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Circles(1)
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- 2. Pi Circle words Rounding Refresher Area Perimeter and Areaof compound shapes Perimeters of sectors Area of Sectors Volumes of Cylinders Volume of Spheres and cones Equation of a circle 2 Simultaneous Equations Finding the radius of sectors Radius and Height of Cylinders Circle theorems Area of Segments Equation of a circle 1 Circle formulae Circumference
- 3. Match the wordsto the definitions •Sector •Segment •Chord •Radius •Arc •Tangent •Diameter •Circumference •The length around the outside of a circle •A line which just touches a circle at one point •A section of a circle which looks like a slice of pizza •A section circle formed with an arc and a chord •The distance from the centre of a circle to the edge •The distance from one side of a circle to the other (through the centre) •A section of the curved surface of a circle •A straight line connecting two points on the edge of a circle HOME
- 4. Think about circles Imagethat line straightened out- this outside of a circle Think about a line around theis the circumference
- 5. Pi People noticed thatif you divide the circumference of a circle by the diameter you ALWAYS get the same answer They called the answer Pi (π) , which is: 3.1415926535897932384626433832795028841971693993751 058209749445923078164062862089986280348253421170679 821480865132823066470938446095505822317253594081284 You can use the 8111745028410270193852110555964462294895493038196 π button on your 442881097566593344612847564823378678316527120190914 calculator 5648566923460348610454326648213393607260249141273 724587006606315588174881520920962829254091715364367 892590360011330530548820466521384146951941511609...
- 6. How many digitscan you memorise in 2 minutes? 3.141592653589793238462643383279502884197169399375 10582097494459230781640628620899862803482534211 70679821480865132823066470938446095505822317253 59408128481117450284102701938521105559644622948 95493038196442881097566593344612847564823378678 31652712019091456485669234603486104543266482133 93607260249141273724587006606315588174881520920 96282925409171536436789259036001133053054882046 6521384146951941511609...
- 7. Write down pi! 3.141592653589793238462643383279502884197169399375 10582097494459230781640628620899862803482534211 70679821480865132823066470938446095505822317253 59408128481117450284102701938521105559644622948 95493038196442881097566593344612847564823378678 31652712019091456485669234603486104543266482133 93607260249141273724587006606315588174881520920 96282925409171536436789259036001133053054882046 6521384146951941511609... Howdid you do? What do you think the world record is?
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- 9. Pi Story • Oneway to memorise Pi is to write a Pi-em (pi poem) where the number of letters in each word is the same as the number in pi. For example: “Now I, even I, would celebrate in rhymes inept, the great immortal Syracusan rivall'd nevermore who in his wondrous lore passed on before left men his guidance how to circles mensurate.” Can you write one of your own? HOME
- 10. Rounding to DecimalPlaces 10 multiple choice questions
