Welcome dear students! Today we are going to learn about Surface Areas And Volumes from Class 10 Maths. From Class IX, you are familiar with basic solids like cuboid, cone, cylinder, and sphere. You have also learnt how to find their surface areas and volumes. In daily life, we encounter solids made of combinations of two or more basic solids. Consider a truck container carrying oil or water. It is made of a cylinder with two hemispheres at its ends. Similarly, a test tube is a combination of a cylinder and a hemisphere. Buildings and monuments also often combine these shapes. To find the surface areas, volumes, or capacities of such composite objects, we cannot use single solid formulas. In this chapter, you will learn how to calculate them. [CHECKPOINT]
Let us consider the container. To find its surface area, we break it into smaller, familiar parts. This solid is a cylinder with two hemispheres attached at either end. When joined, the visible surface consists only of the curved surfaces of the two hemispheres and the curved surface of the cylinder. The flat faces disappear inside. Therefore, the Total Surface Area (TSA) of the new solid is the sum of the Curved Surface Areas (CSA) of each part. Mathematically: TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of the other hemisphere. Now, consider a toy made by joining a hemisphere and a cone base-to-base. The base radius of the cone equals the radius of the hemisphere for a smooth surface. The visible surface area of this toy consists of the CSA of the hemisphere and the CSA of the cone. Thus: Total surface area of the toy = CSA of hemisphere + CSA of cone. Let us now apply this to examples. [CHECKPOINT]
Example 1: Rasheed got a playing top shaped like a cone surmounted by a hemisphere, as shown in Figure 12.6. The entire top is 5 cm in height and the diameter is 3.5 cm. Find the area he has to colour. Take π = 22/7. Solution: The top matches our previous discussion. TSA of the toy = CSA of hemisphere + CSA of cone. Radius r = 3.5/2 cm. CSA of hemisphere = 2πr² = 2 × (22/7) × (3.5/2)² cm². Height of cone = Total height – radius of hemisphere = 5 – (3.5/2) = 3.25 cm. Slant height of cone, l = √(r² + h²) = √((3.5/2)² + (3.25)²) ≈ 3.7 cm. CSA of cone = πrl = (22/7) × (3.5/2) × 3.7 cm². Total surface area = 2πr² + πrl = πr(2r + l) = (22/7) × (3.5/2) × (3.5 + 3.7) = 11 × 3.6 = 39.6 cm² (approx). Note: The TSA of the top is not the sum of the TSAs of the separate cone and hemisphere. [CHECKPOINT]
Example 2: A decorative block, shown in Figure 12.7, is made of a cube and a hemisphere. The cube has an edge of 5 cm. The hemisphere on top has a diameter of 4.2 cm. Find the TSA of the block. Take π = 22/7. Solution: TSA of cube = 6 × (edge)² = 6 × 5² = 150 cm². The part of the cube where the hemisphere attaches is hidden. So, surface area of block = TSA of cube – base area of hemisphere + CSA of hemisphere. This equals 150 – πr² + 2πr² = 150 + πr². Radius r = 4.2/2 = 2.1 cm. Surface area = 150 + (22/7) × (2.1)² = 150 + 13.86 = 163.86 cm². [CHECKPOINT]
