Question

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm2/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 190 cm2? cm/min

Answer 1

Answer:

The base is reducing at a rate of 3.4cm per min.

Step-by-step explanation:

Let The Height of the Triangle, h

Area of the triangle = A

Area of a Triangle, $A= \frac{1}{2}bh$

When $A=190 cm^2, h=10cm$

$190= \frac{1}{2}*b*10\\b=38cm$

$\frac{dA}{dt}=\frac{1}{2}h\frac{db}{dt}+ \frac{1}{2}b\frac{dh}{dt}\\\frac{dh}{dt}=1 cm/min, \frac{dA}{dt}=2cm^2/min,$

$2=\frac{1}{2}*10*\frac{db}{dt}+ \frac{1}{2}*38*1\\2=5\frac{db}{dt}+19\\2-19=5\frac{db}{dt}\\-17=5\frac{db}{dt}\\\frac{db}{dt}=-\frac{17}{5} =-3.4 cm/min$

The base is reducing at a rate of 3.4cm per min.