Discussion
Perimeter of a quadrilateral ABCD is 120 cm. If BC = 48 cm, CD = 17 cm, AD = 40 cm and \(\angle A = \angle B = 90^{\circ}\), then area of quadrilateral ABCD (in cm2) is :
Answer: A
The area is 660 cm2.
First find AB from the perimeter. Since ∠A = ∠B = 90°, AD and BC are both perpendicular to AB, so AD ∥ BC and ABCD is a trapezium with height AB:
\[ \begin{aligned} AB &= 120 - (48 + 17 + 40) = 15 \\ \text{Area} &= \tfrac{1}{2}(AD + BC) \times AB \\ &= \tfrac{1}{2}(40 + 48) \times 15 \\ &= 44 \times 15 = 660 \end{aligned} \]
Check: the parallel sides differ by 8 and the height is 15, and \(\sqrt{8^2 + 15^2} = 17\), which is CD, so the figure is consistent. 720 is what you get by multiplying 48 by 15 as if it were a rectangle.
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