Geometry & Mensuration • Topic 5 of 5

3D Mensuration

Solid formulas: cuboid volume lbh (surface 2(lb+bh+hl)); cube a^3 (surface 6a^2); cylinder pi r^2 h (curved surface 2 pi r h); cone (1/3) pi r^2 h (slant l = root(r^2+h^2)); sphere (4/3) pi r^3 (surface 4 pi r^2); hemisphere (2/3) pi r^3. When a solid is melted and recast, volume is conserved — equate the two volumes.

✅ Solved examples

1. Volume of a cuboid 5 x 4 x 3?
60 cubic units.
2. Volume of a cube of side 6?
6^3 = 216.
3. Volume of a cylinder r = 7, h = 10 (pi 22/7)?
22/7 x 49 x 10 = 1540.
4. A sphere r = 3 is melted into a cylinder of radius 3. Find the cylinder height.
Sphere vol = (4/3) pi 27 = 36 pi; cylinder pi 9 h = 36 pi -> h = 4.

✏️ Practice — try these, take hints as needed

1. Volume of cube side 10?
a^3.
—
—
1000
2. Surface area of cube side 5?
6a^2.
6 x 25.
—
150
3. Cylinder r = 7, h = 5 volume (pi 22/7)?
pi r^2 h.
22/7 x 49 x 5.
—
770
4. Cone r = 3, h = 4 slant height?
root(9+16).
root 25.
—
5
5. Volume of sphere r = 3 (in pi)?
(4/3) pi r^3.
(4/3) pi 27.
—
36 pi

📝 Topic test — 8 questions

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