Topical Revision Paper 13 Unit 13 Circular Measure

Topical Revision Paper 13 Term 3 Unit 13: Circular Measure Question 1: The diagram shows two circles with centres A and B, intersecting at points C and D. The radii of the circles are 8 cm and 12 cm respectively, and angle ACB = angle ADB = 2  radians. (i) Show that angle CBD is approximately 1.18 radians. [2] The shaded region is bounded by minor arcs CD with centres A and B respectively. (ii) Find the perimeter of the shaded region. [3] (iii) Find the area of the shaded region. [3] Solution: 1 8( ) 2 tan 1.176 1.18 12i CBD rad       ()ii arc CD centre A arc CD centre B    8 1.18 12 1.18 29.9 cm     (iii) Area of segment given by 2211 sin22rr . (iii) P1 D C A B

    21 8 1.18 sin 1.18 33.1842 area of segment centre A      [Accept 33.360 if students use 1.176]   2112 1.18 sin1.18 18.3882 area of segment centre B    Area of shaded region = 251.6 cm . Question 2: The diagram shows a quadrant inscribed in the square ABCD of sides 6 cm. ACE is a sector centred at A. F is the point of intersection of the diagonal AC and the arc DB. FG is perpendicular to AE. Find (i) the area of the shaded region BCE, [3] (ii) the perimeter of FGBECF. [4] Solution: (i) Area of shaded region BCE

      22 22 2 11 ( ) 62 4 2 6 6 188 9 18 10.3 cm AC           (ii) Perimeter of FGBEC = GE + CE + FC + FG = AC AG +CE + FC + AG 72 72 72 6 4 17.6 cm      Question 3: The diagram below shows a tangent RST to two semi-circles with centre P and Q. The point R, A, P, B, Q and C lie on the same straight line. It is also given that AB = 2 cm and AC = 8 cm. (i) Show that RA = 1 cm. [2] (ii) Find QRT in radian. [2] (iii) Calculate the perimeter of the shaded region. [4] (iv) Calculate the area of the shaded region. [4] Solution: (i) RTQ and RSP are similar 11 53 RA RA   3RA +3 = RA + 5 2RA = 2 RA = 1 cm  

(ii) 1sin 2 6 CRT CRT    (iii) 222 = (1) (3) 4 233 2 = Perimeter Arc Arc Li 123 5 = 12 or 8.70 ne segment cm3 SB B c T m ST            (iv) 2 2 A 3 rea 1 of [ shade ( d regi 7 on Trap 1 1 2 1= (1 ) 2 ez ) (3 )( )]3 2 3 2 3 3= o 2 12 [ ] 32 11= 2 12 or 1.1 cm6 ium (Sect r Sector ) cm PTQS PBS QBT            Question 4: The diagram below shows a semicircle ADB with centre O and a radius of 6 m. ACE is a sector of a circle with centre C and a radius of 9.8 m. CE intersects the semicircle at D such that CD is 6 m.

(i) Show that 1.89AOD radians. [2] (ii) Calculate the area of the shaded region. [4] Solution: (i) 1 3.8 2cos 6 1.9 cos 6 1.2486 DOC DOC      1.8930 =1.89 (3s.f.) DOA DOC    (ii) Shaded Area = area sector ECA – area DCO – area sector DOA          221 1 19.8 6 3.8 sin 62 2 2 DCO DCO DOA      = 59.95696022 10.81332049 34.07417206 = 15.06946767 = 15.1 2m E D AOB C 9.8 m 6 m 6 m