ABCDEF is an equiangular hexagon of perimeter 21. Given that AB = 3, CD = 4, and EF = 5, compute the area of hexagon ABCDEF.
In order for EF to meet DE, the sides starting at AB need to be 3, 4, 4, 2, 5, 3.
Putting AB and DE at the top and bottom respectively,
The two sides of length 4 along with BD form an isosceles triangle of base 4sqrt(3) and height 2 for an area of 4sqrt(3).
Adjacent to the base of that triangle is rectangle A'BDE, where A' is one unit to the right of A, with dimensions 2 x 4sqrt(3), for an area of 8sqrt(3).
There's a smaller rectangle, with AA' as the top, and going down 3sqrt(3)/2. As its width is 1, its area is also 3sqrt(3)/2.
To the left of that rectangle is the right triangle with hypotenuse 3. It's base is 1.5 and height 3sqrt(3)/2 for an area of 4.5sqrt(3)/4.
Finally the right triangle with hypotenuse 5. Its base is 2.5 and height 2.5sqrt(3) for area (5/2)^2 * sqrt(3) / 2.
That totals to (4 + 8 + 3/2 + 4.5/4 + (5/2)^2 / 2) * sqrt(3)
= 17.75 * sqrt(3)
~= 30.7439018343476
|
Posted by Charlie
on 2021-05-17 12:18:07 |