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This text provides step-by-step calculations for finding the volume and surface area of various metal shapes such as prisms, cylinders, cones, and spheres. It also includes examples of converting shapes and calculating the maximum number of shapes that can be made from given quantities of material.
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1. The volume of the metal prism shown is 1144 cm3. (i) Find the value of x.
1. The volume of the metal prism shown is 1144 cm3. (ii) This prism is melted down and recast as 143 identical cubes. Find the side length of each of these cubes.
2. A solid metal cylindrical rod has a radius of 6 mm and a height of 216 mm. (i) Find the volume of the cylindrical rod, in terms of π. Volume of cylindrical rod = πr2h = π(6)2(216) = 7776π mm3
2. A solid metal cylindrical rod has a radius of 6 mm and a height of 216 mm. (ii) Find the surface area of the rod, in terms of π. Surface area of cylinder = 2πrh + 2πr2 = 2π(6)(216) + 2π(6)2 = 2592π + 72π = 2664π mm2
2. A solid metal cylindrical rod has a radius of 6 mm and a height of 216 mm. This metal rod is melted down and recast as a metallic sphere, with no metal wasted. (iii) Find the radius length of this sphere. Volume of sphere = Volume of cylinder
2. A solid metal cylindrical rod has a radius of 6 mm and a height of 216 mm. (iv) Find the surface area of the sphere, in terms of π. Surface area of sphere = 4πr2 = 4π(18)2 = 1296π mm2
2. A solid metal cylindrical rod has a radius of 6 mm and a height of 216 mm. (v) Find surface area of rod : surface area of the sphere. Surface area of rod : Surface area of sphere 2664π : 1296π 2664 : 1296 37 : 18
3. A solid wax candle, in the shape of a cylinder, has a radius of 8 cm and a height of 15 cm. (i) Find the volume of the candle, in terms of π. Volume of cylindrical candle = πr2h = π(8)2(15) = 960π cm3
3. A solid wax candle, in the shape of a cylinder, has a radius of 8 cm and a height of 15 cm. This candle is melted down and remoulded into four identical conical candles, each of height 12 cm. Find the radius of these candles, correct to two decimal places. (ii) Volume of 1 conical candle = 960π ÷4 = 240π cm3
4. A wax candle is in the shape of a right circular cone.The height of the candle is 7 cm and the diameter of the base is 9 cm. (i) Find the volume of the wax candle, correct to two decimal places. Diameter = 9 cm, so radius = 4·5 cm
4. A wax candle is in the shape of a right circular cone.The height of the candle is 7 cm and the diameter of the base is 9 cm. A rectangular block of wax measuring 30 cm by 15 cm by 12 cm is melted down and used to make a number of these candles. (ii) Find the maximum number of candles that can be made from the block of wax, if 6% of the wax is lost in the process. Volume of rectangular block = L × W × H = 30 × 15 × 12 = 5400
4. A wax candle is in the shape of a right circular cone.The height of the candle is 7 cm and the diameter of the base is 9 cm. A rectangular block of wax measuring 30 cm by 15 cm by 12 cm is melted down and used to make a number of these candles. (ii) Find the maximum number of candles that can be made from the block of wax, if 6% of the wax is lost in the process.
5. (i) A spherical golf ball has a diameter of 4 cm. Find the volume of the golf ball, in terms of π. Diameter = 4 cm, so radius = 2 cm
5. (ii) A cylindrical hole on a golf course is 10 cm in diameter and 12 cm deep. The hole is half full of water. Calculate the volume of water in the hole, in terms of π. Diameter = 10 cm, so radius = 5 cm Volume of cylinder = πr2h = π(5)2(6) = 150π cm3
5. (iii)The golf ball is dropped into the hole. Find the rise in the level of the water, correct to two decimal places. Volume of rise = Volume of ball
6. (i)Find the volume of a solid sphere which has radius of length 2∙1 cm. Give your answer correct to the nearest cm3. Take π = .
6. (ii)A cylindrical container contains water. When this sphere and a solid cube of side length 3 cm are completely submerged in the water in the cylindrical container, the water level rises by 4 cm. Find the radius of the cylinder, correct to one decimal place. Take π = . Volume of cube = L3 = 33 = 27 cm3 Total volume immersed = Volume of cube + Volume of sphere = 27 + 39 = 66 cm3
6. (ii)A cylindrical container contains water. When this sphere and a solid cube of side length 3 cm are completely submerged in the water in the cylindrical container, the water level rises by 4 cm. Find the radius of the cylinder, correct to one decimal place. Take π = .
