# Homework Help: Geometry Problem

1. May 11, 2006

### therisingpower

Hexagon ABCDEF has the following properties:
-diagonals AC, CE and EA are all the same length
-angles ABC and CDE are both 90 degrees
-all the sides of the hexagon have lengths which are different integers

1) what is the minimum perimeter of ABCDEF if AC = squareroot of 85?

2) what is the smallest length of AC for which ABCDEF has all these properties?

3) what is the minimum perimeter in this case (case 2) ?

Last edited: May 11, 2006
2. May 11, 2006

### Curious3141

Looks like homework. Show your thoughts/work first.

Last edited: May 11, 2006
3. May 12, 2006

### Curious3141

OK, here's a hint to start you out. Apply constraints on the sides to meet the conditions. You'll need Pythagoras' Theorem and the triangle inequality for starters.

4. May 29, 2006

### therisingpower

hi, what do you mean by triangle inequality? could you please enlighten me? thanks.

5. May 29, 2006

### Curious3141

Triangle inequality : The sum of the lengths of any two sides of a triangle is always greater than the third.

Here's an outline of how to approach the problem, you must flesh out the details yourself :

For part 1)

1) You're given that all the diagonals are the same length. The 3 diagonals divide the figure into three peripheral triangles and one central triangle. You only need to concern yourself with the three peripheral triangles.

2) You know that two of the angles (at least) opposite the diagonals are right angles. From Pythagoras' Theorem, come up with two equations relating the squares of the lengths of four of the sides.

3) Since you know that all the lengths are different integers, list out the possible values that the four sides can take.

4) That leaves one more peripheral triangle, and you're not restricted to a right triangle here. You do however, know the length of one side (the diagonal). Use the triangle inequality to come up with the minimum sum of lengths of the other two sides. Deduce possible values of the other two sides that sum up to this length, and see if they're admissible. If not, add one to that minimum sum and try again.

For part 2)

Use Pythagoras theorem and listing of squares of integers to see if you can come up with two distinct sets of squares that add up to the same integer. This will be either less than or equal to 85, naturally, given the previous info.

For part 3)

Work thru' the same logic as before and find possible minimal lengths for the last two sides given this new diagonal length.

Last edited: May 29, 2006
6. Jun 22, 2007

### therisingpower

thanks! =) i finally understand..