NECO 2023 · Paper 2 · Q3

In the diagram, $O$ is the centre of the circle, the chord $AB$ is $12\text{ cm}$ long and $\angle ACB = 30^\circ$ .

(a)
Find (i) the value of $\theta$ ; (ii) the radius of the circle.
Separate values with commas, e.g. 3, −2
(b)(i)
Calculate the area of the shaded region, correct to three significant figures. $\left[\pi = \frac{22}{7}\right]$
(b)(ii)
What type of triangle is $\triangle AOB$ ?
Show the answer
Equilateral

Worked solution (try it first)

(a)(i)

The angle at the centre is twice the angle at the circumference on the same arc: $\theta = 2 \times 30^\circ = 60^\circ$ .

(ii)

$OA = OB$ (radii), so triangle $AOB$ is isosceles.
With a $60^\circ$ angle between the equal sides, the other two angles are also $60^\circ$ .
So it is equilateral, and the radius equals the chord: $r = 12$ cm.

(b)(i)

The shaded region is the minor segment: sector minus triangle.
Sector $= \frac{60}{360} \times \frac{22}{7} \times 12^2$
$\approx 75.43\text{ cm}^2$ .
Triangle $= \frac12 \times 12 \times 12 \times \sin 60^\circ$
$\approx 62.35\text{ cm}^2$ .
Segment $\approx 75.43 - 62.35 = 13.08$ , which is $13.1\text{ cm}^2$ to three significant figures.

(ii)

Triangle $AOB$ is equilateral.

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