Question

A triangle has a height that is increasing at a rate of 2 cm/sec and its area is increasing at a rate of 4 cm2/sec. Find the rate at which the base of the triangle is changing when the height of the triangle is 4 cm and the area is 20 cm2 .

Answer 1

Answer:

The value of rate of which the base is changing $\frac{dB}{dt}$ = - 3 $\frac{cm}{s}$

Step-by-step explanation:

Area = 20 $cm^{2}$

height = 4 cm

$\frac{dH}{dt} = 2 \frac{cm}{sec}$

$\frac{dA}{dt} =$  $2 \ \frac{cm^{2} }{sec}$

we know that area of the triangle is given by

$A= \frac{1}{2} B H$

$B = \frac{2A}{H}$

$B = \frac{2 (20)}{4}$

B = 10 cm

Rate of change of area is given by

$\frac{dA}{dT} = \frac{1}{2} [B\frac{dH}{dt} + H \frac{dB}{dt} ]$

4 =  0.5 [10 × 2 + 4 $\frac{dB}{dt}$ ]

$\frac{dB}{dt}$ = - 3 $\frac{cm}{s}$

This is the value of rate of which the base is changing.