Question

# 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

## Let x be the side and A be the area of an equilateral triangle at any time t. It is given that,  $\frac{dx}{dt}$ = 2 cm/sec Area of the equilateral triangle, A = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^{2}$ = $\frac{\sqrt{3}}{4}{x}^{2}$                   $A=\frac{\sqrt{3}}{4}{x}^{2}$ Differentiating both sides with respect to t, we get $\frac{dA}{dt}=\frac{\sqrt{3}}{4}×\frac{d}{dt}{x}^{2}$ $⇒\frac{dA}{dt}=\frac{\sqrt{3}}{4}×2x\frac{dx}{dt}$ $⇒\frac{dA}{dt}=\frac{\sqrt{3}}{2}x\frac{dx}{dt}$ When x = 10 cm and $\frac{dx}{dt}$ = 2 cm/sec, we get $\frac{dA}{dt}=\frac{\sqrt{3}}{2}×10×2$ $⇒\frac{dA}{dt}=10\sqrt{3}$ cm2/sec Thus, the area of the equilateral triangle increases at the rate of $10\sqrt{3}$ cm2/sec. 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 .MathematicsRD Sharma XII Vol 1 (2019)All

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