Surface Areas and Volumes

Welcome dear students! Today we are going to learn about Surface Areas and Volumes from Class 9 Maths. We have already studied the surface areas of a cube, a cuboid, and a cylinder. We will now study the surface area of a cone. So far, we have been generating solids by stacking up congruent figures. Incidentally, such figures are called prisms. Now let us look at another kind of solid which is not a prism. These kinds of solids are called pyramids. Let us see how we can generate them.

Activity: Cut out a right-angled triangle ABC right-angled at B. Paste a long thick string along one of the perpendicular sides, say AB of the triangle. In Figure 11.1(a), you can see the triangle with the string pasted along side AB. Hold the string with your hands on either side of the triangle and rotate the triangle about the string a number of times. What happens? Do you recognize the shape that the triangle is forming as it rotates around the string? In Figure 11.1(b), you can see the solid shape being formed. Does it remind you of the time you had eaten an ice-cream heaped into a container of that shape? In Figures 11.1(c) and (d), you see the familiar ice-cream cone shape. This is called a right circular cone.

In Figure 11.1(c) of the right circular cone, the point A is called the vertex, AB is called the height, BC is called the radius, and AC is called the slant height of the cone. Here B will be the centre of the circular base of the cone. The height, radius, and slant height of the cone are usually denoted by h, r, and l respectively. Once again, let us see what kind of cone we cannot call a right circular cone. In Figure 11.2, you see two shapes. What you see in these figures are not right circular cones. In (a), the line joining its vertex to the centre of its base is not at a right angle to the base. In (b), the base is not circular. As in the case of a cylinder, since we will be studying only about right circular cones, remember that by 'cone' in this chapter, we shall mean a 'right circular cone.'

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Activity: First, cut out a neatly made paper cone that does not have any overlapped paper, straight along its side, and open it out to see the shape of paper that forms the surface of the cone. The line along which you cut the cone is the slant height of the cone, which is represented by l. It looks like a part of a round cake. If you now bring the sides marked A and B at the tips together, you can see that the curved portion of Figure 11.3(c) will form the circular base of the cone. If the paper like the one in Figure 11.3(c) is now cut into hundreds of little pieces, along the lines drawn from the point O, each cut portion is almost a small triangle, whose height is the slant height l of the cone. Now the area of each triangle = 1/2 × base of each triangle × l. So, the area of the entire piece of paper = sum of the areas of all the triangles = 1/2 × l × (b₁ + b₂ + b₃ + ...). This sum of bases makes up the curved portion of the figure. But the curved portion of the figure makes up the perimeter of the base of the cone, and the circumference of the base of the cone = 2πr, where r is the base radius of the cone. So, Curved Surface Area of a Cone = 1/2 × l × 2πr = πrl, where r is its base radius and l is its slant height.

Note that l² = r² + h², as can be seen from Figure 11.4, by applying Pythagoras Theorem. Here h is the height of the cone. Therefore, l = √(r² + h²). Now if the base of the cone is to be closed, then a circular piece of paper of radius r is also required, whose area is πr². So, Total Surface Area of a Cone = πrl + πr² = πr(l + r).

Let us look at Example 1. Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm. Solution: Curved surface area = πrl. Substituting the values, we get 22/7 × 7 × 10 cm² = 220 cm².

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Example 2: The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone. Use π = 3.14. Solution: Here, h = 16 cm and r = 12 cm. So, from l² = h² + r², we have l = √(16² + 12²) cm = √(256 + 144) cm = √400 cm = 20 cm. So, curved surface area = πrl = 3.14 × 12 × 20 cm² = 753.6 cm². Further, total surface area = πrl + πr² = (753.6 + 3.14 × 12 × 12) cm² = (753.6 + 452.16) cm² = 1205.76 cm².

