WAEC 2023 · Paper 2 · Q3

Plane mensuration

In the diagram, $PQR$ is an equilateral triangle of side $18\text{ cm}$ . $M$ is the midpoint of $\overline{QR}$ . An arc of a circle with centre $P$ touches $\overline{QR}$ at $M$ and meets $\overline{PQ}$ at $A$ and $\overline{PR}$ at $B$ . Calculate, correct to two decimal places, the area of the shaded region. $\left[\text{Take } \pi = \frac{22}{7}\right]$

(a)
Area of the shaded region (cm²)

Worked solution (try it first)

(a)

The shaded region is the triangle minus the sector $PAMB$ .
The arc touches $QR$ at $M$ , so the radius is $PM$ , the height of the triangle: $PM = \sqrt{18^2 - 9^2}$
$= \sqrt{243}$
$= 9\sqrt3$
$\approx 15.59$ cm.
The angle at $P$ is $60^\circ$ (equilateral triangle).
Sector $= \frac{60}{360} \times \frac{22}{7} \times 243$
$\approx 127.29\text{ cm}^2$ .
Triangle $PQR = \frac12 \times 18 \times 9\sqrt3$
$\approx 140.30\text{ cm}^2$ .
Shaded area $\approx 140.30 - 127.29$
$= 13.01\text{ cm}^2$ .

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