JAMB 1990 · UME · Q41

Plane mensuration

In the figure, $PQR$ is a semicircle on $PR$ , below the rectangle $OPRS$ with $OP = 11$ cm. The triangle $OTS$ , with $OT = 6$ cm, $TS = 8$ cm and a right angle at $T$ , is not shaded. Calculate the area of the shaded region. $\left[\pi = \frac{22}{7}\right]$

Worked solution (try it first)

In the right-angled triangle $OTS$ , by Pythagoras, $OS = \sqrt{6^2 + 8^2} = 10$ cm.
So $PR = 10$ cm and the semicircle has radius 5 cm.
Rectangle: $10 \times 11 = 110\text{ cm}^2$ .
Triangle $OTS$ : $\frac12 \times 6 \times 8 = 24\text{ cm}^2$ .
Semicircle: $\frac12 \times \frac{22}{7} \times 5^2 = \frac{275}{7}$
$= 39\frac27\text{ cm}^2$ .
Shaded = rectangle − triangle + semicircle: $110 - 24 + 39\frac27 = 125\frac27\text{ cm}^2$ , option A.

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