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- 21. Finding the Circumference Youcan find the circumference of a circle by using the formula- Circumference = π x diameter For Example10cm Area= π x 10 = 31.41592654.... = 31.4 cm (to 1 dp)
- 22. You can findthe circumference of a circle by using the formula- Circumference = π x diameter For Example- 10cm Area= π x 10 = 31.41592654.... = 31.4 cm (to 1 dp) Find the Circumference of a circles with: 1. A diameter of : a) 8cm b) 4cm c) 11cm d) 21cm e) 15cm 2. A radius of : a) 6cm b) 32cm c) 18cm d) 24cm e) 50cm ANSWERS 1a 25.1cm b 12.6cm c 34.6cm d 66.0cm e 47.1cm 2a b c d e 37.7cm 201.1cm 113.1cm 150.8cm 157.1cm HOME
- 23. Finding the Area Youcan find the area of a circle by using the formula- Area= π x Radius2 For Example7cm Area= π x 72 = π x 49 = 153.93804 = 153.9 (to 1dp) cm2
- 24. Finding the Area Youcan find the area of a circle by using the formula- Area= π x Radius2 7cm For ExampleArea= π x 72 = π x 49 = 153.93804 = 153.9 (to 1dp) cm2 ANSWERS 2a b c d e f g h HOME 12.6 78.5 15.2 380.1 314.2 153.9 100.5 28.3
- 25. Finding the Areaof a Sector To find the area of a sector, you need to work out what fraction of a full circle you have, then work out the area of the full circle and find the fraction of that area. For ExampleThe sector here is ¾ of a full circle Find the area of the full circle 7cm Area= π x 72 = π x 49 = 153.93804 = DON’T ROUND YET! Then find ¾ of that area ¾ of 153.93804 = 115.45353 (divide by 4 and multiply by 3)
- 26. Finding the Areaof a Sector Sometimes it is not easy to see what fraction of a full circle you have. You can work it out based on the size of the angle. If a full circle is 360° , and this sector is 216°, the sector is 216/360, which can be simplified to 3/5. For Example- The sector here is 3/5 of a full circle Find the area of the full circle 216° 7cm Area= π x 72 = π x 49 = 153.93804 = DON’T ROUND YET! Then find 3/5 of that area 3/5 of 153.93804 = 92.362824 (divide by 5 and multiply by 3) = 92.4cm2 Sometimes the fraction cannot be simplified and will stay over 360
- 27. Finding the Areaof a Sector The general formula for finding the area is: Area of sector= Angle of Sector x πr2 360 Fraction of full circle that sector covers “of” Area of full circle
- 28. Questions Find the areaof these sectors, to 1 decimal place 1 3 2 10cm 11cm 260° 190° 4 12cm 251° 6 5 87° ANSWERS 5cm 6.5cm 32° 1 2 3 4 5 6 226.9 200.6 315.4 19.0 61.2 80.7 17cm 166° HOME
- 29. Finding the Perimeterof a Sector To find the perimeter of a sector, you need to work out what fraction of a full circle you have, then work out the circumference of the full circle and find the fraction of that circumference. You then need to add on the radius twice, as so far you have worked out the length of the curved edge For Example- The sector here is ¾ of a full circle Find the area of the full circle 7cm Area= π x 14 (the diameter is twice the radius) = π x 49 = 43.982297...... = DON’T ROUND YET! Then find ¾ of that circumference ¾ of 43.982297...... = 32.99 cm (2 dp) Remember to add on 7 twice from the straight sides
