Grade 9 Surface Area and Volume

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Malcolm Sims
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ID : my-9-surface-area-and-volume [1] Grade 9 Surface Area and Volume For more such worksheets visit www.edugain.

1 ID : my-9-surface-area-and-volume [1] Grade 9 Surface Area and Volume For more such worksheets visit Answer the questions (1) If the radii of two spheres are in ratio 5:2, find the ratio of their surface area. (2) If radius of a sphere is 4b, find its volume. () The heights of two cylinders are in the ratio 5:2 and their radii are in the ratio 7:5. Find the ratio of their volumes. (4) The area of a trapezium is 424 cm 2 and the distance between its parallel sides is 16 cm. If one of the parallel sides is of length 24 cm, find the length of the other side. (5) A sphere is perfectly enclosed inside a cube of volume 60 cm. Find the volume of the sphere. (6) A sphere is just enclosed inside a right circular cylinder. If the volume of the cylinder is 570 cm, find the volume of the sphere. (7) The radius of a cylinder is doubled and the height is tripled. What is the area of the curved surface when compared to the same area previously? (8) Find the volume of the biggest cone that can fit inside a cube of side 4 cm. (9) If the radius of a hemisphere is 2y, find its curved surface area. (10) A sphere and a right circular cylinder have the same radius. If the volume of the sphere is half of the volume of the cylinder, what is the ratio of the height of the cylinder to its radius? (11) A sphere is just enclosed inside a right circular cylinder. If the total surface area of the cylinder is 0 cm 2, find the surface area of the sphere. (12) A sphere and a cone have the same radius. If the volume of the sphere is triple of the volume of the cone, find the ratio of the height and radius of the cone. (1) A sphere is expanded to a bigger sphere such that its volume increases by a factor of 8, find the change in its surface area. (14) Find the volume the biggest sphere which can fit in a cube of side 6x.

2 (15) Find the surface area of the biggest sphere which can fit inside a cube of side 4a. ID : my-9-surface-area-and-volume [2] 2017 Edugain ( All Rights Reserved Many more such worksheets can be generated at

3 Answers ID : my-9-surface-area-and-volume [] (1) 25:4 The surface area of a sphere of radius x is given by 4πx 2. Assume the radii of these two spheres are 5x and 2x. Note that this allows us to get the ratio of 5:2, which is the only thing we know about these radii. The surface area of the first sphere will become 4π (5x) 2 or, 100πx 2 and the surface area of the second sphere will become 4π (2x) 2 = 16πx 2. Hence, we can see that the ratio of the surface area of two given spheres is 100πx 2 : 16πx 2 = 100 : 16 = 25:4 (2) 256 πb The volume of a sphere of radius x is given by 4 πx. The above expression is valid when the radius is x. We will have to replace it by 4b as per the question: Volume of sphere = 4 π (4b) = 4 π 64b = 256 πb. This gives us the answer as 256 πb.

4 () 49:10 ID : my-9-surface-area-and-volume [4] The volume of a cylinder is πr 2 h. To compare the ratio of the volumes we can ignore the constant multiplier π as it will be present in both the volumes. We are told that the ratio of the heights of two cylinders is 5:2. So let us represent the height of the first cylinder as 5h and that of the second cylinder as 2h. Similarly, their radii are in the ratio 7:5. So, let us represent the radius of the first cylinder as 7r and that of the second cylinder as 5r. So the ratio of the volumes is (7r) 2 5h : (5r) 2 2h. This can be simplified to 49r 2 5h : 25r 2 2h. Step 6 Simplifying this further, we get the ratio of the volumes of the two cylinders is 49:10.

5 (4) 29 cm ID : my-9-surface-area-and-volume [5] We have been told: Area of the trapezium (A) = 424 cm 2 Distance between its parallel sides (d) = 16 cm Length of one parallel side (a) = 24 cm We know that: Area of a trapezium = 1 2 (a + b) d, where a and b are the lengths of the two parallel sides and d is the distance between them. Let's substitute the known values into the formula, and solve for b: 424 = 1 2 (24 + b) = (24 + b) b = 5-24 b = 29 Therefore, the length of the other side is 29 cm.

6 ID : my-9-surface-area-and-volume [6] (5) 10π cm The volume of a sphere of radius r = 4 πr If it fits perfectly within a cube, this means the length/width/height of the cube is the same as the diameter of the sphere i.e. 2r. The volume of a cube of side 2r = (2r) = 8r. So, 8r = 60 cm This means, r = 7.5 cm We know, volume of the sphere = 4 πr Subsituting the value of r in the above equation,we get the volume = 4 π 7.5 = 10π cm

