If the difference of two numbers is even then so is their sum.
1 answer:
Answer and explanation:
Statement - If the difference of two numbers is even then so is their sum.
Let the two even numbers be '2m' and '2n' with m and n are integers.
The difference of two number is
Now, The sum of the numbers is
Let where k is an integer
Then, which is also an even number as 2 is multiplied with it.
So, If the difference of two numbers is even then so is their sum.
For example -
Let two even number 2 and 4.
The difference is , 2 is even.
The sum is , 6 is even.
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Check the picture below.
so.. simply, use the distance formula, to get their length an add them up, and that's the perimeter of the polygon.
![\bf -------------------------------\\\\ d=\sqrt{[2-(-1)]^2+(4-2)^2}\implies d=\sqrt{(2+1)^2+(2)^2} \\\\\\ d=\sqrt{3^2+2^2}\implies \boxed{d=\sqrt{13}}\\\\ -------------------------------\\\\ d=\sqrt{(3-2)^2+(-2-4)^2}\implies d=\sqrt{1^2+(-6)^2}\implies \boxed{d=\sqrt{37}}\\\\ -------------------------------\\\\ d=\sqrt{(-2-3)^2+[-3-(-2)]^2}\implies d=\sqrt{(-5)^2+(-3+2)^2} \\\\\\ d=\sqrt{(-5)^2+(-1)^2}\implies \boxed{d=\sqrt{26}}](https://tex.z-dn.net/?f=%5Cbf%20-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B2-%28-1%29%5D%5E2%2B%284-2%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%282%2B1%29%5E2%2B%282%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B3%5E2%2B2%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B13%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%283-2%29%5E2%2B%28-2-4%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B1%5E2%2B%28-6%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B37%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-2-3%29%5E2%2B%5B-3-%28-2%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-5%29%5E2%2B%28-3%2B2%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-5%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B26%7D%7D)
![\\\\ -------------------------------\\\\ d=\sqrt{[-1-(-2)]^2+[2-(-3)]^2}\implies d=\sqrt{(-1+2)^2+(2+3)^2} \\\\\\ d=\sqrt{(1)^2+(5)^2}\implies \boxed{d=\sqrt{26}}](https://tex.z-dn.net/?f=%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B-1-%28-2%29%5D%5E2%2B%5B2-%28-3%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-1%2B2%29%5E2%2B%282%2B3%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%281%29%5E2%2B%285%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B26%7D%7D)
so, those are their lengths, sum them all up, that's the polygon's perimeter.
so.. simply, use the distance formula, to get their length an add them up, and that's the perimeter of the polygon.
so, those are their lengths, sum them all up, that's the polygon's perimeter.