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  • Find the area of the trapezium PQRS with height PQ given in Figure.

Find the area of the trapezium PQRS with height PQ given in Figure.

Answers (1)

[114\; cm^{2}]

Firstly\; we\; have\; side \; PS = 12 m, SR = 13, QR = 7 m

                                       

Join RT

So here PT = PS – ST

PT = 12 m – 5 m

PT = 7 m

and ST = PS – PT

ST = \left ( 12-7 \right )m

ST = 5 m

Now, In \DeltaSTR, Using Pythagoras theorem

We\; get, (SR)^{2} = (ST)^{2} + (TR)^{2}

(13)^{2} = (5)^{2} + (TR)^{2}

169 = 25 + (TR)^{2}\; \; \; \; \; \; \; \; \; \; \; [(13)^{2} = 169 \; and\; (5)^{2} = 25]

169 - 25 = (TR)^{2}

144 = (TR)^{2}\; \; \; \; \; \; \; \; \; \; \; \; [ \sqrt{144} = 12]

TR = 12cm\; \; \; \; \; \; \; \; \; \; \; \; [ \sqrt{12\times 12} = 12]

Now, we can find the area of trapezium

Area \; of\; trapezium = \frac{1}{2} \times [sum\; of \; parallel\; sides] \times height

= \frac{1}{2}\times \left [ 12+7 \right ]\times 12

= 19\times 6\; cm^{2}

= 114\; cm^{2}

Hence, the\; area\; of\; trapezium\; is\; 114\; cm^{2}.

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