Example 3: A corn cob, shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and length, meaning height, as 20 cm. If each 1 cm² of the surface of the cob carries an average of four grains, find how many grains you would find on the entire cob. Solution: Since the grains of corn are found only on the curved surface of the corn cob, we would need to know the curved surface area of the corn cob to find the total number of grains on it. In this question, we are given the height of the cone, so we need to find its slant height. Here, l = √(r² + h²) = √(2.1² + 20²) cm = √(4.41 + 400) cm = √404.41 cm = 20.11 cm. Therefore, the curved surface area of the corn cob = πrl = 22/7 × 2.1 × 20.11 cm² = 132.726 cm² ≈ 132.73 cm². Number of grains of corn on 1 cm² of the surface of the corn cob = 4. Therefore, number of grains on the entire curved surface of the cob = 132.73 × 4 = 530.92 ≈ 531. So, there would be approximately 531 grains of corn on the cob.

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Now let us move on to Exercise 11.1. Assume π = 22/7, unless stated otherwise. Question 1: Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area. Solution: Radius r = 10.5/2 = 5.25 cm. Slant height l = 10 cm. Curved surface area = πrl = 22/7 × 5.25 × 10 cm² = 165 cm².

Question 2: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. Solution: Radius r = 24/2 = 12 m. Slant height l = 21 m. Total surface area = πr(l + r) = 22/7 × 12 × (21 + 12) m² = 22/7 × 12 × 33 m² = 1244.57 m².

Question 3: Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone. Solution: Curved surface area = πrl. So, 308 = 22/7 × r × 14. Solving for r, we get r = 308 × 7 / (22 × 14) = 7 cm. Now, total surface area = πrl + πr². We know πrl = 308. πr² = 22/7 × 7² = 154. So total surface area = 308 + 154 = 462 cm².

Question 4: A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent. (ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹ 70. Solution: Height h = 10 m, radius r = 24 m. Slant height l = √(h² + r²) = √(10² + 24²) = √(100 + 576) = √676 = 26 m. Curved surface area of tent = πrl = 22/7 × 24 × 26 m² = 1372.57 m². Cost of canvas = area × rate = 1372.57 × 70 = ₹ 96080.

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Question 5: What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. Use π = 3.14. Solution: Height h = 8 m, radius r = 6 m. Slant height l = √(h² + r²) = √(8² + 6²) = √(64 + 36) = √100 = 10 m. Curved surface area = πrl = 3.14 × 6 × 10 = 188.4 m². Area of tarpaulin = length × width. So length × 3 = 188.4. Length = 62.8 m. Adding extra length for wastage, total length = 62.8 + 0.2 = 63 m.

Question 6: The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m². Solution: Diameter = 14 m, so radius r = 7 m. Slant height l = 25 m. Curved surface area = πrl = 22/7 × 7 × 25 = 550 m². Cost per 100 m² is ₹ 210, so cost per 1 m² is ₹ 2.10. Total cost = 550 × 2.10 = ₹ 1155.

Question 7: A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps. Solution: Radius r = 7 cm, height h = 24 cm. Slant height l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 cm. Curved surface area of one cap = πrl = 22/7 × 7 × 25 = 550 cm². Area for 10 caps = 550 × 10 = 5500 cm².

Question 8: A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹ 12 per m², what will be the cost of painting all these cones? Use π = 3.14 and take √1.04 = 1.02. Solution: Diameter = 40 cm, so radius r = 20 cm = 0.2 m. Height h = 1 m. Slant height l = √(r² + h²) = √(0.2² + 1²) = √(0.04 + 1) = √1.04 = 1.02 m. Curved surface area of one cone = πrl = 3.14 × 0.2 × 1.02 = 0.64056 m². Total area for 50 cones = 50 × 0.64056 = 32.028 m². Cost of painting = 32.028 × 12 = ₹ 384.34.

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Now let us move on to section 11.2, Surface Area of a Sphere. What is a sphere? Is it the same as a circle? Can you draw a circle on a paper? Yes, you can, because a circle is a plane closed figure whose every point lies at a constant distance, called radius, from a fixed point, which is called the centre of the circle. Now if you paste a string along a diameter of a circular disc and rotate it as you had rotated the triangle in the previous section, you see a new solid. In Figure 11.6, you can see this rotation forming a solid ball. What does it resemble? A ball? Yes. It is called a sphere. Can you guess what happens to the centre of the circle, when it forms a sphere on rotation? Of course, it becomes the centre of the sphere. So, a sphere is a three dimensional figure, which is made up of all points in the space, which lie at a constant distance called the radius, from a fixed point called the centre of the sphere. Note: A sphere is like the surface of a ball. The word solid sphere is used for the solid whose surface is a sphere.

