-3
$\begingroup$

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

Progress

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: $$\text{base}\cdot \text{height}/2 =2l^2/2= l^2 = \text{area of the square}$$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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$$C=\arcsin\frac{2S}{ab}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.