WAEC 2023 · Paper 2 · Q13

(a)
In the diagram, the radius of the sector of the circle centre $O$ is $7\text{ cm}$ and $\angle MON = 60^\circ$ ; $NT \perp OM$ . Find, correct to one decimal place, the area of the shaded portion. $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
The $x$ and $y$ intercepts of a straight line are $-\frac34$ and $\frac27$ respectively. Find the equation of the line.

Worked solution (try it first)

(a)

The shaded portion is sector $MON$ minus the right-angled triangle $ONT$ .
Sector $= \frac{60}{360} \times \frac{22}{7} \times 7^2$
$\approx 25.67\text{ cm}^2$ .
In triangle $ONT$ (right angle at $T$ ), $OT = 7\cos 60^\circ = 3.5$ cm and $NT = 7\sin 60^\circ \approx 6.062$ cm, so its area is $\frac12 \times 3.5 \times 6.062 \approx 10.61\text{ cm}^2$ .
Shaded area $\approx 25.67 - 10.61 = 15.1\text{ cm}^2$ .

(b)

The line passes through $\left(-\frac34, 0\right)$ and $\left(0, \frac27\right)$ .
Gradient $= \frac{2/7 - 0}{0 - (-3/4)}$
$= \frac{2}{7} \times \frac43$
$= \frac{8}{21}$ , so $y = \frac{8}{21}x + \frac27$ .
Multiply by 21: $21y = 8x + 6$ .

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