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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?
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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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