Step 1:

The area of a triangle is given by

$A=\large\frac{1}{2}$$bh$-----(i) Where b is the base, h is the altitude.

All three variables depend on time t

Step 2:

Rate of change of area is

$\large\frac{dA}{dt}=\frac{d}{dt}\bigg(\frac{1}{2}$$bh\bigg)$

$\qquad=\large\frac{1}{2}$$\bigg(b\large\frac{dh}{dt}$$+h\large\frac{db}{dt}\bigg)$-----(ii)

Step 3:

$\large\frac{dA}{dt}$$=2 cm^2/min,$

$\large\frac{dh}{dt}$$=1 cm/min$

$h=10 cm, A=100 cm^2$

When $h=10 cm,$ and $ A=100 cm^2$ we have from (i)

$100=\large\frac{1}{2}$$b \times 10$

$b=20 \;cm$

Step 4:

Substituting the values in (ii), we have

$2=\large\frac{1}{2}$$(20 \times 1+10 \large\frac{db}{dt})$

$4=20+10 \large\frac{db}{dt}$

$\large\frac{db}{dt}$$=-1.6 cm/sec$

The base of the triangle is decreasing at $-1.6 \;cm/s$