- 30. Finding the Areaof a Sector Sometimes you will not be able to see easily what fraction of the full circle you have. To find the fraction you put the angle of the sector over 360 250° This sector is 250/360 or two hundred and fifty, three hundred and sixty-ITHS of the full circle Simplify if you can Sometimes the fraction cannot be simplified and will stay over 360
- 31. Finding the Perimeterof a Sector The general formula for finding the area is: Perimeter of sector= (Angle of Sector x πd) + r + r 360 Fraction of full circle that sector covers “of” Circumference of full circle Don’t forget the straight sides This is the same as d of 2r, but I like r +r as it helps me remember why we do it
- 32. Questions Find the perimeterof these sectors, to 1 decimal place 1 3 2 10cm 11cm 260° 190° 4 251° 6 5 87° 12cm 5cm 6.5cm 32° ANSWERS 1 65.4 2 58.5 3 76.6 4 17.6 5 31.8 6 43.5 17cm 166° HOME
- 33. Compound Area andPerimeter Here we will look at shapes made up of triangles, rectangles, semi and quarter circles. Find the area of the shape below: 10cm 8cm 10cm Area of this rectangle= 8 x10 =80cm2 Area of this semi circle = π r2 ÷ 2 = π x 52 ÷ 2 = π x 25 ÷ 2 =39.3 cm2 (1dp) Area of whole shape = 80 + 39.3 = 119.3 cm2
- 34. Compound Area andPerimeter Find the perimeter of the shape below: 10cm 8cm 10cm Perimeter of this rectangle= 8 + 8 + 10 =26cm (don’t include the red side) Circumference of this semi circle = πd ÷ 2 = π x 10 ÷ 2 =15.7 cm (1dp) Perimeter of whole shape = 26 + 15.7 = 31.7 cm
- 35. Compound Area andPerimeter Find the areaof the shape below: Area of this quarter circle = π r2 ÷ 4 = π x 52 ÷ 4 = π x 25 ÷ 4 =19.7 cm2 (1dp) 5cm 10cm 11cm Area of this rectangle 10 x 11=110 Area of whole shape = 110+ 19.7 = 129.7cm2
- 36. Compound Area andPerimeter Work out all missing Find the perimeter of the shape below: sides first ? 5cm 5cm Circumference of this quarter circle = πd ÷ 4 = π x 10 ÷ 4 (if radius is 5, diameter is 10) =7.9 cm (1dp) 6cm 10cm 10cm 11cm Add all the straight sides= 10+10 + 11+ 5 + 6= 42cm Area of whole shape = 42+ 7.9 = 49.9cm
- 37. Questions ANSWERS AREA PERIMETER 1 38.1 23.4 2135.0 61.3 Find the perimeter and area of these shapes, to 1 decimal place 3 181.1 60.8 2cm 3 4 27.3 1 2 12cm 10cm 5 129.3 47.7 20cm 6 128.5 6cm 11cm 4cm 17cm 4cm 4 Do not worry about perimeter here 5cm 12cm 6 6cm 5 Do not worry about perimeter here 10cm 5cm 5cm 10cm 20cm HOME
- 38. Volume of Cylinders Herewe will find the volume of cylinders Cylinders are prisms with a circular cross sections, there are two steps to find the volume 1) Find the area of the circle 1) Multiple the area of the circle by the height or length of the cylinder
- 39. Volume of Cylinders2 EXAMPLE- find the volume of this cylinder 4cm 1) Find the area of the circle π x r2 π x 42 π x 16 = 50.3 cm2 (1dp) 10cm 2) Multiple the area of the circle by the height or length of the cylinder 50.3 (use unrounded answer from calculator) x 10 = 503cm3
- 40. Questions Find the volumeof these cylinders, to 1 decimal place 1 2 4cm 3cm 3 ANSWERS 5cm 10cm 15cm 603.2 2 12cm 1 282.7 3 1178.1 4 142.0 5 2155.1 4 2cm 5 11.3cm 6 7cm 14cm 6 508.9 3cm 18cm HOME
- 41. Volume of Cylinders2 EXAMPLE- find the height of this cylinder 4cm 1) Find the area of the circle π x r2 π x 42 π x 16 = 50.3 cm2 (1dp) h 2) Multiple the area of the circle by the height or length of the cylinder 50.3 x h = 140cm3 Rearrange this to give h= 140 ÷ 50.3 h=2.8 cm Volume= 140cm3