Example 3: A wooden toy rocket is a cone mounted on a cylinder, as shown in Figure 12.8. Total height is 26 cm, conical height is 6 cm. Cone base diameter is 5 cm, cylinder base diameter is 3 cm. The conical part is painted orange, cylindrical part yellow. Find the painted area for each colour. Take π = 3.14. Solution: Let r = cone radius = 2.5 cm, h = cone height = 6 cm. Let r′ = cylinder radius = 1.5 cm, h′ = cylinder height = 26 – 6 = 20 cm. Slant height of cone, l = √(r² + h²) = √(2.5² + 6²) = 6.5 cm. The cone's base rests on the cylinder, but the cone base is larger. A ring-shaped portion of the cone's base is exposed and must be painted orange. Area painted orange = CSA of cone + base area of cone – base area of cylinder = πrl + πr² – π(r′)² = π[rl + r² – (r′)²] = 3.14 × [(2.5 × 6.5) + (2.5)² – (1.5)²] = 3.14 × [16.25 + 6.25 – 2.25] = 3.14 × 20.25 = 63.585 cm². Area painted yellow = CSA of cylinder + area of one base of cylinder = 2πr′h′ + π(r′)² = πr′(2h′ + r′) = 3.14 × 1.5 × (2 × 20 + 1.5) = 4.71 × 41.5 = 195.465 cm². [CHECKPOINT]
Example 4: A bird-bath, shown in Figure 12.9, is a cylinder with a hemispherical depression at one end. Cylinder height is 1.45 m, radius is 30 cm. Find the TSA. Take π = 22/7. Solution: Let h = 1.45 m = 145 cm, r = 30 cm. TSA of bird-bath = CSA of cylinder + CSA of hemisphere = 2πrh + 2πr² = 2πr(h + r) = 2 × (22/7) × 30 × (145 + 30) = (1320/7) × 175 = 33000 cm² = 3.3 m². [CHECKPOINT]
Now, Exercise 12.1. Unless stated otherwise, take π = 22/7. Question 1: Two cubes, each of volume 64 cm³, are joined end to end. Find the surface area of the resulting cuboid. Solution: Side of cube = ∛64 = 4 cm. Joined cuboid dimensions: length = 8 cm, breadth = 4 cm, height = 4 cm. Surface area = 2(lb + bh + hl) = 2(8×4 + 4×4 + 4×8) = 2(32 + 16 + 32) = 160 cm². Question 2: A vessel is a hollow hemisphere mounted by a hollow cylinder. Hemisphere diameter is 14 cm, total height is 13 cm. Find inner surface area. Solution: Radius r = 7 cm. Cylinder height h = 13 – 7 = 6 cm. Inner surface area = CSA of hemisphere + CSA of cylinder = 2πr² + 2πrh = 2πr(r + h) = 2 × (22/7) × 7 × (7 + 6) = 44 × 13 = 572 cm². [CHECKPOINT]
Question 3: A toy is a cone of radius 3.5 cm mounted on a hemisphere of same radius. Total height is 15.5 cm. Find TSA. Solution: r = 3.5 cm. Cone height h = 15.5 – 3.5 = 12 cm. Slant height l = √(r² + h²) = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm. TSA = CSA of cone + CSA of hemisphere = πrl + 2πr² = πr(l + 2r) = (22/7) × 3.5 × (12.5 + 7) = 11 × 19.5 = 214.5 cm². Question 4: A cubical block of side 7 cm is surmounted by a hemisphere. Greatest diameter? Find surface area. Solution: Greatest diameter = edge of cube = 7 cm. Radius r = 3.5 cm. Surface area = TSA of cube – base area of hemisphere + CSA of hemisphere = 6a² + πr² = 6 × 7² + (22/7) × (3.5)² = 294 + 38.5 = 332.5 cm². [CHECKPOINT]
Question 5: A hemispherical depression is cut from one face of a cubical block. Diameter of hemisphere equals cube edge l. Find surface area of remaining solid. Solution: Radius = l/2. Surface area = TSA of cube – base area of hemisphere + CSA of hemisphere = 6l² – π(l/2)² + 2π(l/2)² = 6l² + πl²/4. Question 6: A medicine capsule is a cylinder with two hemispheres at ends, as shown in Figure 12.10. Total length 14 mm, diameter 5 mm. Find surface area. Solution: Radius r = 2.5 mm. Cylinder length h = 14 – 2(2.5) = 9 mm. Surface area = CSA of cylinder + 2 × CSA of hemisphere = 2πrh + 4πr² = 2πr(h + 2r) = 2 × (22/7) × 2.5 × (9 + 5) = (110/7) × 14 = 220 mm². [CHECKPOINT]
