|
| 30 Jul 2011 11:36 PM |
Well?
thelastbunny's new account |
|
|
| Report Abuse |
|
|
| |
|
|
| 30 Jul 2011 11:42 PM |
Is there an equilatera triangle that has both a height and side length divisible by 1, or at least 0.4?
thelastbunny's new account |
|
|
| Report Abuse |
|
|
thegenius
|
  |
| Joined: 03 Oct 2008 |
| Total Posts: 3838 |
|
| |
|
|
| 31 Jul 2011 02:06 PM |
Yes, evenly divisible by one.
For instance, 5, 4, 3, 9001 NOT 9001.123432345432
thelastbunny's new account |
|
|
| Report Abuse |
|
|
sdfgw
|
  |
 |
| Joined: 08 Jan 2009 |
| Total Posts: 41681 |
|
|
| 31 Jul 2011 03:04 PM |
Lemme think about this.
Let's call its side length x. We drop a perpendicular. This gives us two isosceles 30-60-90 triangles with short size x/2, long side is the height, and hypotenuse x.
A well known fact about the 30-60-90 triangle is that the sides are in the ratio 1 : root(3) : 2. Apply this to our triangle, we get x/2 : xroot(3)/2 : x.
Therefore: If the side length = x, the height = x/2 times the square root of 3.
The ratio of side length : height = 1 : root(3) / 2, or 2 : root(3). Since one number is an integer and the other is irrational, it is impossible to find any two integers to fit this ratio.
tl;dr: no. |
|
|
| Report Abuse |
|
|
|
| 31 Jul 2011 03:18 PM |
I thought so, thanks. (I had to find the area of a regular hexagon in seventh grade with only the side length once)
thelastbunny's new account |
|
|
| Report Abuse |
|
|
sdfgw
|
  |
 |
| Joined: 08 Jan 2009 |
| Total Posts: 41681 |
|
|
| 31 Jul 2011 03:21 PM |
| I think I actually have a formula I made somewhere that finds out the area of any regular shape, given just the side length and number of sides. |
|
|
| Report Abuse |
|
|