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?

Answer 1

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)`