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 = $$\frac{1}{2}\times base \times height$$)

= $$\frac{1}{2}\times 6 \times 2$$ + $$\frac{1}{2}\times 6 \times 3$$

= $$\frac{1}{2}\times 6(2 + 3)$$

= $$15 cm^2$$