WAEC 2024 · Paper 2 · Q4

Circle geometry

In the diagram, $O$ is the centre of the circle and $OBCD$ is a rhombus. $\angle ADO = \angle OBA = y$ and $\angle BAD = t$ . Find:

(a)
the value of $t$ ;
(b)
the value of $y$ ;
(c)
$\angle ADC$ .

Worked solution (try it first)

(a)

$OBCD$ is a rhombus, so $OB = BC = CD = DO$ .
Also $OB = OC = OD$ (radii).
So triangles $OBC$ and $OCD$ are equilateral, each with $60^\circ$ angles.
So $\angle BOD = 60^\circ + 60^\circ$
$= 120^\circ$ .
The angle at the circumference is half the angle at the centre: $t = \angle BAD$
$= \frac12 \times 120^\circ$
$= 60^\circ$ .

(b)

$OA = OB$ (radii), so $\angle OAB = \angle OBA = y$ .
Likewise $OA = OD$ gives $\angle OAD = \angle ODA = y$ .
So $t = \angle OAB + \angle OAD = 2y$ , and $2y = 60^\circ$ gives $y = 30^\circ$ .

(c)

$\angle ADC = \angle ADO + \angle ODC$
$= 30^\circ + 60^\circ$
$= 90^\circ$ .
(This fits: $A$ , $O$ and $C$ are on one line, so $AC$ is a diameter and $\angle ADC$ is an angle in a semicircle.)

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