All Formulae — Quick Reference
Cuboid
| Quantity | Formula |
|---|---|
| LSA (Lateral Surface Area) | 2h(l+b) |
| TSA | 2(lb+bh+hl) |
| Volume | l × b × h |
| Diagonal | √(l²+b²+h²) |
Cube (side a)
- TSA = 6a²; Volume = a³; Diagonal = a√3
Cylinder (radius r, height h)
| Quantity | Formula |
|---|---|
| CSA | 2πrh |
| TSA | 2πr(h+r) |
| Volume | πr²h |
Cone (radius r, height h, slant height l = √(r²+h²))
| Quantity | Formula |
|---|---|
| CSA | πrl |
| TSA | πr(l+r) |
| Volume | (1/3)πr²h |
Sphere (radius r)
| Quantity | Formula |
|---|---|
| Surface Area | 4πr² |
| Volume | (4/3)πr³ |
Hemisphere (radius r)
| Quantity | Formula |
|---|---|
| CSA | 2πr² |
| TSA | 3πr² |
| Volume | (2/3)πr³ |
Frustum of Cone (radii r₁, r₂; height h; slant height l)
l = √[h² + (r₁−r₂)²]
| Quantity | Formula |
|---|---|
| CSA | π(r₁+r₂)l |
| TSA | π(r₁+r₂)l + πr₁² + πr₂² |
| Volume | (πh/3)(r₁² + r₂² + r₁r₂) |
Worked Examples — Combination of Solids
Q1. Toy = Cone on top of Hemisphere A toy is in the form of a cone mounted on a hemisphere. The radius of base is 3.5 cm and total height is 15.5 cm. Find TSA.
Hemisphere radius = 3.5 cm; Cone height = 15.5 − 3.5 = 12 cm Slant height l = √(12² + 3.5²) = √(144 + 12.25) = √156.25 = 12.5 cm
TSA = CSA of cone + CSA of hemisphere = πrl + 2πr² = π × 3.5 × 12.5 + 2π × 3.5² = π(43.75 + 24.5) = 68.25π = 68.25 × 22/7 = 214.5 cm²
Q2. Capsule = Cylinder with two hemispheres A capsule is cylinder of radius r and length l, with hemispherical ends. Find total surface area.
TSA = CSA of cylinder + 2 × CSA of hemisphere = 2πrl + 2 × 2πr² = 2πr(l + 2r)
(The flat circular ends of the cylinder are covered by hemispheres — so no πr² terms)
Q3. Solid = Cone removed from Cylinder A solid wooden toy is a cylinder of radius 7 cm and height 10 cm, with a cone of same base removed from top (cone height = 6 cm). Find volume.
Volume = Volume of cylinder − Volume of cone = πr²h − (1/3)πr²h_cone = π(7²)(10) − (1/3)π(7²)(6) = 490π − 98π = 392π = 392 × 22/7 = 1232 cm³
Conversion / Recasting Problems
Rule: When a solid is melted and recast, volume is conserved.
Q1. A metallic sphere of radius 4.2 cm is melted and recast into small spheres of radius 0.3 cm. Find the number of small spheres.
n = R³/r³ = (4.2)³/(0.3)³ = 74.088/0.027 = 2744 spheres
Q2. A cone of radius 5 cm and height 24 cm is melted and cast into a sphere. Find radius of sphere.
Volume of cone = (1/3)π(25)(24) = 200π cm³ Volume of sphere = (4/3)πR³ = 200π R³ = 150 → R = ∛150 ≈ 5.31 cm
Q3. Water flows through a cylindrical pipe of radius 1 cm at 2 m/s. How long to fill a tank 40 cm × 30 cm × 20 cm?
Volume of tank = 40 × 30 × 20 = 24000 cm³ Volume flowing per second = πr²v = π × 1 × 200 = 200π cm³/s
Time = 24000/(200π) = 120/π = 120 × 7/22 ≈ 38.2 seconds
Board Exam Tips
- 8–10 marks from this chapter — very high marks
- Slant height l = √(r²+h²) for cone — never confuse with height h
- For combination problems: draw the figure FIRST, identify which surfaces are exposed
- TSA of combined solid: exclude the circular faces where one solid sits on another
- Recasting: set volumes equal; state "volume is conserved"
- Frustum: the most formula-heavy part — write l formula first, then CSA, TSA, Volume in order
- π = 22/7 with radii that are multiples of 7; π = 3.14 otherwise
📚 RD Sharma Mathematics Class 10 — Surface Areas and Volumes
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