Mensuration (3D Solids)

ICSE Class 10 Mathematics • Chapter 12

1. Master Formula Table

Solid	CSA	TSA	Volume
Cylinder	$2\pi rh$	$2\pi r(r+h)$	$\pi r^2h$
Cone	$\pi rl$	$\pi r(r+l)$	$\frac{1}{3}\pi r^2h$
Sphere	$4\pi r^2$		$\frac{4}{3}\pi r^3$
Hemisphere	$2\pi r^2$	$3\pi r^2$	$\frac{2}{3}\pi r^3$
Hollow Cylinder	$2\pi h(R+r)$	$2\pi(R+r)(h+R-r)$	$\pi h(R^2-r^2)$

Key Relationships:

Cone volume = $\frac{1}{3}$ × Cylinder volume (same base, height)
Hemisphere volume = $\frac{1}{2}$ × Sphere volume
Slant height of cone: $l = \sqrt{r^2 + h^2}$

2. Right Circular Cylinder

[Diagram: Cylinder with radius r, height h]

CSA = $2\pi rh$ (curved surface only)
TSA = $2\pi rh + 2\pi r^2 = 2\pi r(r+h)$ (includes top & bottom)
Volume = $\pi r^2h$

Example 1: A cylinder has radius 7 cm and height 10 cm. Find CSA, TSA, and Volume.

CSA = $2 \times \frac{22}{7} \times 7 \times 10 = 440$ cm²

TSA = $2 \times \frac{22}{7} \times 7 \times (7+10) = 44 \times 17 = 748$ cm²

Volume = $\frac{22}{7} \times 49 \times 10 = 1540$ cm³

3. Right Circular Cone

[Diagram: Cone with radius r, height h, slant height l]

Slant Height: $l = \sqrt{r^2 + h^2}$
CSA = $\pi rl$
TSA = $\pi rl + \pi r^2 = \pi r(r+l)$
Volume = $\frac{1}{3}\pi r^2h$

Example 2: A cone has radius 6 cm and height 8 cm. Find slant height, CSA, and volume.

$l = \sqrt{36 + 64} = \sqrt{100} = 10$ cm

CSA = $\pi \times 6 \times 10 = 60\pi \approx 188.57$ cm²

Volume = $\frac{1}{3} \times \pi \times 36 \times 8 = 96\pi \approx 301.71$ cm³

4. Sphere and Hemisphere

[Diagram: Sphere and Hemisphere with radius r]

	Sphere	Hemisphere
Surface Area	$4\pi r^2$	CSA = $2\pi r^2$, TSA = $3\pi r^2$
Volume	$\frac{4}{3}\pi r^3$	$\frac{2}{3}\pi r^3$

5. Combination of Solids

Strategy:

Volume of combined solid = Sum of individual volumes
Surface Area = Sum of exposed surfaces only (exclude hidden/common parts)

Example 3: A toy is shaped like a cone mounted on a hemisphere. Radius = 3.5 cm, total height = 15.5 cm. Find TSA.

Step 1: Height of cone = 15.5 − 3.5 = 12 cm

Step 2: $l = \sqrt{3.5^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5$ cm

Step 3: TSA = CSA of cone + CSA of hemisphere (base is hidden)

= $\pi rl + 2\pi r^2 = \pi r(l + 2r) = \frac{22}{7} \times 3.5 \times (12.5 + 7)$

= $11 \times 19.5 = 214.5$ cm²

6. Recasting / Melting Problems

Volume of original solid = Volume of new solid(s)

When a solid is melted and recast, only volume is conserved!

Example 4: A metallic sphere of radius 6 cm is melted and recast into a cone of radius 12 cm. Find height of cone.

Volume of sphere = Volume of cone

$\frac{4}{3}\pi (6)^3 = \frac{1}{3}\pi (12)^2 \times h$

$\frac{4}{3} \times 216 = \frac{1}{3} \times 144 \times h$

$288 = 48h$

$h = 6$ cm

Example 5: How many spherical balls of radius 1 cm can be made from a sphere of radius 5 cm?

Volume of big sphere = $\frac{4}{3}\pi (5)^3 = \frac{500\pi}{3}$

Volume of small sphere = $\frac{4}{3}\pi (1)^3 = \frac{4\pi}{3}$

Number of balls = $\frac{500\pi/3}{4\pi/3} = \frac{500}{4} = 125$ balls

7. Conversion Relations

From	To	Multiply by
cm³	litres	÷ 1000
m³	litres	× 1000
cm³	m³	÷ 10⁶

8. Quick Reference

Solid	Volume Formula	Key Relation
Cylinder	$\pi r^2 h$	Base × Height
Cone	$\frac{1}{3}\pi r^2 h$	$\frac{1}{3}$ cylinder
Sphere	$\frac{4}{3}\pi r^3$	4 cones (same r, h)
Hemisphere	$\frac{2}{3}\pi r^3$	Half sphere

Exam Practice (PYQ Trends)

PYQ: 2023

BOARD A solid is in the form of a cone standing on a hemisphere with both their radii being equal to 7 cm. If the height of cone is 5 cm, find volume of the solid.

PYQ: 2022

BOARD A solid metallic cylinder of diameter 12 cm and height 15 cm is melted and recast into toys in shape of cones of radius 3 cm and height 9 cm. Find number of toys.