WAEC 2021 · Paper 2 · Q9✱✱

(a)
A tailor had a piece of cloth in the shape of a trapezium. The perpendicular distance between the two parallel edges was $38\text{ cm}$ . The lengths of the two parallel edges are $46\text{ cm}$ and $60\text{ cm}$ . The tailor cut off a semi-circular piece of the cloth of radius $18\text{ cm}$ from the $60\text{ cm}$ edge. Calculate, correct to one decimal place, the area of the remaining piece of cloth. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

Area of the trapezium $= \frac12(\text{sum of the parallel sides}) \times \text{height}$
$= \frac12(46 + 60) \times 38$
$= 53 \times 38$
$= 2014\text{ cm}^2$ .
Area of the semicircle cut off $= \frac12 \pi r^2$
$= \frac12 \times \frac{22}{7} \times 18^2$
$= \frac{3564}{7}$
$\approx 509.14\text{ cm}^2$ .
Remaining cloth $= 2014 - 509.14 = 1504.86\text{ cm}^2$ .
Correct to one decimal place: $1504.9\text{ cm}^2$ .