How can the water level in a trough be calculated using related rates?

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Homework Statement


12.) A trough is 15ft long and 4ft across the top as shown in the figure. Its ends are isosceles triangles with height 3ft. Water runs into the trough at the rate of 2.5 ft3/min. How fast is the water level rising when it is 2 ft deep?

Homework Equations


The Attempt at a Solution



Not sure how to do this..
dv/dt = 2.5

I tried using the volume of a triangular prism: V=(1/2)bhl but no success.. I think the problem is IDK which numbers to use for the volume equation?
Can someone point me in the correct direction?
 

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If anyone is looking at this, what you have to do is find the area of the water, not the actual triangular prism.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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