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Need solution for RD Sharma maths class 12 chapter 12 Derivative as a rate measure exercise Very short answer type question  3

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Answer: 10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{s}

Hint: Here, we all know the formula about equilateral triangle are increasing at the rate                        

                      Of 2cm/sec, x=10cm. Here, x stands for edge.

Solution:  Let the side of an equilateral triangle be x cm

 Area of equilateral triangle A=\frac{\sqrt{3}}{4} x^{2}    .................(1)


Differentiating equation (1) along with time, we get

        \begin{aligned} &\frac{d A}{d t}=\frac{\sqrt{3}}{4} \times 2 x \times \frac{d x}{d t} \\\\ &=\frac{\sqrt{3} x}{2} \frac{d x}{d t} \end{aligned}

Let's put the value of x where (x=10 cm)

        \frac{d A}{d t}=\frac{\sqrt{3} \times 10}{2} \times 2=10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}



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