WAEC 2023 · Paper 2 · Q3

An isosceles triangle $PQR$ has its vertices on the circumference of a circle. If $|PQ| = |QR| = 17\text{ cm}$ , $|PR| = 16\text{ cm}$ and $M$ is the midpoint of $\overline{PR}$ , calculate:

(a)
$|QM|$ ;
(b)
correct to the nearest whole number, the radius of the circle.

Worked solution (try it first)

(a)

Triangle $PQR$ is isosceles with $|PQ| = |QR|$ , so the line from $Q$ to the midpoint $M$ of $PR$ is perpendicular to $PR$ .
$|PM| = \frac12 \times 16 = 8$ cm.
In the right-angled triangle $QMP$ : $|QM| = \sqrt{17^2 - 8^2}$
$= \sqrt{289 - 64}$
$= \sqrt{225}$
$= 15$ cm.

(b)

The centre $O$ of the circle is on the perpendicular bisector of the chord $PR$ , which is the line $QM$ .
Let the radius be $r$ : $|OQ| = |OP| = r$ and $|OM| = 15 - r$ .
In the right-angled triangle $OMP$ : $r^2 = (15 - r)^2 + 8^2$
$= 225 - 30r + r^2 + 64$ .
The $r^2$ terms cancel: $30r = 289$ , so $r = \frac{289}{30} \approx 9.63$ .
Correct to the nearest whole number, the radius is 10 cm.

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