In this chapter, we study the surface areas and volumes of combinations of solids, and also conversion of solids from one shape to another. These concepts are essential in packaging, construction, and manufacturing.
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Key Formulas
- Cuboid (length l, breadth b, height h):
- Total Surface Area (TSA) = 2(lb + bh + hl)
- Volume = l × b × h
- Cube (side a):
- TSA = 6a2, Volume = a3
- Cylinder (radius r, height h):
- Curved Surface Area (CSA) = 2 pi r h
- TSA = 2 pi r (r + h)
- Volume = pi r2 h
- Cone (radius r, height h, slant height l):
- l = √(r2 + h2)
- CSA = pi r l
- TSA = pi r (r + l)
- Volume = (1/3) pi r2 h
- Sphere (radius r):
- Surface Area = 4 pi r2
- Volume = (4/3) pi r3
- Hemisphere (radius r):
- CSA = 2 pi r2
- TSA = 3 pi r2
- Volume = (2/3) pi r3
- Frustum of a Cone (radii r1, r2; height h; slant height l):
- l = √(h2 + (r1 - r2)2)
- CSA = pi (r1 + r2) l
- TSA = pi (r1 + r2) l + pi r12 + pi r22
- Volume = (1/3) pi h (r12 + r22 + r1 r2)
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Worked Examples
A toy is a combination of a cylinder and a cone. The cylinder has radius 3 cm and height 12 cm. The cone has radius 3 cm and height 4 cm. Find the total surface area.
- Slant height of cone = √(32 + 42) = 5 cm
- TSA = CSA of cylinder + Area of circle base + CSA of cone
- = 2 pi × 3 × 12 + pi × 9 + pi × 3 × 5 = 72 pi + 9 pi + 15 pi = 96 pi sq cm
A copper sphere of radius 3 cm is melted and recast into a wire of diameter 0.4 cm. Find the length of the wire.
- Volume of sphere = (4/3) pi × 27 = 36 pi cu cm
- Volume of wire = pi × 0.04 × l (radius = 0.2 cm)
- 0.04 pi l = 36 pi => l = 900 cm = 9 m
A cone of height 24 cm and radius 6 cm is made of modelling clay. A child reshapes it into a sphere. Find the radius of the sphere.
- Volume of cone = (1/3) pi × 36 × 24 = 288 pi cu cm
- (4/3) pi r3 = 288 pi => r3 = 216 => r = 6 cm
A bucket is in the shape of a frustum. Top radius = 20 cm, bottom radius = 10 cm, height = 30 cm. Find the volume. (pi = 22/7)
- Volume = (1/3) × (22/7) × 30 × (400 + 100 + 200) = 10 × (22/7) × 700 = 22000 cu cm
A drinking glass is in the shape of a frustum with top radius 4 cm, bottom radius 2 cm and height 14 cm. Find the capacity. (pi = 22/7)
- Volume = (1/3) × (22/7) × 14 × (16 + 4 + 8) = (22/7) × 14/3 × 28 = (22 × 14 × 28) / (7 × 3) = 8624/21 = 410.67 cu cm
A hemispherical bowl of radius 9 cm is filled with water. The water is poured into a conical vessel of base radius 6 cm. Find the height of water in the cone.
- Volume of hemisphere = (2/3) pi × 729 = 486 pi cu cm
- (1/3) pi × 36 × h = 486 pi => 12h = 486 => h = 40.5 cm
Find the volume of wood required to make a wooden box (open at the top) with external dimensions 30 cm × 25 cm × 20 cm and wood thickness 2 cm.
- External volume = 30 × 25 × 20 = 15000 cu cm
- Internal dimensions = 26 × 21 × 16 (only sides, no top)
- Internal volume = 26 × 21 × 16 = 8736 cu cm
- Volume of wood = 15000 - 8736 = 6264 cu cm
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Common mistakes
> Common mistakes: Students confuse TSA and CSA — CSA excludes the base(s). For combination solids, never add the areas of the hidden/joined faces. When a solid is melted and recast, always equate volumes (not surface areas). With frustums, remember to use slant height l = √(h2 + (r1-r2)2), not just h.
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Summary
This chapter focuses on combinations of solids and conversion between shapes. Equating volumes is the key when one solid is reshaped into another. For TSA of combination solids, exclude the area of joined surfaces. Memorise all formulas for sphere, cone, cylinder, and frustum — they are used repeatedly.