Topical Revision Worksheet 9 Circular Measure Solution (LX)

Topical Revision Worksheet 9 Topic: Circular Measure Question 1 [2009 S4 EOY, Q21] marks A pulley system in a car comprises two circular discs with circumferences cm and 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 radians. Solution: [2] (ii) Find the length of the belt ABC in contact with the bigger disc. Solution: [3] (iii) Find the area of the shaded region. Solution: Shaded area = trapezium AQPG – area of sectors [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 = radians. (i) Show that angle CBD is approximately 1.18 radians. Solution: [2] The shaded region is bounded by minor arcs CD with centres A and B respectively. (ii) Find the perimeter of the shaded region. Solution: [3] (iii) Find the area of the shaded region. Solution: (iii) Area of segment given by . [Accept 33.360 if students use 1.176] Area of shaded region = . [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: A1 [3] (ii) the area of the segment ADB. Solution: [4]

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 [3] (ii) the perimeter of FGBECF. Solution: Perimeter of FGBEC = GE + CE + FC + FG = +CE + FC + AG [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 is . Solution: and Since triangle is isosceles with [2] (ii) Find the area of the minor segment ACE. Solution: Hence area of minor segment ACE [3] (iii) Find the area of the shaded crescent. Solution: Area of sector ACD Hence area of shaded crescent [3] A B C D O E F

Question 6 [2014 S4 EOY, Q6] marks The diagram shows two circles, circle A, with radius and circle B, with radius having a common internal tangent . If the distance between is , find . Solution: Extend line BQ. Construct line parallel to PQ passing through A to meet extended line BQ at T. PQ = AT = 19.6 cm [3] Using similar triangles, Let AR = x and RB = (22-x) PQ = 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 . Solution: Since BE = BC, [2] (ii) Find the length of line segment CE, Solution: Using cosine rule, [2] (iii) Find the perimeter of the shaded region, Solution: [2]

(iv) Find the area of the shaded region, leave your answer in the form , where a and b are constants to be determined. Solution: [3]