# CHAPTER 28 THE CIRCLE AND ITS PROPERTIES

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1 CHAPTER 8 THE CIRCLE AND ITS PROPERTIES EXERCISE 118 Page Calculate the length of the circumference of a circle of radius 7. cm. Circumference, c = r = (7.) = 45.4 cm. If the diameter of a circle is 8.6 mm, calculate the circumference of the circle. Circumference, c = r = d = (8.6) = 59.5 mm. 3. Determine the radius of a circle whose circumference is 16.5 cm. Circumference, c = r from which, radius, r = c 16.5 = =.69 cm 4. Find the diameter of a circle whose perimeter is cm. If perimeter, or circumference, c = d, then = d and diameter, d = = cm 5. A crank mechanism is shown below, where XY is a tangent to the circle at point X. If the circle radius 0X is 10 cm and length 0Y is 40 cm, determine the length of the connecting rod XY. If XY is a tangent to the circle, then 0XY = , John Bird

2 Thus, by Pythagoras, 0Y = 0X + XY from which, XY = ( Y X ) = = = cm If the circumference of the Earth is km at the equator, calculate its diameter. Circumference, c = r = d from which, diameter, d = c = = 1 73 km = km, correct to 4 significant figures 7. Calculate the length of wire in the paper clip shown below. The dimensions are in millimetres. 1.5 Length of wire = (1.5) + ( ) + ( ) + ( ) = = = mm (3.5 3) ( ) , John Bird

4 (b).69 rad = = or (c) 7.41 rad = = or A car engine speed is 1000 rev/min. Convert this speed into rad/s rev/min = 1000 rev/min rad/rev 60s/min = rad/s , John Bird

5 EXERCISE 10 Page Calculate the area of a circle of radius 6.0 cm, correct to the nearest square centimetre. r = 6.0 = cm Area of circle = ( ). The diameter of a circle is 55.0 mm. Determine its area, correct to the nearest square millimetre. d 55.0 Area of circle = r = = 4 4 = 376 mm 3. The perimeter of a circle is 150 mm. Find its area, correct to the nearest square millimetre. Perimeter = circumference = 150 = r from which, radius, r = = 75 Area of circle = r = = 1790 mm 4. Find the area of the sector, correct to the nearest square millimetre, of a circle having a radius of 35 mm, with angle subtended at centre of 75. θ 75 = = 80 mm Area of sector = ( r ) ( 35 ) 5. An annulus has an outside diameter of 49.0 mm and an inside diameter of 15.0 mm. Find its area correct to 4 significant figures. d d = = = = 1709 mm Area of annulus = r 1 r ( d1 d ) ( ) 6. Find the area, correct to the nearest square metre, of a m wide path surrounding a circular plot of land 00 m in diameter , John Bird

6 = = 169 m 4 4 Area of path = ( d 1 d ) ( ) 7. A rectangular park measures 50 m by 40 m. A 3 m flower bed is made round the two longer sides and one short side. A circular fish pond of diameter 8.0 m is constructed in the centre of the park. It is planned to grass the remaining area. Find, correct to the nearest square metre, the area of grass. 8.0 Area of grass = (50 40) (50 3) (34 3) = = 1548 m 4 8. With reference to the diagram, determine (a) the perimeter and (b) the area. (a) Perimeter = ( 14) = cm (b) Area = (8 17) + ( 14 ) 1 = = cm 9. Find the area of the shaded portion shown. Shaded area = (10 10) (5) = = 1.46 m , John Bird

7 10. Find the length of an arc of a circle of radius 8.3 cm when the angle subtended at the centre is.14 radians. Calculate also the area of the minor sector formed. Arc length, s = rθ = (8.3)(.14) = cm 1 1 r θ = = m Area of minor sector = ( ) ( ) 11. If the angle subtended at the centre of a circle of diameter 8 mm is 1.46 rad, find the lengths of the (a) minor arc, and (b) major arc. If diameter d = 8 mm, radius r = 8 = 41 mm (a) Minor arc length, s = rθ = (41)(1.46) = mm (b) Major arc length = circumference minor arc = (41) = = mm 1. A pendulum of length 1.5 m swings through an angle of 10 in a single swing. Find, in centimetres, the length of the arc traced by the pendulum bob. Arc length of pendulum bob, s = rθ = (1.5) = 0.6 m or 6. cm 13. Determine the shaded area of the section shown , John Bird

9 18. Find the area swept out in 50 minutes by the minute hand of a large floral clock, if the hand is m long r = = m Area swept out = ( ) 19. Determine (a) the shaded area below and (b) the percentage of the whole sector that the area of the shaded area represents. 1 1 (a) Shaded area = (50) (0.75) (38) (0.75) 1 (0.75) = 396 mm = [ ] (b) Percentage of whole sector = (50) (0.75) 100% = 4.4% 0. Determine the length of steel strip required to make the clip shown. Angle of sector = = 30 = 30 rad = rad , John Bird

10 Thus, arc length, s = rθ = (15)(4.0146) = mm Length of steel strip in clip = = mm 1. A 50 tapered hole is checked with a 40 mm diameter ball as shown below. Determine the length shown as x. From the sketch below, tan 5 = 35 AC from which, AC = 35 tan 5 and sin 5 = 0 AB from which, AB = 0 sin 5 = mm = 47.3 mm i.e. AC = = x AB = x and x = i.e. x = 7.74 mm , John Bird