Activity: Have you ever played with a top or have you at least watched someone play with one? You must be aware of how a string is wound around it. Now, let us take a rubber ball and drive a nail into it. Taking support of the nail, let us wind a string around the ball. When you have reached the ‘fullest’ part of the ball, use pins to keep the string in place, and continue to wind the string around the remaining part of the ball, till you have completely covered the ball. In Figure 11.7(a), you can see this process. Mark the starting and finishing points on the string, and slowly unwind the string from the surface of the ball. Now, ask your teacher to help you in measuring the diameter of the ball, from which you easily get its radius. Then on a sheet of paper, draw four circles with radius equal to the radius of the ball. Start filling the circles one by one, with the string you had wound around the ball, as shown in Figure 11.7(b). What have you achieved in all this? The string, which had completely covered the surface area of the sphere, has been used to completely fill the regions of four circles, all of the same radius as of the sphere. So, what does that mean? This suggests that the surface area of a sphere of radius r = 4 × area of a circle of radius r = 4 × πr². So, Surface Area of a Sphere = 4πr², where r is the radius of the sphere. How many faces do you see in the surface of a sphere? There is only one, which is curved.

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Now, let us take a solid sphere, and slice it exactly ‘through the middle’ with a plane that passes through its centre. What happens to the sphere? Yes, it gets divided into two equal parts, as shown in Figure 11.8. What will each half be called? It is called a hemisphere, because ‘hemi’ also means ‘half’. And what about the surface of a hemisphere? How many faces does it have? Two! There is a curved face and a flat face, which is the base. The curved surface area of a hemisphere is half the surface area of the sphere, which is 1/2 of 4πr². Therefore, Curved Surface Area of a Hemisphere = 2πr², where r is the radius of the sphere of which the hemisphere is a part. Now taking the two faces of a hemisphere, its surface area = 2πr² + πr². So, Total Surface Area of a Hemisphere = 3πr².

Let us look at Example 4. Find the surface area of a sphere of radius 7 cm. Solution: The surface area of a sphere of radius 7 cm would be 4πr² = 4 × 22/7 × 7 × 7 cm² = 616 cm².

Example 5: Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm. Solution: The curved surface area of a hemisphere of radius 21 cm would be 2πr² = 2 × 22/7 × 21 × 21 cm² = 2772 cm². (ii) The total surface area of the hemisphere would be 3πr² = 3 × 22/7 × 21 × 21 cm² = 4158 cm².

Example 6: The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding. Solution: Diameter of the sphere = 7 m. Therefore, radius is 3.5 m. So, the riding space available for the motorcyclist is the surface area of the ‘sphere’, which is given by 4πr² = 4 × 22/7 × 3.5 × 3.5 m² = 154 m².

Example 7: A hemispherical dome of a building needs to be painted. In Figure 11.9, you can see the dome. If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is ₹ 5 per 100 cm². Solution: Since only the rounded surface of the dome is to be painted, we would need to find the curved surface area of the hemisphere to know the extent of painting that needs to be done. Now, circumference of the dome = 17.6 m. Therefore, 17.6 = 2πr. So, the radius of the dome = 17.6 × 7 / (2 × 22) m = 2.8 m. The curved surface area of the dome = 2πr² = 2 × 22/7 × 2.8 × 2.8 m² = 49.28 m². Now, cost of painting 100 cm² is ₹ 5. So, cost of painting 1 m² = ₹ 500. Therefore, cost of painting the whole dome = ₹ 500 × 49.28 = ₹ 24640.

