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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
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B
18cm2/sec
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C
33cm2/sec
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D
10cm2/sec
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Solution

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

1080834_1172813_ans_a668cd6f450344bf9c6f2b85159ad359.jpg

Mathematics

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