Circular Measure, Long Questions (Paper 2)


Question 7:
Diagram below shows a semicircle PTS, centre O and radius 8 cm. PTR is a sector of a circle with centre P and Q is the midpoint of OS.

[Use π = 3.142]
Calculate
(a)TOQ, in radians,
(b) the length , in cm , of the arc TR ,
(c) the area, in cm2 ,of the shaded region.

Solution:
(a)
cosTOQ= 4 8 = 1 2  TOQ= 60 o =60× π 180 =1.047 radians

(b)


TPO= 30 o   =30× π 180   =0.5237 P T 2 = 8 2 + 8 2 2( 8 )( 8 )cos120 P T 2 =192 PT= 192 PT=13.86 cm Length of arc TR=13.86×0.5237   =7.258 cm

(c)
Area of sector PTR = 1 2 × 13.86 2 ×0.5237 =50.30  cm 2 Length TQ = P T 2 P Q 2 = 13.86 2 12 2 =6.935 cm Area of  PTQ = 1 2 ×12×6.935 =41.61  cm 2 Area of shaded region =50.3041.61 =8.69  cm 2



Question 8:
Diagram below shows a circle PQT with centre O and radius 7 cm.

QS
is a tangent to the circle at point Q and QSR is a quadrant of a circle with centre Q. Q is the midpoint of OR and QP is a chord. OQR and SOP are straight lines.
[Use π = 3.142]
Calculate
(a) angle θ, in radians,
(b) the perimeter, in cm ,of the shaded region,
(c) the area, in cm2 ,of the shaded region.

Solution:
(a)
OQ=QR=QS=7 cm tanθ=1      θ= 45 o        = 45 o × π 180 o        =0.7855 rad

(b)
Length of arc RS =7×( 0.7855×2 ) π rad= 180 o 45 o =0.7855 rad 90 o =0.7855 ×2 rad =7×1.571 =10.997 cm Length of arc QP =7×( 0.7855×3 ) =7×2.3565 =16.496 cm Length of chord QP = 7 2 + 7 2 2( 7 )( 7 )cos 135 o refer form 4 chapter 10 ( solution of triangle )for cosine rule = 167.30 =12.934 cm Perimeter of the shaded region =7+7+10.997+16.496+12.934 =54.427 cm

(c)
Area of shaded region =( 1 2 × 7 2 ×1.571 )+( 1 2 × 7 2 ×2.3565 ) ( 1 2 ×7×7×sin 135 o ) refer form 4 chapter 10 ( solution of triangle )for area rule =38.4895+57.734317.3241 =78.8997  cm 2

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