Xavier Aptitude Test 2020 Paper

Question : 55 of 100

Marks: +1, -0

In the figure given below, the circle has a chord $\AB$ of length $12 \cm$ , which makes an angle of $60°$ at the center of the circle, $\O$ . $\ABCD$ , as shown in the diagram, is a rectangle. $\OQ$ is the perpendicular bisector of $\AB$ , intersecting the chord $\AB$ at $\P$ , thearc $\AB$ at $\M$ and $\CD$ at $\Q$ . $\OM = \MQ$ . The area of the region enclosed by the line segments $\AQ$ and $\QB$ ,and the arc $\BMA$ , is closest to (in $\cm^2$ ):

Validate

Solution:

In triangle

\OAP

, since

\AP=6

\OA

will be

12

\Area \of \AQBMA = \Area \of \Triangle \ABQ - (\Area \of \minor \arc \AMB- \Area \of \OAB)

Length of

\PQ = \MQ+\PM = \12+(\12- 6√3)=24-6√3

Area of Triangle

\ABQ= 1/2 . 12 . (24-6√3)=6 (24-6√3). =81.64

Area of minor

\arc \AMB- \Area \of \OAB = 60/300 . π . 144-1/2 . 12 . 6√3= 24π -36√3. = 13.07

∴

Area of

\AQBMA = 68.57 = 69

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