WAEC 2020 · Paper 2 · Q13

(a)
The area of a triangle is $54\text{ cm}^2$ . If the height is $3\text{ cm}$ more than the base, find the length of the base.
(b)
In the diagram, $O$ is the centre of the circle. The points $X$ , $Y$ and $Z$ are on the circumference of the circle, $\angle ZXO = 38^\circ$ and $\angle XOY = 130^\circ$ . Find: (i) $\angle OYZ$ ; (ii) $\angle ZOY$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

Let the base be $x$ cm, so the height is $(x + 3)$ cm.
The area is $\frac12 \times \text{base} \times \text{height}$ : $\frac12 x(x + 3) = 54$ .
Multiply by 2: $x^2 + 3x = 108$ , so $x^2 + 3x - 108 = 0$ .
Factorise: $(x + 12)(x - 9) = 0$ .
A length can't be negative, so $x = 9$ .
The base is $9$ cm (and the height is 12 cm: $\frac12 \times 9 \times 12 = 54$ ✓).

(b)(i)

The angle at the centre is twice the angle at the circumference: $\angle XZY = \frac12 \times 130^\circ$
$= 65^\circ$ .
$OX = OY$ (radii), so $\angle OXY = \angle OYX$
$= \frac{180^\circ - 130^\circ}{2}$
$= 25^\circ$ .
In triangle $XYZ$ : $\angle ZXY = 38^\circ + 25^\circ$
$= 63^\circ$ , so $\angle XYZ = 180^\circ - 63^\circ - 65^\circ$
$= 52^\circ$ .
Then $\angle OYZ = 52^\circ - 25^\circ$
$= 27^\circ$ .

(ii)

$OY = OZ$ (radii), so $\angle OZY = \angle OYZ = 27^\circ$ , and $\angle ZOY = 180^\circ - 2 \times 27^\circ$
$= 126^\circ$ .

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