Show me a picture of the grid. (The square with the lines)
isosceles trapezoid
....................................
Answer:
![BE = 4\sqrt2](https://tex.z-dn.net/?f=BE%20%3D%204%5Csqrt2)
Step-by-step explanation:
Given
The attached square
Required
Find BE
First, we calculate BD (the diagonal) using Pythagoras theorem.
![BD^2 = BA^2 + AD^2](https://tex.z-dn.net/?f=BD%5E2%20%3D%20BA%5E2%20%2B%20AD%5E2)
Since the shape is a square, then
![BA = AD = 8](https://tex.z-dn.net/?f=BA%20%3D%20AD%20%3D%208)
So:
![BD^2 = 8^2 + 8^2](https://tex.z-dn.net/?f=BD%5E2%20%3D%208%5E2%20%2B%208%5E2)
![BD^2 = 2(8^2)](https://tex.z-dn.net/?f=BD%5E2%20%3D%202%288%5E2%29)
Take square roots
![BD = \sqrt{2(8^2)](https://tex.z-dn.net/?f=BD%20%3D%20%5Csqrt%7B2%288%5E2%29)
![BD = \sqrt{2} * 8](https://tex.z-dn.net/?f=BD%20%3D%20%5Csqrt%7B2%7D%20%2A%208)
![BD = 8\sqrt{2}](https://tex.z-dn.net/?f=BD%20%3D%208%5Csqrt%7B2%7D)
BE is half BD.
So:
![BE = \frac{1}{2} * BD](https://tex.z-dn.net/?f=BE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20BD)
![BE = \frac{1}{2} * 8\sqrt2](https://tex.z-dn.net/?f=BE%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%208%5Csqrt2)
![BE = 4\sqrt2](https://tex.z-dn.net/?f=BE%20%3D%204%5Csqrt2)
First lets simplify the first term and yield (2a)
The second term:
1. 2a * (-1/2) = -a
2. -a + a = 0
Now we have simplified both terms to: (2a) (0)
Anything multiplied by 0 is 0, so the answer is also 0.
A Division sentence that would represent the following repeated subtraction is 24 divided by 8 = 3