A trough is filled with water at a rate of 1 cubic meter per second. The trough has a trapezoidal cross section with the lower base of length half a meter and one meter sides opening outwards at an angle of $45^\circ$ from the base. The length of the trough is 4 meters. What is the rate at which the water level h is rising when h is one half meter?
The book's solution says that that the volume of the trough is given by $V = 4(h^2 + h/2)$, I want to know how and why?
Thanks,