- 42. Volume of Cylinders EXAMPLE-find the radius of this cylinder r 1) Find the area of the circle π x r2 2) Multiple the area of the circle by the height or length of the cylinder π x r2 x 30 = 250cm3 94.2... x r2 = 250 Rearrange this to give r2 = 250 ÷ 94.2 r2 =2.7 (1dp) r= 1.6 (1dp) cm 30cm Volume= 250cm3
- 43. Questions 1 ANSWERS 1 6.4 2 4.2 3 Findthe volume of these cylinders, to 1 decimal place 1.3 4 2.3 3 4cm 2 5 1.8 3cm 5cm 6 1.9 h h volume= 320cm3 4 volume= 120cm3 5 r volume= 100cm3 6 r 12cm volume= 200cm3 h r 8cm 14cm volume= 150cm3 volume= 90cm3 HOME
- 44. Volume of Spheres Theformula for the volume of a sphere is e.g A= 4/3 x π x 103 A= 4/3 x π x 1000 A=4188.8 cm3 (1 dp) 10cm
- 45. Volume of Cones Theformula for the volume of a cone is e.g A= 1/3 x π x 42 x 10 A= 1/3 x π x 16 x 10 A=167.6 cm3 (1 dp) 10cm 4cm
- 46. Questions 1 ANSWERS 1 4188.8 2 33510.3 3 Findthe volume of these spheres, to 1 decimal place 523.6 4 201.1 3 5 122.5 2 6 1272.3 10cm 4 20cm 5cm 6 5 12cm 13cm 4cm 3cm 15cm 9cm HOME
- 47. Circles Theorems Angle at the centre Angles connectedby a chord Triangles made with a diameter or radii Cyclic Quadrilaterals Tangents
- 48. Double Angle The angleat the centre of a circle is twice the angle at the edge Example Angle x = 50 x 2 x=100° 50° x
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- 50. Triangles inside circles Atriangle containing a diameter, will be a right angled triangle A triangle containing two radii, will be isosceles 90° x x
- 51. 2 1 3 72° x x 60° Answers x 1) X=30 2)x=18 3)x=45 4)X=40y=40 5)x=30 y= 120 6)x=22 y=136 1 x x x y 3 2 x 22° 100° y 30° y x HOME
- 52. Angles connected bya chord Angles connected by a chord are equal x x y y
- 53. 2 1 25° x 15° Answers 1) x=25 y=15 2)x=125y= 40 z=15 3)x=10 y=70 z=100 4)X=105 y=40 z=35 5)x=53 y= 30 z=72 4 6)x=85 y=80 z=17 y x 10° x 40° x y z 3 y z 100° 125° z 15° y 5 6 z y 40° 35° x 25° 30° 53° y x 17° 80° 95° z HOME
- 54. Tangents to acircle A tangent will always meet a radius at 90° 90°
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- 56. Cyclic Quadrilaterals Opposite anglesin a cyclic quadrilaterals add up to 180° 100° 60° x y 100 + y = 180 y=80° 60 + x = 180 x = 120 °
- 57. 2 1 95° 3 54° 110° 20° x y x y Answers 1) x=70 y=85 2)x=126y=105 3)x=100 y=160 4)w=15 x=70 y=65 z= 25 5)a=60 b=36 x 75° 4 80° y 5 15° 25° 70° a w 4b y z x b 2a HOME
- 58. Area of Segments Herewe will look at finding the area of sectors You will need to be able to do two things: 1) Find the area of a sector using the formula- 2) Find the area of a triangle using the formulaArea= ½ absinC Area of sector= Angle of Sector x πr2 360 a C b
- 59. Examplefind the areaof the blue segment Step 1- find the area of the whole sector Area= 100/360 x π x r2 = 100/360 x π x 102 =100/360 x π x 100 =87.3cm2 10cm 100° 10cm Step 2- find the area of the triangle Area= ½ absinC =1/2 x 10 x 10 x sin100 = 49.2cm2 Step 3- take the area of the triangle from the area of the segment 87.3 – 49.3 = 38 cm2
