Calculate the depth of water in the trough when it is exactly half full

Question

I am in my last year of high school and am currently studying for my finals by going over exercises in my coursebook. I came across this exercise and have been stuck on it for some weeks now. I have asked my class mates as well as my math tutor who also wasn't able to figure it out. The question seems to defy logic and it doesn't really make much sense to me. I am beyond figuring it out for any study purposes - I am simply curious as to how on earth this question is supposed to be answered!!

According to my text book, the answer is supposed to be 22.93 m^3.

Answer 1

The equation is $$70*d+\color{blue}{\frac{1}{2}*d^2*\frac{30}{50}}=70*(50-d)+\frac{1}{2}*(50*30-d^2*\frac{30}{50})$$

The blue part: You can draw a orthogonal line from the end of the bottom upward. You get a triangle on the right side. This (big) triangle is divided by the horizontal line in two areas.

One of these areas is a small triangle. The height of the small triangle is d. Then you have a base (y). This base can be expressed in terms of d by using the intercept theory

$\frac{50}{d}=\frac{30}{y} \Rightarrow y=\frac{30}{50}d$

Thus the area of the small triangle is $\frac{1}{2}\cdot d \cdot \frac{30}{50}d$

The remaining area is the area of the (big) triangle minus the area of the small triangle: $\frac{1}{2}*(50*30-d^2*\frac{30}{50})$

Like I commented, your solution is wrong.

Answer 2

Hint: Compute the angle of the slope side by using trigonometry. Then write the constraint that the volumes are equal with an equation in $d$. Finally, solve for $d$.

Answer 3

Another hint: ideally cut off the ("upside-down") right-triangle at the right of the picture, which has sides $30$, $50$ and $\sqrt{30^2+50^2}$. Use then the similarity criteria to find the sides of the water-filled smaller triangle.

Answer 4

(Sorry if this is a bit late for your finals :-P) The question gives a hint that it can be solved by equating the areas of the two trapezia. The area (A) of a trapezium is given by the formula:

$$ A = h (a+b)/2 $$

where a and b are the lengths of the two parallel sides and h is the height of the trapezium. For the total area this gives:

$$ A = 50(70+100)/{2} = 4250 $$

Let's assign c as the length of dividing line parallel to a and b and to the surface of the water.

For the lower trapezium, the half area (A2) is given by:

$$ A2 = A/2 = h (70+c)/2 = 4250/2$$

and solve for c: $ c = 4250/h - 70$

where h is the height that gives the half area, that we are looking for. The area of the upper trapezium is given by:

$$ A2 = A/2 = (50-h) (c + 100)/2 = 4250/2$$

and solve for c: $ c = 4250/(50-h) - 100$

Now we can equate the two equations for c:

$$4250/h - 70 = 4250/(50-h) - 100$$

and solve for h. This turns out to be a quadratic for h and the positive root is $h \approx 27.189$