CUET · MATHS · PYQ PAPER 2025
If the area of an equilateral triangle is increasing at the rate of \(4 \sqrt{3} cm^2 / sec\), then the rate of increase of its perimeter when the side is 4 cm , is
- A 2 cm/sec
- B 3 cm/sec
- C 4 cm/sec
- D 6 cm/sec
Answer & Solution
Correct Answer
(D) 6 cm/sec
Step-by-step Solution
Detailed explanation
\(A = \frac{\sqrt{3}}{4} s^2\) \(\frac{dA}{dt} = \frac{\sqrt{3}}{4} (2s) \frac{ds}{dt} = \frac{\sqrt{3}}{2} s \frac{ds}{dt}\) \(4\sqrt{3} = \frac{\sqrt{3}}{2} (4) \frac{ds}{dt}\) \(\frac{ds}{dt} = \frac{4\sqrt{3}}{2\sqrt{3}} = 2 \, cm/sec\) \(P = 3s\)…
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