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1.5.1 Circular Measure Long Questions (Question 1 & 2)


Question 1:
Diagram shows a circle, centre O and radius 8 cm inscribed in a sector SPT of a circle at centre P.  The straight lines, SP and TP, are tangents to the circle at point Q and point R, respectively.

[Use p= 3.142]
Calculate
(a) the length, in cm, of the arc ST,
(b) the area, in cm2, of the shaded region.


Solution:
(a)
For triangle  O P Q sin 30 = 8 O P O P = 8 sin 30 = 16  cm Radius of sector  S P T = 16 + 8 = 24  cm S P T = 60 × 3.142 180 = 1.047  radian Length of arc  S T = 24 × 1.047 = 25.14  cm


(b)
For triangle  O P Q : tan 30 = 8 Q P P Q = 8 tan 30 = 13.86  cm Q O R = 2 ( 60 ) = 120 Reflex angle  Q O R = 360 120 = 240 240 = 3.142 180 × 240 = 4.189  radian Area of shaded region = (   Area of  sector  S P T ) ( Area of major    sector  O Q R ) ( Area of triangle  O P Q  and  O P R ) = 1 2 ( 24 ) 2 ( 1.047 ) 1 2 ( 8 ) 2 ( 4.189 ) 2 ( 1 2 × 8 × 13.86 ) = 301.54 134.05 110.88 = 56.61  cm 2

Question 2:
Diagram below shows a semicircle PTQ, with centre O and quadrant of a circle RST, with centre R.

[Use π = 3.142]
Calculate
(a) the value of θ, in radians,
(b) the perimeter, in cm, of the whole diagram,
(c) the area, in cm2, of the shaded region.


Solution:
( a ) sinROT= 2.5 5  ROT= 30 o θ= 180 o 30 o = 150 o   =150× π 180   =2.618 rad


( b ) Length of arc PT=rθ    =5×2.618    =13.09 cm Length of arc ST= π 2 ×2.5   =3.9275 cm O R 2 + 2.5 2 = 5 2   O R 2 = 5 2 2.5 2 OR=4.330 Perimeter=13.09+3.9275+2.5+4.330+5    =28.8475 cm


( c ) Area of shaded region =Area of quadrant RSTArea of quadrant RQT Area of quadrant RQT =Area of OQTArea of OTR = 1 2 ( 5 ) 2 ×( 30× π 180 ) 1 2 ( 4.33 )( 2.5 ) =1.1333  cm 2 Area of shaded region =Area of quadrant RSTArea of quadrant RQT = 1 2 ( 2.5 ) 2 ×( 90× π 180 )1.1333 =3.7661  cm 2

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