Explanation:
We have area of triangle,
A = 0.5 bh
Where b is base and h is altitude.
Differentiating with respect to time
[tex]A=0.5bh\\\\\frac{dA}{dt}=0.5\times \left ( b\times \frac{dh}{dt}+h\times \frac{db}{dt}\right )[/tex]
Here area is 80 square centimeters and altitude is 7.5 centimeters,
So we have
A = 0.5 bh
80 = 0.5 x b x 7.5
b = 21.33 cm
We also have
[tex]\frac{dh}{dt}=1.5cm/min\\\\\frac{dA}{dt}=4.5cm^2/min[/tex]
Substituting in differentiated equation
[tex]\frac{dA}{dt}=0.5\times \left ( b\times \frac{dh}{dt}+h\times \frac{db}{dt}\right )\\\\4.5=0.5\times \left ( 21.33\times 1.5+7.5\times \frac{db}{dt}\right )\\\\9= 32+7.5\times \frac{db}{dt}\\\\\frac{db}{dt}=-3.07cm/min[/tex]
Rate at which base is decreasing = 3.07 cm /min
