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

Solution

Correct Option: 2

An equilateral triangle has sides increasing at 2 cm/sec. The rate at which the area increases when the side is 10 cm needs to be found.

For an equilateral triangle with side aa:

A=34a2A = \frac{\sqrt{3}}{4}a^2


Differentiating both sides with respect to time tt:

dAdt=34×2a×dadt\frac{dA}{dt} = \frac{\sqrt{3}}{4} \times 2a \times \frac{da}{dt}

dAdt=32×a×dadt\frac{dA}{dt} = \frac{\sqrt{3}}{2} \times a \times \frac{da}{dt}

The term dadt\frac{da}{dt} appears due to the chain rule, since aa changes with time.


Given:

a=10a = 10 cm

dadt=2\frac{da}{dt} = 2 cm/sec

Substituting these values:

dAdt=32×10×2\frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 10 \times 2

dAdt=103\frac{dA}{dt} = 10\sqrt{3} cm²/sec

Therefore, the area increases at a rate of 10310\sqrt{3} cm²/sec.

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