Ex08 Circular Measure Answer

Topic: Circular Measure Question 1 [2009 S4 EOY, Q21] marks A pulley system in a car comprises two circular discs with circumferences  28 cm and 8 cm  respectively. The two discs are bound tightly by a belt ABCDEFGHA and their centres P and Q  are 20 cm apart.   (i) Show that angle AQP is  3   radians. Solution: 10 1cos 20 2AQP       ()3AQP shown [2] (ii) Find the length of the belt ABC in contact with the bigger disc. Solution:   cmAQ 142 28                                          [3]      C P Q G A BF E D H

422 3 414 58.63 AQC AQP Arc ABC                     (iii) Find the area of the shaded region. Solution: 2220 10 300 2 3 GA GPQ AQP                     Shaded area   = trapezium AQPG – area of sectors  22 2 11 1 22 300(4 14) (14) (4)22 3 2 3 73.0 (3 . .)cm s f                           [4]      Question 2 [2010 S4 EOY, Q16] marks 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. Solution: 1 8( ) 2 tan 1.176 1.18 12iC B D r a d                 [2] P1 D C A B

The shaded region is bounded by minor arcs CD with centres A and B respectively.   (ii) Find the perimeter of the shaded region.  Solution: ()ii arc CD centre A arc CD centre B               8 1.18 12 1.18 29.9 cm                [3] (iii) Find the area of the shaded region. Solution: (iii) Area of segment given by  2211 sin22rr  .         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 .                            [5]

Question 3 [2011 S4 EOY, Q8] marks During a school field trip, Jane discovered a broken circular plate, as shown in the diagram  below, from a deserted historical site. She measured the length of the chord AB as 14 cm  and from C, the mid‐point of AB, she measured the length CD as 1 cm where D is the point  on the circumference nearest to  C.     Find     (i) the radius of the plate,  Solution:       222 17rr      25r   A1  [3] (ii) the area of the segment ADB.  Solution: 11 72t a n 2t a n 24 ACAOB OC                2 11725 2 tan22 4 Area of sector OAB            [4] C A BD r 7

 2 11725 sin 2 tan22 4 Area of OAB             2 11 2 17 725 2 tan sin 2 tan22 4 2 4 9.37 Area of segment Area of sector OAB Area of OAB cm              Question 4 [2012 S4 EOY, Q9] marks 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, Solution: Area of shaded region BCE               22 22 2 11 () 624 2 66 1 88 9 18 10.3 cm AC                      [3]

(ii) the perimeter of FGBECF. Solution: Perimeter of FGBEC          = GE + CE + FC + FG          =  ACA G +CE + FC + AG           72 72 72 6 4 17.6 cm    [4] Question 5 [2013 S4 EOY, Q15] marks Two circles are such that A is the centre of the bigger circle and AB is the diameter of  the smaller circle. The radius of the bigger circle is 7 cm and the diameter AB is of length  10 cm. Points C and D are the intersection of the two circles and O is the centre of the  smaller circle.            (i) Show that cos CAO  is  7 10  .  Solution: 7 cmAC and 5 cmAO Since triangle OAC is isosceles with OA OC 7cos 5 2CAO  7 10 [2] A B C D O  E F

(ii) Find the area of the minor segment ACE. Solution: 222 1 557cos 2(5)(5)AOC   1 1cos 50  Hence area of minor segment ACE 221 (5) sin cm2 AOC AOC   26.89 cm (3 d.p.) [3] (iii) Find the area of the shaded crescent.  Solution: Area of sector ACD 21 217(7) 2 cos cm21 0  238.975 cm (3 d.p.) Hence area of shaded crescent 22(7) 2(6.887) 38.975 cm  2101 cm (3 s.f.) [3]

Question 6 [2014 S4 EOY, Q6] marks The diagram shows two circles, circle A, with radius 4 cmAP and circle B, with radius 6 cmBQ  having a common internal tangent PQ . If the distance between and A B is 22 cm , find PQ . Solution: Extend line BQ. Construct line parallel to PQ passing through A to meet extended line BQ at T.    641 0 c mBT     22 2 22 10 22 22 10 384 19.6 cm (3s.f.) AT AT       PQ = AT = 19.6 cm  [3] 22 cm 6 cm 4 cm Q P A B 22 cm4 cm 6 cm T Q P A B Using similar triangles,   Let AR = x and RB = (22‐x)  4 62 2 x x     8.8x    PQ = 22 228.8 4 13.2 6  R

Question 7 [2015 S4 EOY, Q9] marks The diagram shows a square ABCD of length 2 cm and an equilateral triangle ABE. An arc ABF is drawn with radius AE and centre at E. Points E, C and F lie on a straight line. (i) Show that 5 radians12BEC  . Solution: 236EBC       Since BE = BC,    1 26 5 = radians (shown)12 BEC     [2] (ii) Find the length of line segment CE, Solution: Using cosine rule,   222 2 2(2)(2) cos 1.04 cm (3 s.f.)6CE            [2] F E CD A B

(iii) Find the perimeter of the shaded region, Solution:   55Length of arc 2 cm 12 6 Perimeter of shaded region 5= 2 2 1.035 5.58 cm (3 s.f.)6 BF          [2] (iv) Find the area of the shaded region, leave your answer in the form ab  , where a and b are constants to be determined. Solution:    2 2 Area of shaded region = Area of sector Area of triangle 15 122 2 s i n21 2 2 6 5 1 c m6 EBF EBC         [3]