A, B, C, D are the midpoints of a rectangle. AB = 10 cm. If one of the sides of the rectangle is 12 cm, then area of figure ABCD is -

192

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240

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120

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Solution

The correct option is C
96

$\begin{matrix} P Q = 12 c m, \Rightarrow P B = 6 c m . A P = \sqrt{1 0^{2} - 6^{2}} = 8 P S = 2 \times 8 = 16. A r e a o f r e c t a n g l e P Q R S = P Q \times P S = 12 \times 16 = 192 A r e a o f A B C D = \frac{192}{2} = 96 c m^{2} \end{matrix}$

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