1
$\begingroup$

I have the measurements of the four sides of an irregular polygon and I need to find out the size of each interior angle.

I know the sum of the angles is 360 degrees but because it's not a regular polygon I don't know how to calculate each angle.

The measurements of each side are 882.9, 80, 576.8 and 293.3.

However, it is confirmed that one of the angles is 90 degrees. How do I find the other 3 angles? Please refer to the image below for reference.

polygon

$\endgroup$
0

1 Answer 1

Reset to default
1
$\begingroup$

There is no such quadrilateral. For draw the hypotenuse of the right triangle with legs $882.9$ and $80$. This hypotenuse has length $l$ greater than $882.9$.

But $293.3+576.8\lt 882.9\lt l$, contradicting the triangle inequality: the sum of any two sides of a triangle is greater than the third side.

$\endgroup$
9
  • $\begingroup$ Hi, I am quite unclear about your statement as it is a 4 sided polygon. Can you please explain in layman's term. $\endgroup$
    – Kai
    Feb 26, 2016 at 7:54
  • $\begingroup$ Let us label the vertices. Bottom right, at the right angle, $A$. Top right, $B$. Extreme left $C$. And the last one, close to $A$ but $80$ to the left of it, $D$. Join the points $B$ and $D$ by a straight line. This line $BD$ is the hypotenuse of a right-angled triangle, so it has length that I called $l$ in the post, where $l\gt 882.9$. We could compute $l$ using the Pythagorean Theorem, but don't need to. (Continued) $\endgroup$
    – André Nicolas
    Feb 26, 2016 at 8:02
  • $\begingroup$ (Continued) Now look at "triangle" $BCD$. I put triangle in quotes because it is an impossible triangle. For the sum of two of its supposed sides, $BC$ and $CD$, is less than the third side $BD$. Thus there is no such thing as a quadrilateral with sides as given and right angle at $A$. $\endgroup$
    – André Nicolas
    Feb 26, 2016 at 8:06
  • $\begingroup$ What if the questioner misrecorded the quadrilateral? Maybe the 80 side is supposed to be opposite the 882.9 side, if so then we can get the required triangle inequalities to work. If this is so, we need to have the sides properly ordered and also must locate which vertex angle is the right angle. $\endgroup$
    – Oscar Lanzi
    Feb 27, 2016 at 12:08
  • $\begingroup$ It may be a homework "trick question." If OP modifies the question so that it the numbers describe a real situation, one can answer it by a computation, Pythagorean Theorem and then Cosine Law. $\endgroup$
    – André Nicolas
    Feb 27, 2016 at 14:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .