JAMB 1983 · UME · Q45

Plane mensuration

$PQR$ is the diameter of a semicircle $RSP$ with centre $Q$ and radius 3.5 cm. If $\angle QPT = \angle QRT = 60^\circ$ , find the perimeter of the figure. $\left[\pi = \frac{22}{7}\right]$

Worked solution (try it first)

In triangle $PTR$ the angles at $P$ and $R$ are both $60^\circ$ , so the third angle is $60^\circ$ too: the triangle is equilateral with side $PR = 2 \times 3.5 = 7$ cm.
The semicircular arc is half the circumference: $\frac12 \times 2 \times \frac{22}{7} \times 3.5 = 11$ cm.
The perimeter is the arc plus $PT$ and $TR$ : $11 + 7 + 7 = 25$ cm, option A.

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