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Now let us move on to Exercise 11.2. Assume π = 22/7, unless stated otherwise. Question 1: Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm. Solution for (i): Surface area = 4πr² = 4 × 22/7 × 10.5² cm² = 1386 cm². Solution for (ii): Surface area = 4 × 22/7 × 5.6² cm² = 394.24 cm². Solution for (iii): Surface area = 4 × 22/7 × 14² cm² = 2464 cm².

Question 2: Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m. Solution for (i): Radius = 7 cm. Surface area = 4 × 22/7 × 7² = 616 cm². Solution for (ii): Radius = 10.5 cm. Surface area = 4 × 22/7 × 10.5² = 1386 cm². Solution for (iii): Radius = 1.75 m. Surface area = 4 × 22/7 × 1.75² = 38.5 m².

Question 3: Find the total surface area of a hemisphere of radius 10 cm. Use π = 3.14. Solution: Total surface area = 3πr² = 3 × 3.14 × 10² = 942 cm².

Question 4: The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. Solution: Surface area is proportional to r². So ratio = 7² : 14² = 49 : 196 = 1 : 4.

Question 5: A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹ 16 per 100 cm². Solution: Inner radius = 5.25 cm. Curved surface area of inner hemisphere = 2πr² = 2 × 22/7 × 5.25² = 173.25 cm². Cost per 100 cm² is ₹ 16, so cost per 1 cm² is ₹ 0.16. Total cost = 173.25 × 0.16 = ₹ 27.72.

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Question 6: Find the radius of a sphere whose surface area is 154 cm². Solution: 4πr² = 154. So 4 × 22/7 × r² = 154. r² = 154 × 7 / 88 = 12.25. So r = 3.5 cm.

Question 7: The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas. Solution: Let diameter of earth be D. Diameter of moon = D/4. Radius of earth = D/2. Radius of moon = D/8. Surface area ratio = (D/2)² : (D/8)² = D²/4 : D²/64 = 1/4 : 1/64 = 16 : 1.

Question 8: A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl. Solution: Inner radius = 5 cm. Thickness = 0.25 cm. Outer radius = 5 + 0.25 = 5.25 cm. Outer curved surface area = 2πR² = 2 × 22/7 × 5.25² = 173.25 cm².

Question 9: A right circular cylinder just encloses a sphere of radius r. In Figure 11.10, you can see the sphere inside the cylinder. Find (i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii). Solution: (i) Surface area of sphere = 4πr². (ii) For the cylinder enclosing the sphere, radius of cylinder = r, and height of cylinder = diameter of sphere = 2r. Curved surface area of cylinder = 2πrh = 2πr × 2r = 4πr². (iii) Ratio of surface area of sphere to curved surface area of cylinder = 4πr² : 4πr² = 1 : 1.

Now let us move on to section 11.3, Volume of a Right Circular Cone. In earlier classes we have studied the volumes of cube, cuboid, and cylinder. In Figure 11.11, you can see a right circular cylinder and a right circular cone of the same base radius and the same height. Activity: Try to make a hollow cylinder and a hollow cone like this with the same base radius and the same height. Then, we can try out an experiment that will help us to see practically what the volume of a right circular cone would be. In Figure 11.12, you can see the setup. So, let us start like this. Fill the cone up to the brim with sand once, and empty it into the cylinder. We find that it fills up only a part of the cylinder, as shown in Figure 11.12(a). When we fill up the cone again to the brim, and empty it into the cylinder, we see that the cylinder is still not full, as shown in Figure 11.12(b). When the cone is filled up for the third time, and emptied into the cylinder, it can be seen that the cylinder is also full to the brim, as shown in Figure 11.12(c). With this, we can safely come to the conclusion that three times the volume of a cone makes up the volume of a cylinder, which has the same base radius and the same height as the cone, which means that the volume of the cone is 1/3 the volume of the cylinder. So, Volume of a Cone = 1/3 πr²h, where r is the base radius and h is the height of the cone.

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Example 8: The height and the slant height of a cone are 21 cm and 28 cm respectively. Find the volume of the cone. Solution: From l² = r² + h², we have r = √(l² - h²) = √(28² - 21²) cm = √(784 - 441) cm = √343 cm = 7√7 cm. So, volume of the cone = 1/3 πr²h = 1/3 × 22/7 × (7√7)² × 21 cm³ = 1/3 × 22/7 × 343 × 21 cm³ = 7546 cm³.

