Given the figure of a regular pyramid
The base of the pyramid is a hexagon with a side length = 6
The lateral area is 6 times the area of one of the side triangles
So, the side triangle has a base = 6
The height will be:
![\begin{gathered} h^2=6^2+(\frac{\sqrt[]{3}}{2}\cdot6)^2=36+27=63 \\ h=\sqrt[]{63} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20h%5E2%3D6%5E2%2B%28%5Cfrac%7B%5Csqrt%5B%5D%7B3%7D%7D%7B2%7D%5Ccdot6%29%5E2%3D36%2B27%3D63%20%5C%5C%20h%3D%5Csqrt%5B%5D%7B63%7D%20%5Cend%7Bgathered%7D)
so, the lateral area =
![6\cdot\frac{1}{2}\cdot6\cdot\sqrt[]{63}=18\sqrt[]{63}](https://tex.z-dn.net/?f=6%5Ccdot%5Cfrac%7B1%7D%7B2%7D%5Ccdot6%5Ccdot%5Csqrt%5B%5D%7B63%7D%3D18%5Csqrt%5B%5D%7B63%7D)
Okay, so first you draw a picture and let x be the distance from point D to the rest stop. Then the distance from point to the rest stop is 8 - x
You know that the length of the new trail is y + z, where y is the distance from Ancaster to the rest stop and z is the distance from Dundas to the rest stop.
Now by the Pythagorean theorem, y^2 = 4^2 + x^2 and z^2 = 6^2 + (8 - x) ^2
So take square roots of these, add them, and minimize.
Note: I am assuming the path is perfectly straight, otherwise this approach fails.
To expand you must first distribute (multiply) the 2 to the numbers inside the parentheses
2(x - 5)
(2 * x) + (2 * (-5) )
2x + (-10)
2x - 10
There are no like terms, therefore (x - 5)2 expanded is 2x - 10
Hope this helped!
~ Just a girl in love with Shawn Mendes
