Elven Old Traditional Roof Problem?

Let us think about the rectangles that we obtain when we look at a cross section of the EOTR at a height $h$ . We will find a general formula for its sides and integrate over the height of the EOTR in order to find the total volume. Wlog, let us call the side $a$ and $b$ , where $a \geq b$ and the total height of the EOTR will be called $H$ .

Lemma: The difference between the sides of the cross-sectional rectangles remains constant as $h$ goes from $0$ to $H$ .

Proof of the lemma:

In the case of the shortest side, $b$ , we can see that a half-pyramid is formed if we cut vertically through points $P$ and $Q$ , so we can say that the length of the short side changes as $h$ becomes bigger by a factor $(1-\frac{h}{H})$ . In the case of the longer side, $a$ , it is possible to extend the lines $AP$ and $BQ$ , for example, until they meet at a point at height $K$ , and we can do the same analysis to say that the length of the long side changes by a factor $(1-\frac{h}{K})$ . Now, at height $H$ , we now that the short side becomes zero and the long side becomes $a-b$ . Therefore:

$a(1-\frac{H}{K}) = a-b \Rightarrow a=b(\frac{K}{H})$

Now, for any height $h$ , we would have:

$a(1-\frac{h}{K})-b(1-\frac{h}{H})= a-b(\frac{K}{H})(\frac{h}{K})-b(1-\frac{h}{H}) = a-b$

as we wanted to show.

Now that we proved our lemma, we can write the are of the cross-sectional rectangles as:

$A(h) = short side * (a-b+short side) = b(1-\frac{h}{H})((a-b)+b(1-\frac{h}{H})$ $A(h) = ab-(b^2+ab)(\frac{h}{H})+b^2(\frac{h^2}{H^2}$

The volume would be:

$\int_{0}^{H} A(h) dx = abH-(b^2+ab)(\frac{H}{2})+b^2(\frac{H}{3})$

$a$ and $b$ are calculated using the values of $PQ$ and $\frac{AB}{BC}$ . Then, $H$ can be easily calculated using the value of $S$ and doing some Pythagoras.

Inputting these values in the previous expression we get that the volume is $228$ .

From the given data we get the following values as shown. The sketch shows naming of various point .
$\dfrac{AB}{BC}\!=\!3.5, \ so\ AB\!=\!3.5*BC.\ But\ PQ\!=\!AB\! -\! BC\!=\!3.5*BC \!- \!BC\!=2.5*BC\!=\!5*\sqrt3..given.\\ \therefore\ BC=2*\sqrt3,\ \ AB=3.5\!*\!BC=3.5*2*\sqrt3=7*\sqrt3.\ But\ PQ=5*\sqrt3,\\ \implies AE+FB=2*\sqrt3.\ \ \ \ \ So \ AE=FB=\sqrt3.\\ \text{All these dimensions are shown in the sketches. H also is calculated on the sketches. So :-}\\ Total\ volume \ = \boxed{1+3}+\boxed{2}=\ \frac 1 3*2*(\sqrt3*2\sqrt3)*12+\frac 1 2*(5*\sqrt3*2\sqrt3)*12.\\ Total\ volume=\ \ \ \ \color{#D61F06}{228}$

Solution:

Let's introduce new labels: $AB = b$ , $BC = a$ and $m = b - a$ .

So we need to find volume of pyramid whose base is a square with side $a$ + volume of prism of height $m$ and base as isosceles triangle whose unique side is $a$ . ( Enlarge image bellow )

Imagine that we pull out the middle piece and bring leftmost and rightmost pieces together. That is the earlier mentioned "hidden" pyramid. The middle piece is the prism

Firstly we solve the following equation: $b - a = 5\sqrt{3}; \space \frac{b}{a} = 3.5$ So the ( $a$ , $b$ ) = ( $2\sqrt{3}$ , $7\sqrt{3}$ ).

The orthogonal projection of point $P$ is $P'$ and $|PP'| = H$ . $P'$ is equally distanced from $AB$ and $BC$ , so the $|AP'| = \sqrt{6}$ . From here, $|PP'|$ is calculated using Pythagoras' theorem, and it is $12$ . Volume of the pyramid is then $\frac{1}{3}a^{2}H = 48$ . [1]

For the volume of the prism we need area of base, which is $\frac{1}{2}aH$ multiplied by height of the prism, which is in this case $b - a$ . Then the volume is $180$ . [2]

Using [1] and [2], we get that the volume of this 3D object is $\boxed{228}$