Question 7: A tent is a cylinder surmounted by a cone. Cylinder height 2.1 m, diameter 4 m. Cone slant height 2.8 m. Find canvas area and cost at ₹500/m². Base not covered. Solution: Radius r = 2 m. Cylinder height h = 2.1 m. Cone slant height l = 2.8 m. Canvas area = CSA of cylinder + CSA of cone = 2πrh + πrl = πr(2h + l) = (22/7) × 2 × (4.2 + 2.8) = (44/7) × 7 = 44 m². Cost = 44 × 500 = ₹22,000. Question 8: From a solid cylinder (height 2.4 cm, diameter 1.4 cm), a conical cavity of same dimensions is hollowed out. Find TSA of remaining solid to nearest cm². Solution: Radius r = 0.7 cm, height h = 2.4 cm. Cone slant height l = √(0.7² + 2.4²) = 2.5 cm. TSA = CSA of cylinder + CSA of cone + area of circular base = 2πrh + πrl + πr² = πr(2h + l + r) = (22/7) × 0.7 × (4.8 + 2.5 + 0.7) = 2.2 × 8 = 17.6 cm² ≈ 18 cm². [CHECKPOINT]
Question 9: A wooden article is a cylinder with a hemisphere scooped out from each end, as shown in Figure 12.11. Cylinder height 10 cm, radius 3.5 cm. Find TSA. Solution: r = 3.5 cm, h = 10 cm. TSA = CSA of cylinder + 2 × CSA of hemisphere = 2πrh + 4πr² = 2πr(h + 2r) = 2 × (22/7) × 3.5 × (10 + 7) = 44 × 17 = 748 cm². [CHECKPOINT]
Now, Section 12.3: Volume of a Combination of Solids. Unlike surface area, where joining solids hides some faces, volume is additive. The volume of a composite solid is simply the sum of the volumes of its constituent parts. Let us see examples. [CHECKPOINT]
Example 5: A shed is a cuboid surmounted by a half cylinder, as shown in Figure 12.12. Base dimensions 7 m × 15 m. Cuboid height 8 m. Find volume of air. Machinery occupies 300 m³, 20 workers occupy 0.08 m³ each. Find remaining air volume. Take π = 22/7. Solution: Cuboid volume = 15 × 7 × 8 = 840 m³. Half cylinder: diameter = 7 m (radius = 3.5 m), length = 15 m. Volume = (1/2) × πr²h = (1/2) × (22/7) × (3.5)² × 15 = 288.75 m³. Total air volume = 840 + 288.75 = 1128.75 m³. Space for machinery = 300 m³. Space for workers = 20 × 0.08 = 1.6 m³. Remaining air = 1128.75 – (300 + 1.6) = 827.15 m³. [CHECKPOINT]
Example 6: A cylindrical glass, shown in Figure 12.13, has inner diameter 5 cm, height 10 cm. It has a hemispherical raised bottom. Find apparent and actual capacity. Use π = 3.14. Solution: Radius r = 2.5 cm, height h = 10 cm. Apparent capacity = πr²h = 3.14 × (2.5)² × 10 = 196.25 cm³. Volume of hemisphere = (2/3)πr³ = (2/3) × 3.14 × (2.5)³ = 32.71 cm³. Actual capacity = 196.25 – 32.71 = 163.54 cm³. [CHECKPOINT]
Example 7: A toy is a hemisphere surmounted by a cone, as shown in Figure 12.14. Cone height 2 cm, base diameter 4 cm. Find toy volume. If a cylinder circumscribes the toy, find volume difference. Take π = 3.14. Solution: Radius r = 2 cm. Cone height h = 2 cm. Volume of toy = Volume of cone + Volume of hemisphere = (1/3)πr²h + (2/3)πr³ = (1/3) × 3.14 × 4 × 2 + (2/3) × 3.14 × 8 = 8.37 + 16.75 = 25.12 cm³. Circumscribing cylinder: radius = 2 cm, height = 2 + 2 = 4 cm. Cylinder volume = πr²H = 3.14 × 4 × 4 = 50.24 cm³. Difference = 50.24 – 25.12 = 25.12 cm³. [CHECKPOINT]
