Well, honey, let me drop some knowledge on you. In a rectangle, the diagonals are indeed longer than the sides, thanks to good ol' Pythagoras and his theorem. But in a square, the diagonals are the same length as the sides because all sides are equal. So, it really depends on the shape you're working with.
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Oh, dude, not always! Diagonals are longer than the sides in some shapes, like rectangles and squares, but not in all shapes. In a square, for example, the diagonal is longer than the side because Pythagoras said so, but in a rhombus, they're the same length, so it's like a "who cares" situation. Just depends on the shape, man.
On squares and rectangles, yes.
But on parallelograms and rhombus the one diagonal can be shorter than one of the sides.
Using the formula 0.5(n^2 -3n) whereas n is number of sides, altogether there are 104 diagonals in a 16 sided polygon
Yes, because the sides connected to the right angle cannot extend longer than the distance between their end points.
No. A diagonal goes from corner to opposite corner, which will always be a longer distance than the side length. You can use Pythagoras' theorem to work out the length of the diagonals. It will be the square root of (a2+b2) where a and b are the long and short side lengths of the rectangle respectively. The result will clearly be greater than either a or b.
No it can't. The hypotenuse of a right triangle will always be longer than either one of the other two sides.
12. There is actually a better answer and much larger answer than 12. The real answer is 54. Just because a dodecagon has 12 sides doesn't mean it has 12 diagonals because you can go either which way. To get a answer for this you can: 1. Multiply 12 and 9. You should get 108. 2. Then divid it by 2. You should get 54. That is correct: The general rule for a polygon with n sides is that it has n*(n-3)/2 diagonals.