WAEC 2020 · Paper 2 · Q12

(a)
In the diagram, $\overline{AB}$ is a tangent to the circle with centre $O$ and $COEB$ is a straight line. If $\overline{CD} \parallel \overline{AB}$ and $\angle ABE = 40^\circ$ , find $\angle ODE$ .
(b)
$ABCD$ is a parallelogram in which $|CD| = 7\text{ cm}$ , $|AD| = 5\text{ cm}$ and $\angle ADC = 125^\circ$ . (i) Illustrate the information in a diagram. (ii) Find, correct to one decimal place, the area of the parallelogram.
(c)
If $x = \frac12(1 - \sqrt2)$ , evaluate $(2x^2 - 2x)$ .

Worked solution (try it first)

$CD \parallel AB$ , so alternate angles are equal: $\angle OCD = \angle ABE = 40^\circ$ .
$OC = OD$ (radii), so $\angle ODC = \angle OCD = 40^\circ$ .
$CE$ is a diameter, so the angle in a semicircle is a right angle: $\angle CDE = 90^\circ$ .
Then $\angle ODE = 90^\circ - 40^\circ$
$= 50^\circ$ .

Draw $DC$ (7 cm) along the bottom and $DA$ (5 cm) at $125^\circ$ to it, then complete the parallelogram.

Area $= DC \times DA \times \sin\angle ADC$
$= 7 \times 5 \times \sin 125^\circ$
$\approx 35 \times 0.8192$
$\approx 28.7\text{ cm}^2$ .

$2x^2 - 2x = 2x(x - 1)$ .
Here $2x = 1 - \sqrt2$ and $x - 1 = \frac{1 - \sqrt2}{2} - 1$
$= \frac{-1 - \sqrt2}{2}$ .
So $2x(x - 1) = \frac{(1 - \sqrt2)(-1 - \sqrt2)}{2}$
$= \frac{-(1 - \sqrt2)(1 + \sqrt2)}{2}$
$= \frac{-(1 - 2)}{2}$
$= \frac12$ .