Match List-I with List-II.
| List-I | List-II | ||
|---|---|---|---|
| (A) | Perimeter of a right-angled isosceles triangle with side ‘a’ | (I) | \(\pi\left(r_2^2-r_1^2\right)\) |
| (B) | Area of a parallelogram | (II) | base × height |
| (C) | Area of a ring with \(r_1\) and \(r_2\) as the radii of the inner and outer circles | (III) | \(2a+\sqrt{2}a\) |
| (D) | Area of an equilateral triangle with side ‘a’ | (IV) | \(\frac{\sqrt{3}}{4}a^2\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-II, C-I, D-IV
Each quantity is matched with its standard formula.
(A) Perimeter of a right-angled isosceles triangle with side ‘a’ — this matches (III) \(2a+\sqrt{2}a\). The two equal sides are \(a\) each, and by Pythagoras’ theorem the hypotenuse is \(\sqrt{a^2+a^2}=\sqrt{2}a\), so the perimeter is \(a+a+\sqrt{2}a\).
(B) Area of a parallelogram — this matches (II) base × height. The height is the perpendicular distance between the base and the opposite side, not the slanting side.
(C) Area of a ring — this matches (I) \(\pi\left(r_2^2-r_1^2\right)\). The ring is the outer circle with the inner circle removed: \(\pi r_2^2-\pi r_1^2\).
(D) Area of an equilateral triangle with side ‘a’ — this matches (IV) \(\frac{\sqrt{3}}{4}a^2\). Its height is \(\frac{\sqrt{3}}{2}a\), and half of base × height gives \(\frac{1}{2}\times a\times\frac{\sqrt{3}}{2}a\).