Example 9: Monica has a piece of canvas whose area is 551 m². She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m², find the volume of the tent that can be made with it. Solution: Since the area of the canvas = 551 m² and area of the canvas lost in wastage is 1 m², therefore the area of canvas available for making the tent is 551 - 1 = 550 m². Now, the surface area of the tent = 550 m² and the required base radius of the conical tent = 7 m. Note that a tent has only a curved surface, the floor of a tent is not covered by canvas. Therefore, curved surface area of tent = 550 m². That is, πrl = 550. So, 22/7 × 7 × l = 550. Solving for l, we get l = 550 / 22 = 25 m. Now, l² = r² + h². Therefore, h = √(l² - r²) = √(25² - 7²) m = √(625 - 49) m = √576 m = 24 m. So, the volume of the conical tent = 1/3 πr²h = 1/3 × 22/7 × 7² × 24 m³ = 1232 m³.

Now let us move on to Exercise 11.3. Assume π = 22/7, unless stated otherwise. Question 1: Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm. Solution for (i): Volume = 1/3 πr²h = 1/3 × 22/7 × 6² × 7 cm³ = 264 cm³. Solution for (ii): Volume = 1/3 × 22/7 × 3.5² × 12 cm³ = 154 cm³.

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Question 2: Find the capacity in litres of a conical vessel with (i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm. Solution for (i): Radius r = 7 cm, slant height l = 25 cm. Height h = √(l² - r²) = √(625 - 49) = √576 = 24 cm. Volume = 1/3 πr²h = 1/3 × 22/7 × 49 × 24 cm³ = 1232 cm³. Since 1 litre = 1000 cm³, capacity = 1.232 litres. Solution for (ii): Height h = 12 cm, slant height l = 13 cm. Radius r = √(l² - h²) = √(169 - 144) = √25 = 5 cm. Volume = 1/3 × 22/7 × 25 × 12 cm³ = 2200/7 cm³ ≈ 314.29 cm³. In litres, capacity ≈ 0.314 litres.

Question 3: The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. Use π = 3.14. Solution: Volume = 1/3 πr²h. So 1570 = 1/3 × 3.14 × r² × 15. Simplifying, 1570 = 15.7 × r². So r² = 1570 / 15.7 = 100. Therefore, r = 10 cm.

Question 4: If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base. Solution: Volume = 1/3 πr²h. So 48π = 1/3 πr² × 9. Simplifying, 48 = 3r². So r² = 16. Therefore, r = 4 cm. Diameter = 2r = 8 cm.

Question 5: A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? Solution: Diameter = 3.5 m, so radius r = 1.75 m. Height h = 12 m. Volume = 1/3 πr²h = 1/3 × 22/7 × 1.75² × 12 m³ = 38.5 m³. Since 1 kilolitre = 1 m³, capacity = 38.5 kilolitres.

Question 6: The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone. Solution: Diameter = 28 cm, so radius r = 14 cm. Volume = 1/3 πr²h. So 9856 = 1/3 × 22/7 × 14² × h. Simplifying, 9856 = 616/3 × h. So h = 9856 × 3 / 616 = 48 cm. Slant height l = √(r² + h²) = √(14² + 48²) = √(196 + 2304) = √2500 = 50 cm. Curved surface area = πrl = 22/7 × 14 × 50 cm² = 2200 cm².

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Question 7: A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained. Solution: When revolved about the 12 cm side, it forms a cone with height h = 12 cm and radius r = 5 cm. Volume = 1/3 πr²h = 1/3 × 22/7 × 25 × 12 cm³ = 2200/7 cm³ ≈ 314.29 cm³.

