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Question
If equal sides of an isosceles triangle with fixed base 10 cm are increasing at the rate of 4 cm/sec, how fast is the area of triangle increasing at an instant when all sides become equal?
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Solution
Given, side is increasing at the rate of 4 cm/sec.
Let equal sides are of length x cm.
∴ `dx/dt` = 4 cm/sec
∵ Area of isosceles triangle is `b/2 sqrt(a^2 - b^2/4)`
∵ b = 10 cm and a = x
`\implies` A = `10/2 sqrt(x^2 - (10)^2/4`
`\implies` A = `5sqrt(x^2 - 25)`
`\implies (dA)/dt = 5/2 (2x)/sqrt(x^2 - 25) dx/dt`
`(dA)/dt = (5x)/sqrt(x^2 - 25) dx/dt`
∵ x = 10 ...(when all sides become equal)
`\implies (dA)/dt = (5 xx 10)/sqrt(100 - 25) xx 4`
= `(50 xx 4)/sqrt(75) = (50 xx 4)/(5sqrt(3))`
`(dA)/dt = 40/sqrt(3)`
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