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The height of a triangle is increasing at a rate of \( 1 \mathrm{~cm} / \mathrm{sec} \) while the area of the triangle is increasing at a rate of \( 2 \mathrm{~cm}^{2} / \mathrm{sec} \). At what rate is the base of the triangle decreasing when the altitude of the triangle is \( 10 \mathrm{~cm} \) and the area is \( 100 \mathrm{~cm}^{2} \). (Hint: The answer is negative. For a triangle, \( A=\frac{1}{2} b h \) )

We are given the area as A=12bh First we have to find the base when the altitude is 10 cm and the area is 100 cm2 100=12×b×1010=b2b=20cm We are asked