sqrt (2^2 + 5^2) = sqrt 29 agreed
the distance from -5,5 to 5,5 is 10
the distance from 0,-7 to 7,0
is 7
Oh my :( not really regular
I guess you will have to make a model with 8 equal sides and see if you can sqwish it :)
I think you can.
An octagon is formed by joining the points (7,0), (5,5), (0,7), (-5,5), (-7,0), (-5,-5), (0,-7), (5,-5) and (7,0). The octagon is regular.
I have used proof by exhaustion and got that all sides are square root of 29. Then sketched it, and all sides were also same length. But the book says the conjecture's false...
I believe a regular octagon has all sides same side and all angles are the same, which should fallow from all length being the same.
Help!
3 answers
IF A (0,-7) and B (7,0)
AB = Sqrt of (0-7)^2 + (-7-0) and that is sqrt of 7^2 + 7^2 not 7?
Okay but a regular octagon is meant to have all sides of the same length. (7,0) to (5,-5) is a side and (5,-5) to (0,-7) is a side. (7,0) to (-7,0) is not a side, so why would it matter? I don't get it...
AB = Sqrt of (0-7)^2 + (-7-0) and that is sqrt of 7^2 + 7^2 not 7?
Okay but a regular octagon is meant to have all sides of the same length. (7,0) to (5,-5) is a side and (5,-5) to (0,-7) is a side. (7,0) to (-7,0) is not a side, so why would it matter? I don't get it...
If you can squish it, it is no longer regular. The sides may be the same but not the angles.
Think about a square with hinges at the corners
if you squish it into a parallelogram, it still has 4 equal sides but is no longer a regular polygon.
Think about a square with hinges at the corners
if you squish it into a parallelogram, it still has 4 equal sides but is no longer a regular polygon.
1 answer