JAMB - Mathematics (1979 - No. 48)
In the parallelogram PQRS, PE is perpendicular to QR. Find the area of the parallelogram.
60cm2
65cm2
72cm2
132cm2
156cm2
Wyjaśnienie
By Pythagoras, PE\(^2\) = 12\(^2\) - 5\(^2\)
= 144 - 25 = 119
h = PE\(^2\) = √119 = 10.9 ≈ 11cm,
Area of 11gm = b x h
QR = b =(5 + 7)cm = 12cm
area = 12 x 11
= 132cm\(^2\)