- 60. Examplefind the areaof the blue segment Step 1- find the area of the whole sector Area= 120/360 x π x r2 = 120/360 x π x 122 =120/360 x π x 144 =150.8cm2 12cm 120° 12cm Step 2- find the area of the triangle Area= ½ absinC =1/2 x 12 x 12 x sin120 = 62.4cm2 Step 3- take the area of the triangle from the area of the segment 150.8 – 62.4 = 88.4 cm2
- 61. Questions 1 ANSWERS 1 75.1 2 29.5 3 Findthe area of the blue segments, to 1 decimal place 201.1 4 8.3 3 2 5 51.8 10cm 6 33.0 85° 11cm 170° 12cm 130° 4 6 5 95° 6.5cm 5cm 160° 65° 17cm HOME
- 62. Finding the Radiusor angle of a Sector r 100° Area=200cm2 Area= 100 x π x r2 360 200= 100 x π x r2 360 200x360 = r2 100 x π 229.2=r2 15.1cm =r 10cm x Area=150 Area= θ x π x r2 360 150= θ x π x 102 360 150x360 = θ 102 x π 117.9°= θ
- 63. Questions ANSWERS 1 7.6 2 8.3 Findthe missing radii and angles of these sectors, to 1 decimal 3 5.4 place 4 160.4 3 1 2 5 122.1 r r r 6 47.6 200° 175° Area=100cm2 4 250° Area=120cm2 6 5 5cm θ Area=35cm2 6.5cm Area=50cm2 17cm θ Area=120cm2 θ Area=45cm2 HOME
- 64. The Equation ofa Circle The general equation for a circle is (x-a)2 + (y-b)2=r2 This equation will give a circle whose centre is at (a,b) and has a radius of r For example a circle has the equation (x-2)2 + (y-3)2=52 This equation will give a circle whose centre is at (2,3) and has a radius of 5
- 65. The Equation ofa Circle A circle has the equation (x-5)2 + (y-7)2=16 This equation will give a circle whose centre is at (5,7) and has a radius of 4 (square root of 16 is 4) For example a circle has the equation (x+2)2 + (y-4)2=100 This equation will give a circle whose centre is at (-2,4) and has a radius of 10 You could think of this as (x - -2)2
- 66. The Equation ofa Circle A circle has the equation (x-5)2 + (y-7)2=16 What is y when x is 1? (1-5)2 + (y-7)2=16 12+ (y-7)2=16 1+ (y-7)2=16 (y-7)2=15 y-7= ±3.9 (square root of 15 to 1 dp) y= 7±3.9 y= 10.9 or 3.1 There are two coordinates on the circle with x=1, one is (1,10.9) and the other is (1,3.1)
- 67. The Equation ofa Circle 1) Write down the coordinates of the centre point and radius of each of these circles: Answers a) (x-5)2 + (y-7)2=16 b) (x-3)2 + (y-8)2=36 c) (x+2)2 + (y-5)2=100 d) (x+2)2 e) (x-6)2 + (y+4)2=144 f) x2 + y2=4 g) x2 + (y+4)2=121 h) (x-1)2 + (y+14)2 -16=0 i) 1a) r=4 centre (5,7) b) r=6 centre (3,8) c) r=4 centre (-2,5) d) r=10 centre (-2,-5) e) r=7 centre (6,-4) f) r=12 centre (0,0) g) r=411centre (0,-4) h) r=4 centre (1,-14) i) r=5 centre (5,9) (x-5)2 + (y-9)2 -10=15 + (y+5)2=49 2) What is the diameter of a circle with the equation (x-1)2 + (y+3)2 =64 Answers 2) 16 3)Circumference = 25.1 Area=50.3 4)Circumference = 25.1 Area=50.3 5) Circles have the same radius but different centres, they are translations 3) Calculate the area and circumference of the circle with the equation (x-5)2 + (y-7)2=16 4) Calculate the area and perimeter of the circle with the equation (x-3)2 + (y-5)2=16 5) Compare your answers to question 3 and 4, what do you notice, can you explain this? 6 ) A circle has the equation (x+2)2 + (y-4)2=100, find: a) x when y=7 b) y when x=6 6a) x= 11.5 or -7.5 b) y=11.3 or -3.3 HOME
- 68. The Equation ofa Circle 2 Here we will look at rearranging equations to find properties of the circle they represent Remember- The general equation for a circle is (x-a)2 + (y-b)2=r2 The skill you will need is called completing the square, you may have used it to solve quadratic equations
