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 ?

Explanation

Let `Delta ABC` be an isosceles triangle when BC is the base of definite length b.
Let each of equal sides of `Delta ABC` be x.

Draw ` AD bot BC`.
Now, in ` Delta ADC`, from Pythagoras theorem, `AD = sqrt(x^(2)-(b/2)^(2))=sqrt(x^(2) - b^(2)/4)`
`:." area of triangle (A)" = 1/2 xx" base" xx"height"`
` = b/2 sqrt(x^(2)-b^(2)/4)`
`rArr (dA)/(dt) = 1/2 b xx 1/2 (2x)/sqrt(x^(2)-b^(2)/4)xx (dx)/(dt)`
`=(xb)/sqrt(4x^(2)-b^(2))(dx)/(dt)`
Given, the two equal sides of triangle are decreasing at the rate of 3m/sec.
` rArr (dx)/(dt) =-3` cm/sec
` :. (dA)/(dt) = (-3x b)/sqrt(4x^(2)-b^(2)) cm^(2)//sec`
when x = b, then ` (dA)/(dt) = (-3b^(2))/sqrt(3b^(2))=-sqrt(3b cm^(2)//sec`
`:. ` When the equal sides of triangle are equal to its base then its area is decreasing at the rate of `sqrt(3) b cm^(2)//sec`.