Geometry: Find Min Length & Perimeter of Hexagon ABCDEF

  • Thread starter Thread starter therisingpower
  • Start date Start date
  • Tags Tags
    Geometry
Click For Summary

Homework Help Overview

The problem involves a hexagon ABCDEF with specific properties regarding its diagonals, angles, and side lengths. The original poster seeks to determine the minimum perimeter of the hexagon given that the length of diagonal AC is the square root of 85, as well as the smallest length of AC that satisfies all conditions while ensuring all sides are different integers.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss applying constraints on the side lengths and using Pythagoras' Theorem and triangle inequality to explore the relationships between the sides and diagonals. Questions arise about the triangle inequality and its implications for the problem.

Discussion Status

Some participants have provided hints and guidance on how to approach the problem, emphasizing the need to apply mathematical principles to derive relationships between the sides. The discussion reflects a collaborative effort to clarify concepts and explore potential methods without reaching a consensus or final solution.

Contextual Notes

The original poster's inquiry is framed within homework constraints, and there is an emphasis on the requirement that all side lengths must be different integers. This adds complexity to the problem and influences the approaches discussed.

therisingpower
Messages
5
Reaction score
0
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:
Physics news on Phys.org
Looks like homework. Show your thoughts/work first.
 
Last edited:
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.
 
hi, what do you mean by triangle inequality? could you please enlighten me? thanks.
 
therisingpower said:
hi, what do you mean by triangle inequality? could you please enlighten me? thanks.

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:
thanks! =) i finally understand..
 

Similar threads

Replies
18
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K