Question

# The side of an equilateral triangle expands at the rate of $$\sqrt{3}cm/sec$$.When the side is $$12cm$$, the rate of increase of its area is ______

A
12cm2/sec
B
18cm2/sec
C
33cm2/sec
D
10cm2/sec

Solution

## The correct option is D $$18{cm}^{2}/sec$$Area = $$A=\dfrac{\sqrt{3}}{4}a^{2}$$$$\dfrac{dA}{dt}=\dfrac{\sqrt{3}}{4}2a\dfrac{da}{dt}=\dfrac{\sqrt{3}}{2}a\dfrac{da}{dt}$$$$\dfrac{dA}{dt}=\dfrac{\sqrt{3}}{2}\times 12\times \sqrt{3}cm^{2}/s$$$$= 18 cm^{2}/s$$Mathematics

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