CUET · MATHS · PYQ PAPER 2023
The side of an equilateral triangle increases at the rate of \(3 cm / sec\), then the rate at which the area of the triangle increases when the side is 4 cm , is :
- A \(6 cm^2 / sec\)
- B \(6 \sqrt{3} cm^2 / sec\)
- C \(\sqrt{3} cm^2 / sec\)
- D \(6 / \sqrt{3} cm^2 / sec\)
Answer & Solution
Correct Answer
(B) \(6 \sqrt{3} cm^2 / sec\)
Step-by-step Solution
Detailed explanation
\(A = \frac{\sqrt{3}}{4} s^2\) \(\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2s \frac{ds}{dt}\) \(\frac{dA}{dt} = \frac{\sqrt{3}}{2} s \frac{ds}{dt}\) \(\frac{dA}{dt} = \frac{\sqrt{3}}{2} (4)(3)\) \(\frac{dA}{dt} = 6 \sqrt{3} \, cm^2/sec\)
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