NTSE Stage 2 SAT 2015 Question Paper

Question : 49 of 100

Marks: +1, -0

A sector with acute central angle is cut from a circle of diameter 14cm. The area (in

\cm^2

) of the circle circumscribing the sector is

Solution:

∆=\abc/{4\R}

∴

R=\abc/{4∆}

={2×7×7×14\sin\(\θ/2)}/{4×49\sin\θ }

={7\sin(\θ /2)}/{2\sin(\θ /2)\cos(\θ /2)}

=7/2\sec(\θ /2)

Now, area of circle

\π\R^2

=22/7×[7/2\sec(\θ /2)]^2

=22/7×49/2\sec^2(\θ /2)

=77/2\sec^2(\θ /2)

∴ Option (d) is correct.

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