O is the centre of the circle passing through B, E, D and C. Angle $DOC = 84^\circ$ and angle $CBO = 54^\circ$. $PQ$ is a tangent to the circle at D. Find, giving reasons for each answer, angle BAE, angle QDC; angle ECB, angle CBE. The diagram shows 3 circles with diameters d cm, 2d cm and 3d cm respectively. Find, in terms of π and d, the difference in area between the unshaded region and the shaded region.

1. Find the angle between radius $OD$ and chord $DC$.

Since $OD=OC$ radii of the circle, $\triangle DOC$ is isosceles.

$\angle ODC=\angle OCD=\frac{180^\circ-84^\circ}{2}=48^\circ$

2. Use the tangent-radius property to find $\angle QDC$.

Since $PQ$ is a tangent at $D$, $OD\perp QD$.

Hence
$\angle QDC=90^\circ-\angle ODC=90^\circ-48^\circ=42^\circ.$

1. Write the radii of the three circles from their diameters.

The diameters are $d$, $2d$ and $3d$.

So the radii are
$\frac d2,\quad d,\quad \frac{3d}{2}.$

2. Find the total unshaded area.

Unshaded area = area of inner circle + area of outer annulus.

$\pi\left(\frac d2\right)^2+\left[\pi\left(\frac{3d}{2}\right)^2-\pi d^2\right]$

$=\frac14\pi d^2+\left(\frac94\pi d^2-\pi d^2\right)$

$=\frac14\pi d^2+\frac54\pi d^2$

$=\frac32\pi d^2$

3. Find the shaded area and subtract it from the unshaded area.

Shaded area = middle circle area $-$ inner circle area.

$\pi d^2-\pi\left(\frac d2\right)^2$

$=\pi d^2-\frac14\pi d^2$

$=\frac34\pi d^2$

Difference between unshaded and shaded area:

$\frac32\pi d^2-\frac34\pi d^2=\frac34\pi d^2$