7 ID : my-9-surface-area-and-volume [7] (6) 80 cm There are three equations we need to know in this type of question - the volume of a cylinder, the volume a sphere, and the remaining volume of the gap between the sphere and the cylinder. The volume of a cylinder of radius 'r' and height 'h' is πr 2 h. Here, we know the sphere will fit in exactly in the cylinder, so h=2r, and the formula now becomes 2πr. The sphere will have the radius 'r' so its volume is 4 πr. The volume of the gap between the cylinder and the sphere is all the volume inside the cylinder not taken up by the sphere. This is the difference between the volume of the cylinder and the volume of the sphere. i.e. volume of the gap = 2πr - 4 πr Simplifying, volume of the gap = 2 πr So we have equations: Volume of the cylinder = 2πr Volume of the sphere = 4 πr Volume of the gap = 2 πr Step 6 Here, we know that volume of the cylinder is 570 cm. We need to find the volume of the sphere. Step 7 Substituting from the equation above, we get volume of the sphere = 80 cm

8 (7) 6 times ID : my-9-surface-area-and-volume [8] The curved surface area of a cylinder is 2πrh. Here, we doubled the radius and tripled the height. Putting this into the formula, we see that the curved surface area becomes 6 times. (8) 16π cm The volume of a cone is of radius r and height h = 1 πr 2 h. Since we have to fit it inside a cube of side 4 cm, we see that the diameter of the cone will be 4 cm, and the height will be 4 cm (A cone larger than this in diameter or height will not fit inside the cube). So, the radius of this cone = 4 2 cm = 2 cm Putting these values into the equation of the volume, we get the volume of the cone = 1 π ( 4 2 ) 2 4 cm On solving, we get the volume of the cone = 16π cm

9 ID : my-9-surface-area-and-volume [9] (9) 8πy 2 The surface area of a sphere of radius x is given by 4πx 2. The curved surface area of a hemisphere of same radius x, is half of that of a sphere, i.e. 2π x 2. Although, the above written expressions are valid when the radius is x. We will have to replace it by 2y as per the question. The curved surface area of the hemisphere = 2π (2y) 2 = 2π 4y 2 = 8πy 2. Hence, curved surface area of the hemisphere is 8πy 2. (10) 8: The volume of a sphere of radius 'r' = 4 πr The volume of a cylinder of radius 'r' and height 'h' = πr 2 h Here, we are told the the volume of the sphere is half of the volume of the cylinder. So, 4 πr = 1 2 (πr 2 h) Solving the above equation, we get h = 8r. Therefore, the ratio of the height of the cylinder to its radius is 8:.

10 ID : my-9-surface-area-and-volume [10] (11) 20 cm 2 There are three equations we need to know in this type of question - the total surface area of a cylinder, the curved surface area of a cylinder, and the surface area of a sphere. The curved surface area of a cylinder of radius 'r' and height 'h' is 2πrh. Here we know the sphere is enclosed in the cylinder, so h=2r, and the formula now becomes 4πr 2. The total surface area of the same cylinder will be the sum of the curved area and the surface area of the two circles at top and bottom. So, total surface area of the cylinder = 4πr 2 + 2πr 2 = 6πr 2 And of course, the sphere will have the radius 'r'. So, its surface area is 4πr 2. From these equations, we see that for this case, the surface area of the sphere is the same as the curved surface area of the cylinder, and 2 of the total surface area of the cylinder. Step 6 Here, we know that the total surface area of the cylinder is 0 cm 2. We need to find the surface area of the sphere. Step 7 Using step 5, we get the surface area of the sphere = 20 cm 2

11 (12) 4: ID : my-9-surface-area-and-volume [11] We know that the volume of a cone with radius r and height h = 1 πr 2 h. We also know that the volume of a sphere with radius r = 4 πr. We have been told that the volume of the sphere is triple the volume of the cone. Therefore, 4 πr = ( 1 πr 2 h) or, h r = 4: Thus, the ratio of the height and radius of the cone is 4:. (1) 4 times The volume of a sphere of radius x = 4 πx. The surface area of a sphere of radius x = 4πx 2. This means that the surface area will increase as a square of the increase in radius and the volume will increase as a cube of the increase in radius. Here, we know that the volume is increased by a factor of 8. This means that the surface area grew by 2 power of this value i.e. 8 2/. Solving this, we get the answer as 4 times.

12 ID : my-9-surface-area-and-volume [12] (14) 6 π x The biggest sphere that can fit inside a cube of side 6x will have a diameter of 6x (anything larger will not fit in, as opposite sides are separated by a distance of 6x. This means that the radius of this sphere is (1/2)6x The volume of a sphere of radius x is (4/)πx Therefore the volume of this sphere is (4/)π((1/2)6x) Solving for this gives us 6 π x (15) 16πa 2 The biggest sphere that can fit inside a cube of side 4a will have a diameter of 4a (anything larger will not fit in, as opposite sides are separated by a distance of 4a). This means that the radius of this sphere is 1 2 4a = 2a. We know that the surface area of a sphere of radius x is 4πx 2. So, the surface area of the given sphere of radius 2a is 4π(2a) 2 = 4π 4a 2 = 16πa 2

Grade 9 Surface Area and Volume

ID : ae-9-surface-area-and-volume [1] Grade 9 Surface Area and Volume For more such worksheets visit www.edugain.com Answer the questions (1) The radius of a cylinder is halved and the height is tripled.