Question 8: If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8. Solution: When revolved about the 5 cm side, it forms a cone with height h = 5 cm and radius r = 12 cm. Volume = 1/3 πr²h = 1/3 × 22/7 × 144 × 5 cm³ = 5280/7 cm³ ≈ 754.29 cm³. Ratio of volumes from Question 7 to Question 8 = (2200/7) : (5280/7) = 2200 : 5280 = 5 : 12.

Question 9: A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required. Solution: Diameter = 10.5 m, so radius r = 5.25 m. Height h = 3 m. Volume = 1/3 πr²h = 1/3 × 22/7 × 5.25² × 3 m³ = 86.625 m³. Slant height l = √(r² + h²) = √(5.25² + 3²) = √(27.5625 + 9) = √36.5625 ≈ 6.05 m. Area of canvas required = curved surface area = πrl = 22/7 × 5.25 × 6.05 m² ≈ 99.825 m².

Now let us move on to section 11.4, Volume of a Sphere. First, take two or three spheres of different radii, and a container big enough to be able to put each of the spheres into it, one at a time. Also, take a large trough in which you can place the container. Then, fill the container up to the brim with water, as shown in Figure 11.13(a). Now, carefully place one of the spheres in the container. Some of the water from the container will overflow into the trough in which it is kept, as shown in Figure 11.13(b). Carefully pour out the water from the trough into a measuring cylinder and measure the water overflowed, as shown in Figure 11.13(c). Suppose the radius of the immersed sphere is r. Then evaluate 4/3 πr³. Do you find this value almost equal to the measure of the volume overflowed? Once again repeat the procedure done just now, with a different size of sphere. Find the radius R of this sphere and then calculate the value of 4/3 πR³. Once again this value is nearly equal to the measure of the volume of the water displaced by the sphere. What does this tell us? We know that the volume of the sphere is the same as the measure of the volume of the water displaced by it. By doing this experiment repeatedly with spheres of varying radii, we are getting the same result, namely, the volume of a sphere is equal to 4/3 π times the cube of its radius. This gives us the idea that Volume of a Sphere = 4/3 πr³, where r is the radius of the sphere. Later, in higher classes it can be proved also. But at this stage, we will just take it as true.

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Since a hemisphere is half of a sphere, can you guess what the volume of a hemisphere will be? Yes, it is 1/2 of 4/3 πr³, which equals 2/3 πr³. So, Volume of a Hemisphere = 2/3 πr³, where r is the radius of the hemisphere.

Let us take some examples to illustrate the use of these formulae. Example 10: Find the volume of a sphere of radius 11.2 cm. Solution: Required volume = 4/3 πr³ = 4/3 × 22/7 × 11.2³ cm³ = 5887.32 cm³.

Example 11: A shot-putt is a metallic sphere of radius 4.9 cm. If the density of the metal is 7.8 g per cm³, find the mass of the shot-putt. Solution: Since the shot-putt is a solid sphere made of metal and its mass is equal to the product of its volume and density, we need to find the volume of the sphere. Now, volume of the sphere = 4/3 πr³ = 4/3 × 22/7 × 4.9³ cm³ ≈ 493 cm³. Further, mass of 1 cm³ of metal is 7.8 g. Therefore, mass of the shot-putt = 7.8 × 493 g = 3845.44 g ≈ 3.85 kg.

Example 12: A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain? Solution: The volume of water the bowl can contain = 2/3 πr³ = 2/3 × 22/7 × 3.5³ cm³ = 89.8 cm³.

Now let us move on to Exercise 11.4. Assume π = 22/7, unless stated otherwise. Question 1: Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m. Solution for (i): Volume = 4/3 πr³ = 4/3 × 22/7 × 7³ cm³ = 1437.33 cm³. Solution for (ii): Volume = 4/3 × 22/7 × 0.63³ m³ = 1.0478 m³ ≈ 1.05 m³.

Question 2: Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m. Solution for (i): Diameter = 28 cm, so radius = 14 cm. Volume displaced = 4/3 πr³ = 4/3 × 22/7 × 14³ cm³ = 34496/3 cm³ ≈ 11498.67 cm³. Solution for (ii): Diameter = 0.21 m, so radius = 0.105 m. Volume displaced = 4/3 × 22/7 × 0.105³ m³ ≈ 0.00485 m³.

