Since the question is for dh/dt when h=4ft at the 9-ft deep end, then the volume of a trapezoidal prism won't work. Because at h=4ft, you only have a triangular prism for the volume of water in the tank.

So,

Draw the figure, a longitudinal section of the pool, such that the walls are vertical. It is a trapezoid, (trapezium).

Draw horizontal line passing through the 4-ft deep.

Draw another horizontal line lower than tha first line.

Two similar right triangles are formed.

One has:

----vertical leg = (9-4) = 5ft

----horizontal leg = 40ft

----hypotenuse---unknown.

The smaller one has:

---vertical leg = h

---horizontal leg = x

---hypotenuse----unknown too.

By proportion,

5/40 = h/x

Cross multiply,

5x = 40h

x = 40h/5 = 8h

So, volume of water in the pool when depth at deep end is h ft,

V = (1/2)(h)(x) *(20)

V = 10h(8h)

V = 80h^2

dV/dt = 80(2h *dh/dt)

dV/dt = 160h (dh/dt)

When h=4, and dV/dt is always 10 cuft/min,

10 = 160(4)(dh/dt)

1 = 64(dh/dt)

dh/dt = 1/64

Therefore, at that instant, the water is rising at the rate of 1/64 ft/min. ----answer.