Given,

⇒ Diameter of the moon (d) = $\dfrac{1}{4}$ × diameter of the earth (D)

⇒ $\dfrac{\text{d}}{2} = \dfrac{1}{4} \times \dfrac{D}{2}$

⇒ The radius of the moon = $\dfrac{1}{4}$ × radius of the earth

Let radius of moon be r and radius of earth be R.

⇒ r = $\dfrac{1}{4}\times R$

⇒ r = $\dfrac{R}{4}$ .....(1)

Since, earth and moon are spherical.

Volume of the earth (V) = $\dfrac{4}{3}πR^3$

Volume of the moon (v) = $\dfrac{4}{3}πr^3$

Substituting value of r from equation (1) in above equation, we get :

$\Rightarrow v = \dfrac{4}{3}\times π \times \Big(\dfrac{R}{4}\Big)^3 (\text{From equation (1)}) \\[1em] \Rightarrow v = \dfrac{4}{3} \times π \times \dfrac{R}{4} \times \dfrac{R}{4} \times \dfrac{R}{4} \\[1em] \Rightarrow v = \dfrac{1}{64} \times \dfrac{4}{3}π R^3 \\[1em] \Rightarrow v = \dfrac{1}{64} \times V.$

Hence, the volume of the moon is $\dfrac{1}{64}$ times the volume of the earth.

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CBSE Class IX | Academic Year 2026-2027

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Surface Areas and Volumes — Interactive Study Guide

Formula Master Table

Solid	CSA	TSA	Volume
Cube (a)	4a²	6a²	a³
Cuboid (l,b,h)	2h(l+b)	2(lb+bh+hl)	lbh
Cylinder (r,h)	2πrh	2πr(r+h)	πr²h
Cone (r,h,l)	πrl	πr(l+r)	⅓πr²h
Sphere (r)	4πr²		&frac43;πr³
Hemisphere (r)	2πr²	3πr²	⅔πr³

Common Traps

Cone CSA uses slant height l, NOT height h! l = √(r²+h²)

Hemisphere TSA = CSA + base circle = 2πr² + πr² = 3πr²

Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check

1. Volume of cube with side 5 cm? (125 cm³)
2. CSA of cylinder with r=7, h=10? (2×22/7×7×10 = 440 cm²)
3. Volume of cone with r=3, h=4? (⅓×π×9×4 = 12π cm³)