A point Q lies inside regular hexagon ABCDEF such that [QAB]=10, [QCD]=17 and [QEF]=5, where the square brackets indicate the area of the enclosed triangle. Find the side of the hexagon.
The 3 areas given are for the non-adjacent triangles in
ABCDEF.
It's generally true for all Q that the areas of the two triplets of non-adjacent triangles are equal.
17+10+5= 32, so the area of ABCDEF is 64.
A = (1.5 × √3) × s^2
64 = (3*(3)^(1/2)/2)s^2, so s= (8 sqrt(2))/3^(3/4), or ≈4.96322591521120
Edited on September 25, 2019, 3:30 am
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Posted by broll
on 2019-09-25 03:29:44 |
Possible different solution
