Compute the area of trapezium PQRS in the figure.

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Solution

In $△ T Q R, ∠ R T Q = 90^{\circ} ∴ Q R^{2} = T Q^{2} + R T^{2} \Rightarrow (17)^{2} = T Q^{2} + R T^{2} 289 = 64 + R T^{2} \Rightarrow R T^{2} = 289 - 64 = 225 = (15)^{2} ∴ R T = 15 c m a n d P Q = 8 + 8 = 16 c m$

Now areA of trapezium PQRS,
$= \frac{1}{2} (s u m o f p a r a l l e l s i d e s) \times h e i g h t = \frac{1}{2} (16 + 8) \times 15 c m^{2} = \frac{1}{2} \times 24 \times 15 = 180 c m^{2}$

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