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Question 3: The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³? Solution: Diameter = 4.2 cm, so radius = 2.1 cm. Volume = 4/3 πr³ = 4/3 × 22/7 × 2.1³ cm³ = 38.808 cm³. Mass = density × volume = 8.9 × 38.808 g = 345.39 g.

Question 4: The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon? Solution: Let diameter of earth be D. Diameter of moon = D/4. Radius of earth = D/2. Radius of moon = D/8. Volume ratio = (D/2)³ : (D/8)³ = D³/8 : D³/512 = 1/8 : 1/512 = 1 : 64. So the volume of the moon is 1/64 of the volume of the earth.

Question 5: How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold? Solution: Diameter = 10.5 cm, so radius = 5.25 cm. Volume of hemisphere = 2/3 πr³ = 2/3 × 22/7 × 5.25³ cm³ = 303.1875 cm³. Since 1 litre = 1000 cm³, capacity = 0.303 litres.

Question 6: A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank. Solution: Inner radius r = 1 m = 100 cm. Thickness = 1 cm. Outer radius R = 100 + 1 = 101 cm. Volume of iron = volume of outer hemisphere - volume of inner hemisphere = 2/3 π(R³ - r³) = 2/3 × 22/7 × (101³ - 100³) cm³ = 2/3 × 22/7 × (1030301 - 1000000) cm³ = 2/3 × 22/7 × 30301 cm³ ≈ 63487.33 cm³ ≈ 0.0635 m³.

Question 7: Find the volume of a sphere whose surface area is 154 cm². Solution: Surface area = 4πr². So 154 = 4 × 22/7 × r². Solving for r², we get r² = 12.25. So r = 3.5 cm. Volume = 4/3 πr³ = 4/3 × 22/7 × 3.5³ cm³ = 179.67 cm³.

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Question 8: A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹ 4989.60. If the cost of white-washing is ₹ 20 per square metre, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome. Solution: Total cost = ₹ 4989.60. Rate = ₹ 20 per m². Inside curved surface area = total cost / rate = 4989.60 / 20 = 249.48 m². Curved surface area of hemisphere = 2πr². So 249.48 = 2 × 22/7 × r². Solving for r², we get r² = 39.69. So r = 6.3 m. Volume of air inside = 2/3 πr³ = 2/3 × 22/7 × 6.3³ m³ = 523.908 m³.

Question 9: Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the (i) radius r′ of the new sphere, (ii) ratio of S and S′. Solution: Volume of one small sphere = 4/3 πr³. Volume of 27 spheres = 27 × 4/3 πr³. Volume of new sphere = 4/3 πr′³. Equating volumes, 27 × 4/3 πr³ = 4/3 πr′³. So r′³ = 27r³. Therefore, r′ = 3r. Surface area S = 4πr². Surface area S′ = 4πr′² = 4π(3r)² = 36πr². Ratio of S to S′ = 4πr² : 36πr² = 1 : 9.

Question 10: A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine in mm³ is needed to fill this capsule? Solution: Diameter = 3.5 mm, so radius = 1.75 mm. Volume = 4/3 πr³ = 4/3 × 22/7 × 1.75³ mm³ ≈ 22.46 mm³.

Finally, let us review the summary of this chapter. In this chapter, you have studied the following points. First, Curved surface area of a cone = πrl. Second, Total surface area of a right circular cone = πrl + πr², which is πr(l + r). Third, Surface area of a sphere of radius r = 4πr². Fourth, Curved surface area of a hemisphere = 2πr². Fifth, Total surface area of a hemisphere = 3πr². Sixth, Volume of a cone = 1/3 πr²h. Seventh, Volume of a sphere of radius r = 4/3 πr³. Eighth, Volume of a hemisphere = 2/3 πr³. Here, letters l, b, h, a, r, etc. have been used in their usual meaning, depending on the context. Make sure you memorize these formulas and practice applying them to various problems. Understanding the derivation through activities and experiments will help you retain these concepts for your examinations.

Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]