We have the area of the triangle [tex] A(t)= \frac{bh}{2} [/tex].
[tex] b=\frac{2A}{h} [/tex]
Differentiating with respect to time t,
[tex] \frac{db}{dt}=\frac{d}{dt}(\frac{2A}{h} ) [/tex]
[tex] \frac{db}{dt}=(\frac{2}{h} )\frac{dA}{dt}-(\frac{2A}{h^2} )\frac{dh}{dt} [/tex]
Substituting numerical values, the rate of change of base length is
[tex] \frac{db}{dt}=(\frac{2}{10} )2-(\frac{200}{10^2} )1\\
\frac{db}{dt}=-1.6 cm/min [/tex]
