If the area of an isosceles right angle triangle equals (A)cm2 anf if the rate of change of its area is 32cm/sec at the instant at which the length of each of the equal legs is 8 cm. find (at the instant)the time rate of change of the perimeter (p) of the triangle

Write down your formulae:

A = \frac{l^2}{2}

l is the length of one side the 'leg'.

P = 2l + h

h is the hypotenuse and is obtained through Pythagoras' Theorem;

h^2 = l^2 + l^2

P = 2l + \sqrt{2}l = l(2+\sqrt2)

You are given \frac{dA}{dt} = 32

At that time, l = 8

You must first the rate at which the length is changing and from there you find the rate at which the perimeter is changing.

You can find dl/dt from the chain rule:

\frac{dA}{dt} = \frac{dA}{dl} \times \frac{dl}{dt}

Then you can get the rate of change of perimeter:

\frac{dP}{dt} = \frac{dP}{dl} \times \frac{dl}{dt}