{2}(32 π)+(16 π-32)=32(π-1) {\cm}^2

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