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1.5.2 Circular Measure Long Questions (Question 3 & 4 )


Question 3:
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 cm2, of the shaded region.



Solution:
(a)

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

(b)
Angle QBR = 3.142 – 1.32 = 1.822 rad
Length 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 cm2


Question 4:
Diagram below shows a sector QPR with centre P and sector POQ, with centre O.

It is given that OP = 17 cm and PQ = 8.8 cm.
[Use π = 3.142]
Calculate
(a) angle OPQ, in radians,
(b) the perimeter, in cm, of sector QPR,
(c) the area, in cm2, of the shaded region.


Solution:
( a ) ∠OPQ=∠OQP x+x+30=180    2x=150   x=75 ∠OPQ= 75×3.142 180    =1.3092 radians


( b ) Length of arc QR=rθ    =8.8×1.3092    =11.52 cm Perimeter of sector QPR =11.52+8.8+8.8 =29.12 cm


( c ) 30 o = 30×3.142 180 =0.5237 rad Area of segment PQ = 1 2 r 2 ( θ−sinθ ) = 1 2 × 17 2 ×( 0.5237−sin30 ) = 1 2 ×289×( 0.5237−0.5 ) =3.4247  cm 2 Area of sector QPR = 1 2 r 2 θ = 1 2 × 8.8 2 ×1.3092 =50.692  cm 2 Area of shaded region =3.4247+50.692 =54.1167  cm 2

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