Build a triangle with same surface as a given square, if two of its sides are known.


I know that it suffices to build a rectangle with area twice the area of the square. Once I build a rectangle, I choose one side to be one side of the triangle. Then I choose wherever on the parallel side an apex of the triangle. Because, the area will be: $$\text{base}\cdot \text{height}/2 =2l^2/2= l^2 = \text{area of the square}$$

  • $\begingroup$ Given a square, and a side, you know how to construct a rectangle of the same area as the square with one side being that given side, right? $\endgroup$ – Gina Jan 4 '14 at 11:55
  • $\begingroup$ Thanks for your response, Gina. I don't understand the above explanation. In the problem'context, two sides of the square are known, no? I didn't understand it well $\endgroup$ – Akhtubir Jan 4 '14 at 11:58
  • $\begingroup$ If you have the square, do you know how to construct a rectangle when one side of the rectangle is known? $\endgroup$ – Gina Jan 4 '14 at 12:03
  • $\begingroup$ The wording is sloppy: I think the intention here is two sides of the triangle are known, since two sides of a square would be pretty boring and a little nonsensical...Thus, it seems to be a question about the angle between the two given sides. Trigonometry? $\endgroup$ – DonAntonio Jan 4 '14 at 12:03
  • $\begingroup$ Yes, DonAntonio, two sides of the triangle are known. I will add another tag. $\endgroup$ – Akhtubir Jan 4 '14 at 12:07

Given any rectangle $ABCD$ and we want to construct a rectangle with the same area and one side being a fixed length $l$. We do as follow: put a point $E$ on the ray $AB$ such that $AE=l$. Draw $DE$. Then draw a line through $B$ parallel to $DE$ and let it intercept the line $AD$ at $F$. Then $AE$, $AF$ would be the length of the side of the rectangle. Prove this using similar triangle.

So how to use that to do the problem? Let $a,b$ be the required length of 2 side of the triangle. First, double the square by making a copy of itself, so you get a rectangle. Now using the method above, acquire a rectangle $ABCD$ of the same area as twice the square, where $AB=a$. Let the circle centre $A$ radius $b$ intercept the line $CD$ at $E$. Then $ABE$ is the required triangle. This is because the height of the triangle with apex $E$ is the same as $AD$ and we already know the rectangle is twice the original square.

In fact, the fact that you started with a square is completely inconsequential.

  • $\begingroup$ The circle centre A radius = B?? $\endgroup$ – Akhtubir Jan 4 '14 at 12:44
  • $\begingroup$ How did you start to note the ABCD rectangle? $A$ in the top left apex and $B$ in the top right apex and so on... Or? $\endgroup$ – Akhtubir Jan 4 '14 at 13:45
  • $\begingroup$ It does not matter whether one is top right or top left or anything. And the $B$ was a typo it's $b$ the length of the side of the triangle you need. $\endgroup$ – Gina Jan 4 '14 at 18:53

Let $S$ be the area of square and $a,b$ sides of triangle we need the angle $C$ of triangle such that $$S=\frac{1}{2}ab\sin C$$from above it is$$C=\arcsin\frac{2S}{ab}$$


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