WAEC 2014 · Paper 2 · Q4

(a)
In the diagram, $O$ is the centre of the circle of radius $r\text{ cm}$ and $\angle XOY = 90^\circ$ . If the area of the shaded segment $XKY$ is $504\text{ cm}^2$ , calculate the value of $r$ . $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
Two isosceles triangles $PQR$ and $PQS$ are drawn on opposite sides of a common base $PQ$ . If $\angle PQR = 66^\circ$ and $\angle PSQ = 109^\circ$ , calculate the value of $\angle RQS$ .

Worked solution (try it first)

(a)

The shaded segment is the quarter-circle sector $XOY$ minus the right-angled triangle $XOY$ .
Sector $= \frac{90}{360} \times \frac{22}{7} \times r^2$
$= \frac{11r^2}{14}$ .
Triangle $= \frac12 r^2$ .
Segment $= \frac{11r^2}{14} - \frac{7r^2}{14}$
$= \frac{4r^2}{14}$
$= \frac{2r^2}{7}$ .
So $\frac{2r^2}{7} = 504$ , $r^2 = 1764$ and $r = 42$ cm.

(b)

Triangle $PQS$ is isosceles with the $109^\circ$ angle at $S$ between the equal sides, so its base angles are $\frac12(180^\circ - 109^\circ) = 35.5^\circ$ : $\angle PQS = 35.5^\circ$ .
The triangles are on opposite sides of $PQ$ , so at $Q$ the angles add: $\angle RQS = \angle PQR + \angle PQS$
$= 66^\circ + 35.5^\circ$
$= 101.5^\circ$ .

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