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Premium member Presentation Transcript PowerPoint Presentation: Army public school Jaipur, Rajasthan POWERPOINT PRESENTATION MATHEMATICS TOPIC SURFACE AREA OF CYLINDERPowerPoint Presentation: C ontentsPowerPoint Presentation: objectivesPowerPoint Presentation: A cylinder can be defined as a solid figure that is bound by : A curved surface Curved surface area Two flat surfaces called its bases bases The bases made of congruent circles Congruent circles Introduction to a cylinder 1PowerPoint Presentation: The radius of the two bases is the radius of the cylinder Equal radius The line that passes through the centres of the two circular bases Is perpendicular and is the height of the cylinder height radius & height Introduction to a cylinder 1PowerPoint Presentation: HEIGHT CURVED SURFACE AREA RADIUSPowerPoint Presentation: In the above figures , some of the cylinders are straight while some are slant In case of straight cylinders , the height is at right angle to the base Such cylinder are known as right circular cylinders Right circular cylinder 2PowerPoint Presentation: A right circular cylinder is a hollow right circular cylinder The part of the space enclosed by a right circular cylinder is called its interior A right circular cylinder along with its interior is called a right circular Cylindrical region , which is generally referred to as a solid right circular cylinder Right circular cylinder 2 Points on right circular cylinderPowerPoint Presentation: Surface area of a cylinder 3 height height Length How to find the surface area of a cylinderPowerPoint Presentation: Surface area of a cylinder 3 Judge each individual figure which together forms a cylinder Take a rectangular sheet breadth length The length is just enough to go round the cylinder and whose breadth is equal to the height of the cylinder as shown heightPowerPoint Presentation: Surface area of a cylinder 3 Thus , the area of the rectangular sheet gives us the curved surface area of the cylinder The length of the sheet is equal to the circumference of the circular base which is equal to 2 π r So the curved surface area of a cylinder = area of the rectangular sheet =length X breadth =perimeter of the base X height = 2 π r X height therefore , CSA=2 π rhPowerPoint Presentation: Surface area of a cylinder 3 Area of flat surfaces of a cylinder = area of the bases of the cylinder We know that a cylinder is made of two congruent circles , let us assume the radii of the circle as r How to find the total surface area of a cylinderPowerPoint Presentation: Surface area of a cylinder 3 Since, Area of the circle is π r² area of the flat surfaces= π r² + π r² therefore , area of the flat surfaces = 2 π r²PowerPoint Presentation: Surface area of a cylinder 3 = 2 π rh + 2 π r² =2 π r(r+h) Where , r the radius of the cylinder and h is its height Total surface area (tSA) = curved surface area (CSA ) + area of flat surfacesPowerPoint Presentation: Surface area of a cylinder 3 In a cylinder , if the top and the bottom is also to be covered , then we need two circles (in fact circular regions)to do that , and thus having an area of π r² , giving us the total surface area as 2 π rh + 2 π r² = 2 π r (r+h ) RADIUS heightPowerPoint Presentation: examples 4 The difference between outside and inside surface areas of a metallic open pipe 7 cm long is 44 cm². Find the difference between the outer and inner radii of the pipe. Length of the pipe (h) = 7 cm Let the inner radius is R and outer radius r Figure (B ) shows the cross section of the pipe The difference in outer and inner surface area = 2 π R h – 2 π r h = 2 π h ( R– r) Now, 2 π h ( R – r ) = 44 cm² (R– r )= 44 × 7/ 2 × 22 × 7 = 1 The difference between outer and inner radii is 1 cm height =7 inches Solution Example no:1 (b) (a)PowerPoint Presentation: examples 4 . Curved surface area of a right circular cylinder is 4.4 m² . If the radius of the base of the cylinder is 0.7 m. Find its height. ( Given, π = 22/7) Curved surface area of the cylinder = 4.4 m² Radius of the cylinder (r) = 0.7m Height of the cylinder (h) = ? We know (CSA) of a cylinder = 2 π rh therefore , 2 π rh = 4.4 m² =2 × 22/7 × 0.7 × h = 4.4 m² = 4.4 × 0.1 × h=4.4 m² h= 4.4/4.4 = 1 Therefore the height of the cylinder is 1 m . 4.4 m² o.7 m Required Height = 1m Example no:2 SolutionPowerPoint Presentation: examples 4 The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a play ground. Find the area of the playground in m². (Given, π = 22/7) Diameter of the roller = 84 cm radius of the roller (r) = 84/2 = 42 cm Length of the roller (h) = 120 cm Only the lateral surface area is considered when it is moved on the ground. Lateral surface area of the roller = 2 π rh = 2 × 22 /7 × 42 × 120 = 31680 cm² In one revolution the area of the ground covered by the roller = 31680 cm² The roller takes 500 complete revolutions to move to level the ground Therefore, the area of the playground = 31680 × 500 = 1584000 cm² Area of the play ground in m² is = 1584000/10000 m² = 1584 m² 84 cm 120 cm Solution Example no:3PowerPoint Presentation: examples 4 An iron pipe is 49 cm long and its diameter measures 6 cm. Find the cost of painting the lateral surface of the pipe at the rate of Rs 70/- per cm². (Given, = 22/7) A pipe is cylindrical in shape Length of the iron pipe (h) = 49 cm Diameter of the pipe = 6 cm Radius of the pipe (r) = 3 cm Lateral surface area (LSA) of the pipe = 2 π rh = 2 × 22/7 × 3 × 49 = 924 cm² Rate of painting = 70/- per cm² Now , Cost of painting the pipe of area = 924 × 70 =6465 therefore , It cost R s 6465/- to paint the iron pipe. 49 cm 6 cm Example no:3 SolutionPowerPoint Presentation: thanks You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.