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Question
The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. How far is the area increasing when the side is 10 cms?
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Solution
\[\text { Let x be the side andAbe the area of the equilateral triangle at any timet.Then, }\]
\[A = \frac{\sqrt{3}}{4} x^2 \]
\[ \Rightarrow \frac{dA}{dt} = 2 \times \frac{\sqrt{3}}{4} x^{} \frac{dx}{dt}\]
\[\Rightarrow\frac{dA}{dt}=\frac{\sqrt{3}}{2}\times2\times10\]
\[\Rightarrow\frac{dA}{dt}=10\sqrt{3} \text {cm}^2 /sec\]
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