The side of an equilateral triangle is increasing at the rate of 2 cm/s. Match List-I with List-II, where List-II gives each value at the instant when the side is 6 cm (in cm/s, cm2/s or cm).
| List-I | List-II | ||
| (A) | Rate of increase of the perimeter with time | (I) | \(\sqrt{3}\) |
| (B) | Rate of increase of the area with time | (II) | 6 |
| (C) | Rate of increase of the altitude with time | (III) | \(3\sqrt{3}\) |
| (D) | Rate of change of the area with respect to the side | (IV) | \(6\sqrt{3}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Let a be the side at time t, with \(\frac{da}{dt} = 2\) and a = 6. The perimeter, area and altitude are
\( \displaystyle P = 3a,\ A = \frac{\sqrt{3}}{4}a^{2},\ h = \frac{\sqrt{3}}{2}a \)
(A) Rate of increase of the perimeter with time
This matches (II) 6.
\( \displaystyle \frac{dP}{dt} = 3\frac{da}{dt} = 3 \times 2 = 6 \)
(B) Rate of increase of the area with time
This matches (IV) \(6\sqrt{3}\).
\( \displaystyle \frac{dA}{dt} = \frac{\sqrt{3}}{2}a\frac{da}{dt} = \frac{\sqrt{3}}{2} \times 6 \times 2 = 6\sqrt{3} \)
(C) Rate of increase of the altitude with time
This matches (I) \(\sqrt{3}\).
\( \displaystyle \frac{dh}{dt} = \frac{\sqrt{3}}{2}\frac{da}{dt} = \frac{\sqrt{3}}{2} \times 2 = \sqrt{3} \)
(D) Rate of change of the area with respect to the side
This matches (III) \(3\sqrt{3}\).
\( \displaystyle \frac{dA}{da} = \frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{2} \times 6 = 3\sqrt{3} \)