**Question 2**:

Diagram below shows two circles. The larger circle has centre

*A*and radius 20 cm. The smaller circle has centre*B*and radius 12 cm. The circles touch at point*R*. The straight line*PQ*is a common tangent to the circles at point*P*and point*Q*.[Use π = 3.142]

Given that angle

*PAR*= θ radians,**(a)**show that θ = 1.32 ( to two decimal places),

**(b)**calculate the length, in cm, of the minor arc

*QR*,

**(c)**calculate the area, in cm

^{2}, of the shaded region.

*Solution:***(a)**

$\begin{array}{l}\text{In}\u25b3BSA\\ \mathrm{cos}\theta =\frac{8}{32}=\frac{1}{4}\\ \text{}\theta =1.32\text{rad(2d}\text{.p}\text{.)}\end{array}$

**(b)**

Angle

*QBR*= 3.142 – 1.32 = 1.822 radLength of minor arc

*QR*= 12 × 1.822

**= 21.86 cm**

**(c)**

$PQ=\sqrt{{32}^{2}-{8}^{2}}=30.98\text{cm}$

Area of the shaded region

= Area of trapezium

*PQBA*– Area of sector*QBR*– Area of sector*PAR*= ½ (12 + 20) (30.98) – ½ (12)

^{2}(1.822) – ½ (20)^{2}(1.32)= 495.68 – 131.18 – 264

=

**100.5 cm**^{2}