- 69. The Equation ofa Circle 2 Example x2 + y2 -6x – 8y =0 Create two brackets and put x in one and y in the other (x ) 2 + (y )2 = 0 Half the coefficients of x and y and put them into the brackets, and then subtract those numbers squared (x -3) 2 + (y - 4) 2 – 32 - 42= 0 Tidy this up (x -3) 2 + (y - 4) 2 – 25= 0 (x -3) 2 + (y - 4) 2 = 25 This circle has a radius of 5 and centre of (3,4)
- 70. The Equation ofa Circle 2 Example x2 + y2 -10x – 4y- 7 =0 Create two brackets and put x in one and y in the other (x ) 2 + (y )2 = 0 Half the coefficients of x and y and put them into the brackets, and then subtract those numbers squared (x -5) 2 + (y - 2) 2 – 52 – 22 - 7= 0 Tidy this up (x -5) 2 + (y - 2) 2 – 36= 0 (x -5) 2 + (y - 2) 2 = 36 This circle has a radius of 6 and centre of (5,2)
- 71. The Equation ofa Circle 2 You must always make sure the coefficient of x2 and y2 is 1 You may have to divide through 2x2 + 2y2 -20x – 8y- 14 =0 Divide by 2 to give x2 + y2 -10x – 4y- 7 =0 Then put into the form x2 + y2 -10x – 4y- 7 =0
- 72. Questions Put this equationsinto the form (x-a)2 + (y-b)2=r2 then find the centre and radius of the circle 1. 2. 3. 4. 5. 6. 7. 8. x2 + y2 -8x – 4y- 5 =0 x2 + y2 -12x – 6y- 4 =0 x2 + y2 -4x – 10y- 20 =0 x2 + y2 -10x – 14y- 7 =0 x2 + y2 -12x – 2y- 62 =0 2x2 +2y2 -20x – 20y- 28 =0 3x2 + 3y2 -42x – 24y- 36 =0 5x2 + 5y2 -100x – 30y- 60 =0 Answers 1) r=5 centre (4,2) 2) r=7 centre (6,3) 3) r=7 centre (2,5) 4) r=9 centre (5,7) Answers 5) r=10 centre (6,1) 6) r=8 centre (5,5) 7) r=8 centre (6,4) 8) r=11 centre (10,3) HOME
- 73. Simultaneous Equations A circlehas the equation (x-5)2 + (y-7)2=16 and a line has an equation of y=2x+1, at what points does the line intercept the circle? We need to substitute into the equation of the circle so that we only have x’s or y’s Because y=2x +1 we can rewrite the equation of the circle but instead of putting “y” in we’ll write “2x+1” Ways to solve quadratic equationsCompleting the square Factorising The Quadratic formula So, (x-5)2 + (2x-1-7)2=16 (x-5)2 + (2x-8)2=16 expand the brackets x2-10x + 25 + 4x2 – 32x +64 = 16 simplify and make one side 0 5x2 -42x + 73=0 solve this quadratic equation to find x, Put the value / values of x into y=2x+1 to find the coordinates of the intercept / intercepts to answer the question
- 74. Simultaneous Equations A quadraticequation can give 1,2 or no solutions, a line can cross a circle at 1,2 or no points 1 solution to the quadraticThe line is a tangent 2 solutions to the quadratic 0 solutions to the quadratic the circle and the line never meet
- 75. Intercepts between linesand circles 1) Find out whether these circles and lines intercept, if they do find the coordinates of the interceptions a) (x-5)2 + (y-7)2=16 and b) (x-3)2 + (y-8)2=36 and c) (x+2)2 + (y-5)2=100 and d) (x+2)2 + (y+5)2=49 and e) (x-6)2 + (y+4)2=144 and ANSWERS (all have been rounded) y=2x-2 (3.6,9.8) and (2.2,5.6) y=3x + 3 (7.2,12.3) and (2,2) 2y+4=x (3.5,13.4) and (-2.7,-5) y -3x =5 (-0.4,3.3) and (-6.4,-8.9) (-4.6,6.9) and (-4.6,-8.9) y=3x-1 HOME
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