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Question
The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is ______.
Options
`10 "cm"^(2/"s")`
`sqrt(3) "cm"^(2/"s")`
`10sqrt(3) "cm"^(2/"s")`
`10/3 "cm"^(2/"s")`
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Solution
The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is `10sqrt(3) "cm"^2/"s"`.
Explanation:
Let the length of each side of the given equilateral triangle be x cm.
∴ `"dx"/"dt" = 2 "cm"/sec`
Area of equilateral triangle A = `sqrt(3)/4 x^2`
∴ `"dA"/"dt" = sqrt(3)/4 * 2x * "dx"/"dt"`
= `sqrt(3)/2 xx 10 xx 2`
= `10sqrt(3) "cm"^2/sec`
Hence, the rate of increasing of area = `10sqrt(3) "cm"^2/sec`.
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