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Find the area of the shaded region in the figure, if PQ=24cm, PR=7cm and O is the center of the circle.

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101.98 cm2

161.54 cm2

101.54 cm2

None of these

answer is B.

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Detailed Solution

We know that any angle made by the diameter QR in the semicircle is 90∘
Hence ∠RPQ=90∘
In the right angled ΔRPQ

R Q^{2} = P Q^{2} + P R^{2}

(By Pythagoras Theorem)

= 24^{2} + 7^{2}

= 576 + 49

= 625

➔

RQ = \sqrt{625}

➔

RQ = 25 cm

We know that RQ is the diameter of circle so the radius of circle is

OQ = \frac{RQ}{2} = \frac{25}{2} cm

The area of right angled

ΔRPQ :

= \frac{1}{2} \times base \times height

= \frac{1}{2} \times RP \times PQ

= \frac{1}{2} \times 7 \times 24

= 7 \times 12

= 84 c m^{2}

Now, The area of semicircle:

= \frac{π r^{2}}{2}

= \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2} \times 12

= \frac{11 \times 25 \times 25}{28}

= \frac{6875}{28} c m^{2}

The area of shaded region:
= Area of semicircle -Area of right angled
=

\frac{6875}{28} - 84

= \frac{6875 - 2352}{28}

= \frac{4523}{28}

= 161.54 c m^{2}

Hence the area of the shaded region is

161.54 c m^{2} .

The correct answer is Option (2).

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