Assertion (A) : If S be the area of a circle having radius x and A the area of an equilateral trianlge
having side $π x$ at any instant, then $\frac{d A}{d t} > \frac{d s}{d t}$
Reason(R) : A > S

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Both A and R are true and R is the correct explanation of A

Both A and R are true and R is not the correct explanation of A

A is true but R is false

A is false but R is true

answer is A.

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Detailed Solution

$∵ S = π x^{2} \Rightarrow \frac{d S}{d t} = 2 π x \frac{d x}{d t} and$ $A = \frac{\sqrt{3}}{4} π^{2} x^{2} \Rightarrow \frac{d A}{d t} = 2 π^{2} \frac{\sqrt{3}}{4} x \frac{d x}{d t}$

$then \frac{d S}{d A} = \frac{\frac{d S}{d t}}{\frac{d A}{d t}} = \frac{2 π x \frac{d x}{d t}}{π^{2} \frac{\sqrt{3} x}{2} \frac{d x}{d t}} = \frac{4}{π \sqrt{3}} < 1$ $∴ \frac{d S}{d t} < \frac{d A}{d t} or \frac{d A}{d t} > \frac{d S}{d t}$