WAEC 2013 · Paper 2 · Q10

(a)
A segment of a circle is cut off from a rectangular board $22\text{ cm}$ by $12\text{ cm}$ as shown in the diagram, leaving $5\text{ cm}$ and $3\text{ cm}$ of the base on either side of the chord. If the radius of the circle is $1\frac12$ times the length of the chord, calculate, correct to 2 decimal places, the perimeter of the remaining portion. $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
Evaluate, without using calculators or tables, $\dfrac{3}{\sqrt3}\left(\dfrac{2}{\sqrt3} - \dfrac{\sqrt{12}}{6}\right)$ .

Worked solution (try it first)

The chord is what is left of the 22 cm base: $22 - 5 - 3 = 14$ cm.
The radius is $1\frac12 \times 14 = 21$ cm.
Find the angle $x$ at the centre: the perpendicular from the centre halves the chord, so $\sin\frac{x}{2} = \frac{7}{21} = \frac13$ , giving $\frac{x}{2} \approx 19.47^\circ$ and $x \approx 38.94^\circ$ .
Arc $= \frac{38.94}{360} \times 2 \times \frac{22}{7} \times 21$
$\approx 14.28$ cm.
The remaining board's edge is the top (22), the two sides (12 and 12), the two pieces of base (5 and 3) and the arc instead of the chord: $22 + 12 + 12 + 5 + 3 + 14.28 = 68.28$ cm.

$\frac{3}{\sqrt3} = \sqrt3$ and $\frac{\sqrt{12}}{6} = \frac{2\sqrt3}{6}$
$= \frac{\sqrt3}{3}$ .
So the expression is $\sqrt3\left(\frac{2}{\sqrt3} - \frac{\sqrt3}{3}\right) = 2 - 1$
$= 1$ .