Geometry

Solids

Easy

Match List-I with List-II

List-I List-II
(Figures, etc.) (Area/Volume/Diagonal etc.)
(A) Surface area of Cube (I) $\pi(R^2 - r^2)$
(B) Volume of Right pyramid (II) $a\sqrt{2}$
(C) Area of Circular Ring (III) $6a^2$
(D) Diagonal of Square (IV) $\frac{1}{3}$ (area of base) × height

Choose the correct answer from the options given below:

List-I	List-II
(Figures, etc.)	(Area/Volume/Diagonal etc.)
(A) Surface area of Cube	(I) $\pi(R^2 - r^2)$
(B) Volume of Right pyramid	(II) $a\sqrt{2}$
(C) Area of Circular Ring	(III) $6a^2$
(D) Diagonal of Square	(IV) $\frac{1}{3}$ (area of base) × height

(A) - (I), (B) - (II), (C) - (III), (D) - (IV)

(A) - (II), (B) - (III), (C) - (IV), (D) - (I)

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)

(A) - (IV), (B) - (I), (C) - (II), (D) - (III)

✅ Correct Option: 3

A cube has 6 faces (top, bottom, and 4 sides).

Each face is a square with area $a^2$ .

Total surface area $= 6 \times a^2$

$= 6a^2$

(A) matches with (III)

The volume formula for any pyramid is:

Volume $= \dfrac{1}{3} \times (\text{area of base}) \times \text{height}$

(B) matches with (IV)

A circular ring is the region between two circles.

The larger circle has radius $R$ and the smaller circle has radius $r$ .

Area $= \pi R^2 - \pi r^2$

$= \pi(R^2 - r^2)$

For a square with side $a$ , using the Pythagorean theorem:

$\text{diagonal}^2 = a^2 + a^2$

$\text{diagonal}^2 = 2a^2$

$\text{diagonal} = \sqrt{2a^2}$

$= a\sqrt{2}$

(D) matches with (II)

The correct matching is:

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)

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Solution

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