WAEC 2014 · Paper 2 · Q10

(a)
Solve: $(x - 2)(x - 3) = 12$ .
Separate values with commas, e.g. 3, −2
(b)
In the diagram, $M$ and $N$ are the centres of two circles of equal radii $7\text{ cm}$ . The circles intersect at $P$ and $Q$ . If $\angle PMQ = \angle PNQ = 60^\circ$ , calculate, correct to the nearest whole number, the area of the shaded portion (the overlap). $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

Expand and bring everything to one side: $x^2 - 5x + 6 = 12$ , so $x^2 - 5x - 6 = 0$ .
Factorise: $(x - 6)(x + 1) = 0$ , so $x = 6$ or $x = -1$ .

The overlap is two equal segments back to back, one in each circle, on the common chord $PQ$ .
Each is a $60^\circ$ sector minus the triangle on the chord.
Sector $= \frac{60}{360} \times \frac{22}{7} \times 7^2$
$\approx 25.667\text{ cm}^2$ .
Triangle $= \frac12 \times 7 \times 7 \times \sin 60^\circ$
$\approx 21.218\text{ cm}^2$ .
One segment $\approx 4.449\text{ cm}^2$ .
Shaded area $= 2 \times 4.449 \approx 8.9$ , which is $9\text{ cm}^2$ to the nearest whole number.