WAEC - Mathematics (2001 - No. 45)
PQRS is a trapezium in which |PS| = 9cm, |QR| = 15cm, |PQ| = \(2\sqrt{3}, \angle PQR = 90^o and \angle QRS = 30^o\). Calculate the area of the trapezium
\(24\sqrt{3} cm^2\)
\(36\sqrt{3} cm^2\)
\(42\sqrt{3} cm^2\)
\(72\sqrt{3} cm^2\)
Explanation
Area of trapezium = \(\frac{1}{2}\)( a + b) x h
a = 9cm, b = 15cm, c = (15 - 9) = 6cm since height was not given, so we calculate for it using trig . ratio
h = 2\(\sqrt{3}\)cm
Area of trapezium = \(\frac{1}{2}\)( a + b ) x h = \(\frac{1}{2}\)( 9 + 15 ) x 2\(\sqrt{3}\)
= 24 x \(\sqrt{3}\) = 24\(\sqrt{3}\)cm\(^2\)
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