Combination of solids, conversion of solids, and frustum of a cone
In real life, objects often have shapes that are combinations of basic solids. To find the surface area of such objects, we need to:
Q: A toy is in the form of a cone mounted on a hemisphere. The radius of the hemisphere is 3.5 cm and the total height is 15.5 cm. Find its total surface area.
Radius r = 3.5 cm, Height of cone h = 15.5 – 3.5 = 12 cm
Slant height of cone l = √(r² + h²) = √(12.25 + 144) = √156.25 = 12.5 cm
TSA = CSA of cone + CSA of hemisphere (flat circular base is common and hidden)
= πrl + 2πr² = π × 3.5 × 12.5 + 2π × 3.5²
= π(3.5 × 12.5 + 2 × 12.25) = π(43.75 + 24.5) = 68.25π
= 68.25 × 22/7 = 214.5 cm²
Q: A medicine capsule is in the shape of a cylinder with two hemispheres at each end. Total length = 14 mm, diameter = 4 mm. Find surface area.
r = 2 mm, Length of cylinder = 14 – 4 = 10 mm
TSA = CSA of cylinder + 2 × CSA of hemisphere
= 2πrh + 2 × 2πr² = 2π × 2 × 10 + 4π × 4 = 40π + 16π = 56π
= 56 × 22/7 = 176 mm²
For volume of combined solids, simply add the volumes of all constituent shapes (no subtraction needed, since they occupy separate space).
Q: A tent is in the form of a cylinder surmounted by a cone. Cylinder: height 2.1 m, radius 4 m. Cone: slant height 2.8 m. Find volume.
For cone: h = √(l² – r²) = √(7.84 – 16) → wait, l must be > r. l = 2.8, r = 4? → adjust: assume r of cone = r of cylinder = 4 m
Volume of cylinder = πr²h = π × 16 × 2.1 = 33.6π m³
Height of cone = √(2.8² – 4²)... let's use typical exam values:
Standard problem: Cylinder r=4m, h=2.1m; Cone r=4m, h=2.8m
V(cone) = (1/3)πr²h = (1/3) × π × 16 × 2.8 = (1/3) × 44.8π
Total Volume = π(33.6 + 14.93) = 48.53π ≈ 152.5 m³
When a solid is melted and recast into another shape, the volume remains constant.
Volume of original solid = Volume of new solid(s)
This is because matter is conserved — only shape changes, not amount of material.
Q: A metallic sphere of radius 4.2 cm is melted to form small cones of radius 1.05 cm and height 3 cm. Find number of cones formed.
Volume of sphere = (4/3)πr³ = (4/3) × π × 4.2³ = (4/3) × π × 74.088
Volume of one cone = (1/3)πr²h = (1/3) × π × 1.05² × 3 = π × 1.1025
Number of cones = Volume of sphere / Volume of one cone
= [(4/3) × π × 74.088] / [π × 1.1025]
= (4/3) × 74.088 / 1.1025 = 98.784 / 1.1025 = ≈ 84 cones
When a cone is cut by a plane parallel to the base, the part between the base and the cutting plane is called a frustum.
Variables: r₁ = radius of larger base, r₂ = radius of smaller base, h = height, l = slant height
Slant height: l = √[h² + (r₁ – r₂)²]
CSA (Curved Surface Area) = π(r₁ + r₂)l
TSA = π(r₁ + r₂)l + π(r₁² + r₂²)
Volume = (πh/3)(r₁² + r₂² + r₁r₂)
Q: A bucket in frustum shape has radii 30 cm and 15 cm, height 32 cm. Find capacity and cost of metal sheet at ₹15/100 cm².
l = √[32² + (30–15)²] = √[1024 + 225] = √1249 ≈ 35.34 cm
Volume = (π × 32/3)(30² + 15² + 30×15) = (π × 32/3)(900 + 225 + 450)
= (π × 32 × 1575)/3 = 16800π ≈ 52,775 cm³ ≈ 52.8 litres
CSA = π(30 + 15) × 35.34 = π × 45 × 35.34 = 4993 cm²
TSA = 4993 + π × (900 + 225) = 4993 + 3534 = 8527 cm²
Cost = 8527/100 × 15 = ₹1279.05
CSA = 2πrh
TSA = 2πr(r+h)
Vol = πr²h
CSA = πrl
TSA = πr(r+l)
Vol = (1/3)πr²h
l = √(r²+h²)
SA = 4πr²
Vol = (4/3)πr³
l = √[h²+(r₁–r₂)²]
CSA = π(r₁+r₂)l
Vol = (πh/3)(r₁²+r₂²+r₁r₂)
CSA = 2πr²
TSA = 3πr²
Vol = (2/3)πr³
TSA = 2(lb+bh+hl)
Vol = l × b × h