WAEC 2010 · Paper 2 · Q11

(a)
In the diagram, $\angle PTQ = \angle PSR = 90^\circ$ , $|PQ| = 10\text{ cm}$ , $|PS| = 14.4\text{ cm}$ and $|TQ| = 6\text{ cm}$ . Calculate the area of quadrilateral $QRST$ .
(b)
Two opposite sides of a square are each decreased by $10\%$ while the other two are each increased by $15\%$ to form a rectangle. Find the ratio of the area of the rectangle to that of the square.

Worked solution (try it first)

Pythagoras in triangle $PTQ$ : $|PT| = \sqrt{10^2 - 6^2}$
$= \sqrt{64}$
$= 8\text{ cm}$ .
$TQ \parallel SR$ (both are perpendicular to $PS$ ), so triangles $PTQ$ and $PSR$ are similar: $\dfrac{|SR|}{|TQ|} = \dfrac{|PS|}{|PT|}$ .
So $|SR| = 6 \times \dfrac{14.4}{8}$
$= 10.8\text{ cm}$ .
$QRST$ is a trapezium with parallel sides $6$ and $10.8$ , and height $|TS| = 14.4 - 8 = 6.4\text{ cm}$ .
Area $= \frac12(6 + 10.8) \times 6.4$
$= 53.76\text{ cm}^2$ .