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Question
The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?
Solution
Let us take either of the equal side AB and AC be x
⇒ `dx/dt = -3` cm/sec.
If A is the corresponding area of ΔABC
`A = 1/2` base × height = `1/2 b sqrt (AB^2 - BD^2)`
`= b/2 sqrt (x^2 - (b/2)^2)`
`= b/4 sqrt (4x^2 - b^2)`
`(dA)/dt = b/4 xx 1/2 xx (8x)/ sqrt (4x^2 - b^2) dx/dt`
`= (bx)/ sqrt (4x^2 - b^2) dx/dt`
`((dA)/dt)_(x = b) = ((b xxb)/ sqrt (4b^2 - b^2)) (-3)`
`= - sqrt(3b)`
Area is decreasing at the rate of `b sqrt3` cm2 sec
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