Now, Exercise 12.2. Unless stated otherwise, take π = 22/7. Question 1: A solid is a cone on a hemisphere. Both radii = 1 cm. Cone height = radius = 1 cm. Find volume in terms of π. Solution: Volume = (1/3)πr²h + (2/3)πr³ = (1/3)π(1)²(1) + (2/3)π(1)³ = π/3 + 2π/3 = π cm³. Question 2: Model: cylinder with two cones at ends, as shown in Figure 12.15. Diameter 3 cm, total length 12 cm. Each cone height 2 cm. Find volume of air. Solution: Radius r = 1.5 cm. Cylinder height h = 12 – 2(2) = 8 cm. Volume = πr²h + 2 × (1/3)πr²h_cone = πr²(h + 2h_cone/3) = (22/7) × (1.5)² × (8 + 4/3) = (22/7) × 2.25 × (28/3) = 66 cm³. [CHECKPOINT]
Question 3: Gulab jamun: cylinder with two hemispherical ends, as shown in Figure 12.15. Length 5 cm, diameter 2.8 cm. Syrup is 30% of volume. Find syrup in 45 gulab jamuns. Solution: Radius r = 1.4 cm. Cylinder length h = 5 – 2.8 = 2.2 cm. Volume of one = πr²h + (4/3)πr³ = (22/7) × (1.4)² × 2.2 + (4/3) × (22/7) × (1.4)³ = 13.552 + 11.4987 ≈ 25.05 cm³. Syrup in one = 30% of 25.05 = 7.515 cm³. Syrup in 45 = 45 × 7.515 = 338.175 cm³. Question 4: Pen stand: cuboid 15×10×3.5 cm with four conical depressions (r=0.5 cm, depth=1.4 cm), as shown in Figure 12.16. Find wood volume. Solution: Cuboid volume = 15 × 10 × 3.5 = 525 cm³. Volume of one depression = (1/3)πr²h = (1/3) × (22/7) × (0.5)² × 1.4 ≈ 0.3667 cm³. Volume of four = 1.4667 cm³. Wood volume = 525 – 1.4667 = 523.533 cm³. [CHECKPOINT]
Question 5: Inverted cone vessel: height 8 cm, top radius 5 cm. Filled with water. Lead shots (spheres, r=0.5 cm) dropped, 1/4 water flows out. Find number of shots. Solution: Water volume = (1/3)πR²H = (1/3) × (22/7) × 25 × 8 = 200π/3 cm³. Volume displaced = 1/4 × (200π/3) = 50π/3 cm³. Volume of one shot = (4/3)πr³ = (4/3)π(0.5)³ = π/6 cm³. Number of shots = (50π/3) ÷ (π/6) = 100. Question 6: Iron pole: cylinder 1 (h=220 cm, d=24 cm) surmounted by cylinder 2 (h=60 cm, r=8 cm). Mass? 1 cm³ iron = 8 g. Use π = 3.14. Solution: Cylinder 1: r₁ = 12 cm, h₁ = 220 cm. Vol₁ = 3.14 × 12² × 220 = 99475.2 cm³. Cylinder 2: r₂ = 8 cm, h₂ = 60 cm. Vol₂ = 3.14 × 8² × 60 = 12057.6 cm³. Total vol = 111532.8 cm³. Mass = 111532.8 × 8 = 892262.4 g = 892.26 kg. [CHECKPOINT]
Question 7: Solid: cone (h=120, r=60) on hemisphere (r=60). Placed in cylinder (r=60, H=180) full of water. Find water left. Solution: Cylinder vol = π(60)²(180) = 648000π. Solid vol = (1/3)π(60)²(120) + (2/3)π(60)³ = 144000π + 144000π = 288000π. Water left = 648000π – 288000π = 360000π. Using π = 22/7: 360000 × 22/7 = 1131428.57 cm³. Question 8: Spherical vessel with cylindrical neck (8 cm long, 2 cm diameter). Sphere diameter 8.5 cm. Child measures 345 cm³. Check. Use π = 3.14. Solution: Neck: r₁ = 1 cm, h = 8 cm. Vol = 3.14 × 1² × 8 = 25.12 cm³. Sphere: r₂ = 4.25 cm. Vol = (4/3) × 3.14 × (4.25)³ = 321.39 cm³. Total vol = 25.12 + 321.39 = 346.51 cm³. Child's measurement: 345 cm³. Since 346.51 ≠ 345, the child is incorrect. [CHECKPOINT]
Finally, let us review the summary. In this chapter, you studied: 1. How to determine the surface area of an object formed by combining any two basic solids: cuboid, cone, cylinder, sphere, and hemisphere. 2. How to find the volume of objects formed by combining these solids. Remember, for surface area, we only add exposed curved surfaces, while for volume, we sum the volumes of all parts. Practice these concepts regularly for your examinations. [CHECKPOINT]
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]