Question 75
In quadrilateral PQRS, RM $$\perp$$ QS, PN $$\perp$$ QS and QS = 6 cm. If RM = 3 cm and PN = 2 cm, then the area of PQRS is
Solution
Area of PQRS = area of $$\triangle$$PQS + area of $$\triangle$$RQS
(Area of triangle = $$\frac{1}{2}\times base \times height$$)
= $$\frac{1}{2}\times 6 \times 2$$ + $$\frac{1}{2}\times 6 \times 3$$
= $$\frac{1}{2}\times 6(2 + 3)$$
= $$15 cm^2$$
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