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
B
10cm2/s
C
103cm2/s
D
103cm2/s

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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