Question 93

In the given figure, area of isosceles triangle PQT is $$72cm^2$$. If QT = PQ, PQ = 2 PS and PTIISR, then what is the area $$(in cm^2)$$ of the trapezium PQRS?

Solution

Here, QT = PQ and PQ = 2 PS

Area of $$\triangle$$ PQT = $$\frac{1}{2}\times(PQ)\times(QT)=72$$

=> $$(PQ)^2=72\times2=144$$

=> $$PQ=\sqrt{144}=12$$ cm

=> PS = $$\frac{12}{2}=6$$ cm

Area of parallelogram PSRT = base $$\times$$ height

= $$(PS)\times(PQ)=6\times12=72$$ $$cm^2$$

$$\therefore$$ Area of trapezium PQRS = $$72+72=144$$ $$cm^2$$

=> Ans - (A)

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