Find the area (in $m^{2}$ ) of the trapezium PQRS with height PQ given in the figure.

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Solution

$G i v e n i s a t r a p e z i u m P Q R S w h e r e P S ∥ Q R a n d ∠ P & ∠ Q a r e 90^{o} . D r a w a ⊥ R T f r o m p o i n t R o n t o S P . N o w R T = P Q a n d R T ∥ P Q . H e n c e S T = 12 - 7 = 5 m A p p l y i n g P y t h a g o r a s^{'} T h e o r e m i n △ S T R, T R = \sqrt{(13^{2} - 5^{2})} = \sqrt{(169 - 25)} = \sqrt{144} = 12 m S o, t h e P Q = h e i g h t o f t h e t r a p e z i u m = 12 m T h e a r e a o f t h e t r a p e z i u m P Q R S = \frac{1}{2} \times P Q \times (R Q + S P) = \frac{1}{2} \times 12 \times (7 + 12) = 6 \times 19 = 114.$

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In the trapezium PQRS shown below, PQ is parallel to RS. Find its area.

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