11 EXERCISE 11 Page Determine (a) the radius, and (b) the coordinates of the centre of the circle given by the equation: x + y 6x + 8y + 1 = 0 Method 1: The general equation of a circle ( x a) + ( y b) = r is x + y + ex + fy + c = 0 where coordinate a = e f, coordinate b = and radius r = a + b c Hence, if x + y 6x+ 8y+ 1 = 0 then a = e = 6 = 3, b = f 8 = = 4 (3) + ( 4) (1) = ( ) = 4 = and radius, r = [ ] i.e. the circle x + y 6x+ 8y+ 1 = 0 has (a) radius and (b) centre at (3, 4), as shown below. Method : x + y 6x+ 8y+ 1 = 0 may be rearranged as: ( x 3) + ( y+ 4) 4= 0 i.e. ( x 3) + ( y+ 4) = which has a radius of and centre at (3, 4). Sketch the circle given by the equation: x + y 6x + 4y 3 = 0 Method 1: The general equation of a circle ( x a) + ( y b) = r is x + y + ex + fy + c = 0 where coordinate a = e f, coordinate b = and radius r = a + b c Hence, if x + y 6x+ 4y 3= 0 then a = e = 6 = 3, b = f 4 = = and radius, r = [ ] (3) + ( ) ( 3) = ( ) = 16 = , John Bird

12 i.e. the circle x + y 6x+ 4y 3= 0 has centre at (3, ) and radius 4, as shown below. Method : x + y 6x+ 4y 3= 0 may be rearranged as: ( x 3) + ( y+ ) 16 = 0 i.e. ( x 3) + ( y+ ) = 4 which has a radius of 4 and centre at (3, ) 3. Sketch the curve: x + (y 1) 5 = 0 x + (y 1) 5 = 0 i.e. x + (y 1) = 5 i.e. x + (y 1) = 5 which represents a circle, centre (0, 1) and radius 5 as shown in the sketch below. 4. Sketch the curve: x = 6 y 1 6 If y x = then x y = and x y = , John Bird

13 i.e. x y + = 1 and x + y = which is a circle of radius 6 and co-ordinates of centre at (0, 0), as shown below , John Bird

14 EXERCISE 1 Page A pulley driving a belt has a diameter of 300 mm and is turning at 700/ revolutions per minute. Find the angular velocity of the pulley and the linear velocity of the belt, assuming that no slip occurs. Angular velocity, ω = n = = 90 rad/s Linear velocity v = ωr = (90) 3 = 13.5 m/s. A bicycle is travelling at 36 km/h and the diameter of the wheels of the bicycle is 500 mm, Determine the angular velocity of the wheels of the bicycle and the linear velocity of a point on the rim of one of the wheels Linear velocity, v = 36 km/h = m/s = 10 m/s 3600 (Note that changing from km/h to m/s involves dividing by 3.6) Radius of wheel, r = 500 = 50 mm = 0.5 m Since, v = ωr, then angular velocity, ω = v r = = 40 rad/s 3. A train is travelling at 108 km/h and has wheels of diameter 800 mm. (a) Determine the angular velocity of the wheels in both rad/s and rev/min. (b) If the speed remains constant for.70 km, determine the number of revolutions made by a wheel, assuming no slipping occurs. (a) Linear velocity, v = 108 km/h = Radius of wheel = 800 = 400 mm = 0.4 m Since, v = ωr, then angular velocity, ω = m/s = 30 m/s v 30 r = 0.4 = 75 rad/s , John Bird

15 rev 60s 75 rad/s = 75 rad/s = 716. rev/min rad min (b) Linear velocity, v = s t hence, time, t = s 700m v = 30 m/s = 90 s = = 1.5 minutes Since a wheel is rotating at 716. rev/min, then in 1.5 minutes it makes 716. rev/min 1.5 min = 1074 rev/min , John Bird

16 EXERCISE 13 Page Calculate the tension in a string when it is used to whirl a stone of mass 00 g round in a horizontal circle of radius 90 cm with a constant speed of 3 m/s. Tension in string = centripetal force = mv r 0.0 kg (3m/s) = = N 0.90m. Calculate the centripetal force acting on a vehicle of mass 1 tonne when travelling around a bend of radius 15 m at 40 km/h. If this force should not exceed 750 N, determine the reduction in speed of the vehicle to meet this requirement. Centripetal force = mv r velocity, v = 40 km/h = m/s Hence, centrifugal force = where mass, m = 1 tonne = 1000 kg, radius, r = 15 m and 40 (1000) If centrifugal force is limited to 750 N, then = 988 N (1000) v 750 = 15 from which, velocity, v = (750)(15) 1000 = m/s = = km/h Hence, reduction in speed is = 5.14 km/h 3. A speed-boat negotiates an S-bend consisting of two circular arcs of radii 100 m and 150 m. If the speed of the boat is constant at 34 km/h, determine the change in acceleration when leaving one arc and entering the other. Speed of the boat, v = 34 km/h = m/s , John Bird

17 For the first bend of radius 100 m, acceleration = v r = = 0.89 m/s 100 For the second bend of radius 150 m, acceleration = = = m/s, the negative sign indicating a change in direction Hence, change in acceleration is 0.89 ( 0.595) = 1.49 m/s , John Bird

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