There is no such shape.
A polygon with n (n >2) sides has n*(n-3)/2 diagonals.
Therefore n*(n-3)/2 = 2 implies n^2 - 3n = 4 or n^2 - 3n - 4 = 0
This can be factorised to (n+1)*(n-4) = 0 which implies that n = -1 or n = 4
Now, a polygon cannot have -1 sides and therefore it MUST have 4 sides. That is, it must be a quadrilateral.
quadrilateral
quadrilateral
A quadrilateral has 2 diagonals.
It is a rhombus.
A quadrilateral has four sides and can be represented as AB-CD. To calculate the number of diagonals in any polygon, you can use the formula ( \frac{n(n-3)}{2} ), where ( n ) is the number of sides. For a quadrilateral, ( n = 4 ), so the number of diagonals is ( \frac{4(4-3)}{2} = 2 ). Therefore, a quadrilateral has two diagonals, regardless of whether it has right angles or not.
its rhombus.rectangle,square
well it depends.... will or do the lines cross? if they do then yes they are
2 diagonals
The answer is no the diagonals do not have to bisect each other. That does not necessarily mean that there isn't a quadrilateral that has bisecting diagonals. A quadrilateral is a four-sided polygon which is a close plane figure; in this case with four sides. Meaning that all ends of the straight lines that configure a polygon must connect someway, somehow.
A quadrilateral has 2 diagonals. It does not matter whether it is convex or not.
4 corners so 2 diagonals.
Yes