7. Liquid is poured into a closed bottom cylinder, of radius 6 cm, at a volume of 18πcm3 per second. Find the depth of the water after 33 seconds. Volume of liquid in 33 second = 18π × 33 = 594π cm3
8. A cylindrical container has a radius of 6 cm and contains a volume of water.Water is poured into the container at a rate of 21 cm3 per second. Using the π button on your calculator, find the rise in the level of the water in the container after 14 seconds. Give your answer to one decimal place. Volume in 1 second = 21 cm3 in 14 seconds = 21 × 14 = 294 cm3
9. Water flows through a cylindrical pipe, of internal diameter 4 cm, at a rate of 12∙5 cm per second. (i) Find, in terms of π, how much water flows out of the pipe in one second. Diameter = 4 cm, so radius = 2 cm Height = 12·5 cm Volume of cylinder = πr2h = π(2)2(12·5) = π(4)(12·5) = 50π cm3/sec
9. Water flows through a cylindrical pipe, of internal diameter 4 cm, at a rate of 12∙5 cm per second. (ii) Find how long it takes to fill a hemispherical bowl, of radius 15 cm, using this pipe.
10. Water flows through a cylindrical pipe, of radius 2 cm, at a rate of 5 cm/sec. (i) Find, in terms of π, the volume of water flowing out of the pipe per second. Volume of cylinder in 1 second = πr2h = π(2)2(5) = 20π cm3/second
10. Water flows through a cylindrical pipe, of radius 2 cm, at a rate of 5 cm/sec. (ii) • Water flows through this pipe into a conical-shaped cup, of base diameter 10 cm. If it takes 15 seconds for the cup to fill, find the depth of the cup. Diameter = 10 cm, so radius = 5 cm In 1 sec = 20π cm3 15 sec = 20π × 15 = 300π cm3
10. Water flows through a cylindrical pipe, of radius 2 cm, at a rate of 5 cm/sec. (ii) • Water flows through this pipe into a conical-shaped cup, of base diameter 10 cm. If it takes 15 seconds for the cup to fill, find the depth of the cup.
11. Water flows through a cylindrical garden hose, of internal diameter 8 cm, at a rate of 52 cm per second. (i) • Find, in terms of π, how much water flows out of the hose in one second. Diameter = 8 cm, so radius = 4 cm Height = 52 cm/sec Volume in one second = πr2h = π(4)2(52) = 832π cm3/second
11. Water flows through a cylindrical garden hose, of internal diameter 8 cm, at a rate of 52 cm per second. (ii) • The pipe is used to fill a cylindrical paddling pool of internal diameter 1∙2 m. Find the depth of the water in the pool after 4∙5 minutes.
11. Water flows through a cylindrical garden hose, of internal diameter 8 cm, at a rate of 52 cm per second. (ii) • The pipe is used to fill a cylindrical paddling pool of internal diameter 1∙2 m. Find the depth of the water in the pool after 4∙5 minutes. • Therefore, after 4·5 minutes, the water is at a depth of 62·4 cm in the paddling pool.
12. A sphere has a radius of 6 cm. A larger sphere has a radius of 18 cm. (i) Find surface area of smaller sphere : surface area of larger sphere.
12. A sphere has a radius of 6 cm. A larger sphere has a radius of 18 cm. (ii) Find volume of smaller sphere : volume of larger sphere.
13. A hollow cylinder has a height, h, and radiusr. A second hollow cylinder has dimensions which are double those of the first cylinder. (i) • Find the ratio of the volume of the first cylinder to the volume of the second cylinder. 1st cylinder: radius = r, height = h 2nd cylinder: radius = 2r, height = 2h
13. A hollow cylinder has a height, h, and radiusr. A second hollow cylinder has dimensions which are double those of the first cylinder. (ii) • Find the ratio of the curved surface area of the first cylinder to the curved surface area of the second cylinder.
14. A fuel storage tank is in the shape of a cylinder with a hemisphere at each end, as shown.The capacity (volume) of the tank is 81πm3. The ratio of the capacity of the cylindrical section to the sum of the capacities of the hemispherical ends is 5 : 4. (i) Calculate the internal radius length of the tank. Capacity of cylindrical section : Capacity of hemispherical ends 5 : 4 Divide the total volume, 81π m3 in the ratio of 5 : 4 Capacity of cylindrical section = Capacity of hemispherical ends =
14. A fuel storage tank is in the shape of a cylinder with a hemisphere at each end, as shown.The capacity (volume) of the tank is 81πm3. The ratio of the capacity of the cylindrical section to the sum of the capacities of the hemispherical ends is 5 : 4. (i) Calculate the internal radius length of the tank. Volume of 2 hemispherical ends = Volume of a sphere
14. A fuel storage tank is in the shape of a cylinder with a hemisphere at each end, as shown.The capacity (volume) of the tank is 81πm3. The ratio of the capacity of the cylindrical section to the sum of the capacities of the hemispherical ends is 5 : 4. (ii) Find the height of the cylindrical part of the tank. Volume of cylinder = πr2h
14. A fuel storage tank is in the shape of a cylinder with a hemisphere at each end, as shown.The capacity (volume) of the tank is 81πm3. The ratio of the capacity of the cylindrical section to the sum of the capacities of the hemispherical ends is 5 : 4. (iii) The storage tank is to be painted. Find the curved surface area of the tank, correct to one decimal place. Curved surface area of tank = Curved surface area of cylinder + Curved surface area of hemispheres = 2πrh+ 2(2πr2) = 2π(3)(5) + 2(2π)(3)2 = 30π+ 36π = 66π = 207·345 = 207·3 m3
14. A fuel storage tank is in the shape of a cylinder with a hemisphere at each end, as shown.The capacity (volume) of the tank is 81πm3. The ratio of the capacity of the cylindrical section to the sum of the capacities of the hemispherical ends is 5 : 4. (iv) The paint used is a special paint. One litre of the paint covers 2∙5 m2. Find how many litres of paint are used, correct to the nearest litre.
14. A fuel storage tank is in the shape of a cylinder with a hemisphere at each end, as shown.The capacity (volume) of the tank is 81πm3. The ratio of the capacity of the cylindrical section to the sum of the capacities of the hemispherical ends is 5 : 4. (v) The paint is sold in five litre tins. Each tin costs €75. Find the total cost of the paint.
15. A sphere, of radius r, fits exactly inside an open-ended cylinder, as shown. (i) • Find the volume of the cylinder, in terms of r. Height of cylinder = 2r Volume of cylinder = πr2h = πr2(2r) = 2πr3
15. A sphere, of radius r, fits exactly inside an open-ended cylinder, as shown. (ii) • Find the ratio of the volume of the sphere to the volume of the cylinder.
15. A sphere, of radius r, fits exactly inside an open-ended cylinder, as shown. (iii) • Find the ratio of the curved surface area of the sphere to the curved surface area of the cylinder. Curved surface area of sphere : Curved surface area of cylinder 4πr2 : 2πrh 4πr2 : 2πr(2r) 4πr2 : 4πr2 1 : 1
16. A cone has a height, h, and radius r. A second cone has the same volume as the first, but with a height of 2h. Find the ratio of the radius of the first cone to the radius of the second cone. • So
17. You have a funnel with the dimensions shown. (i) Find the approximate volume of the funnel.
17. You have a funnel with the dimensions shown. (ii) You use the funnel to put oil in a car. Oil flows out of the funnel at a rate of 45 millilitres per second. How long will it take to empty the funnel, when it is full of oil? (1 ml = 1 cm3)
17. You have a funnel with the dimensions shown. (iii) How long would it take to empty a funnel with a radius of 10 cm and a height of 6 cm, if oil flows out of the funnel at the same rate as in part (ii)? Find the volume of the funnel or radius = 10 cm and height = 6 cm:
17. You have a funnel with the dimensions shown. (iii) How long would it take to empty a funnel with a radius of 10 cm and a height of 6 cm, if oil flows out of the funnel at the same rate as in part (ii)?
17. You have a funnel with the dimensions shown. (iv) Explain why you can claim that the time calculated in part (iii)is greater than the time calculated in part (ii), without doing any calculations. Time calculated in part (iii) is greater than the time calculated in part (ii). Since the second funnel has a greater volume than the first. This is due to the fact that the radius is squared, and a larger radius results in a much larger volume. Since (10)2 6 > (6)2 (10)
