**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 p = 3.142]

Given that Ð

*PAR*= q radians,**(a)**show that q = 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)**

In △BSA cosθ= 8 32 = 1 4 θ=1.32 rad (2 d.p.)

**(b)**

Ð

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

*QR*= 12 × 1.822

**= 21.86 cm**

**(c)**

PQ= 32 2 − 8 2 =30.98 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}