Six triangles
As shown, six triangles sharing a vertex have been tiled together to form a hexagon. The three orange, equilateral triangles have respective side lengths 5, 11, and 17, and the other three triangles also have all integer side lengths.
Find the perimeter of the hexagon.
What's the perimeter of this hexagon made of 6 triangles with integer side lengths?
The answer is 76.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
It adds up to 76, not 78.
Log in to reply
Thanks. I've updated the answer to 76.
In the future, if you have concerns about a problem's wording/clarity/etc., you can report the problem. See how here .
S i d e s 5 a n d 1 1 : − a c o s ( 2 ∗ 5 ∗ 1 1 5 2 + 1 1 2 − 9 2 ) = 5 3 . 7 7 8 5 o . S i d e s 1 1 a n d 1 7 : − a c o s ( 2 ∗ 1 1 ∗ 1 7 1 1 2 + 1 7 2 − 2 1 2 ) = 9 4 . 7 5 4 6 o S i d e s 1 7 a n d 5 : − a c o s ( 2 ∗ 1 7 ∗ 5 1 7 2 + 5 2 − 1 3 2 ) = 3 1 . 4 6 7 o J u s t d e t a i l c a l c u l a t i o n s .
Hexagon discovered by a friend of mine at my suggestion to find such hexagons. Confirmed by drawing with Geometer's Sketchpad.
I made a program on my graphing calculator. Here's some pseudocode
for BC,13,21: for DE,7,15: for FA,7,27
use law of cosines to find angles BOC, DOE, FOA
check if they sum to 180 degrees. If they do print BC, DE, FA
next BC: next DE: next FA
End
The program's sole output is [ 1 3 , 9 , 2 1 ]
So the perimeter is 1 7 + 1 3 + 5 + 9 + 1 1 + 2 1 = 7 6