WAEC 2024 · Paper 2 · Q9

The diameter of a circle centre $O$ is $26\text{ cm}$ . A chord $PQ$ is drawn such that its midpoint is $5\text{ cm}$ from $O$ . Calculate, correct to the nearest whole number:

(a)
$\angle POQ$ (degrees);
(b)
the area of the minor segment cut off by $PQ$ (cm²). $\left[\text{Take } \pi = \frac{22}{7}\right]$

Worked solution (try it first)

The radius is $\frac{26}{2} = 13$ cm.
Let $M$ be the midpoint of $PQ$ .
$OM \perp PQ$ and $OM = 5$ cm.

(a)

In the right-angled triangle $OMP$ : $\cos\angle MOP = \frac{5}{13}$ , so $\angle MOP \approx 67.38^\circ$ and $\angle POQ = 2 \times 67.38^\circ$
$\approx 134.76^\circ$ , which is $135^\circ$ to the nearest degree.
Also $PM = \sqrt{13^2 - 5^2} = 12$ cm, so $PQ = 24$ cm.

(b)

Sector $= \frac{134.76}{360} \times \frac{22}{7} \times 13^2$
$\approx 198.8\text{ cm}^2$ .
Triangle $OPQ = \frac12 \times 24 \times 5$
$= 60\text{ cm}^2$ .
Segment $\approx 198.8 - 60 = 138.8$ , which is $139\text{ cm}^2$ to the nearest whole number.

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