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Question

The sides of an equilateral triangle are increasing at the rate of $$2{ cm }/{ s }$$. The rate at which the area increases when the side is $$10 cm$$, is


A
3cm2/s
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B
10cm2/s
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C
103cm2/s
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D
103cm2/s
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Solution

The correct option is B $$10\sqrt { 3 } { { cm }^{ 2 } }/{ s }$$
Let $$x$$ be the side of an equilateral triangle and $$A$$ be the area.

$$\therefore A=\dfrac { \sqrt { 3 }  }{ 4 } { x }^{ 2 }$$

On differentiating both sides with respect to $$t$$, we get

$$\dfrac { dA }{ dt } =\dfrac { \sqrt { 3 }  }{ 4 } 2x\dfrac { dx }{ dt }$$

Given, $$ x=10 cm$$

and $$\dfrac { dx }{ dt } =2{ cm }/{ s }$$

$$\therefore \dfrac { dA }{ dt } =\dfrac { \sqrt { 3 }  }{ 4 } 2\times \left( 10 \right) \times 2$$
        
            $$=10\sqrt { 3 } { { cm }^{ 2 } }/{ s }$$

Mathematics

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