An equiangular hexagon has side lengths 3,4,5,3,4 and 5 units in that order. What is the area?
An equiangular hexagon has side lengths 3,4,5,3,4 and 5 units in that order. What is the area? Express your answer as a common fraction in simplest radical form.
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$\begingroup$Hint...draw a rectangle which encloses the hexagon and which has a pair of sides coincident with the sides of length $3$, and whose other sides touch the vertices of the hexagon between the sides $4$ and $5$.
You can easily work out the area of the rectangle and subtract the four right angled triangles...
$\endgroup$ 1 $\begingroup$If you draw an equilateral triangle containing the hexagon, you find that the hexagon is just the equlateral triangle of lengths 3+4+5 minus 3 equalateral traingles of length 3,4, and 5. So, using the equilateral triangle formula of area= side squared times root 3/4 we find it is just 144 root 3/4 minus ((9+16+25) root 3/4) which simplifies to 47 root 3/2
Thus, 47 sqrt. 3 all over 2 is the answer
$\endgroup$ $\begingroup$Divide the hexagon up like this:
The total angles in a hexagon are $720^0 ((6-2)\cdot 180)$ so each one is $120^0$ By the cosine rule($a^2=b^2+c^2-2bc\cos(A)$), the side marked $k$ is $\sqrt{37}$. Hence the area of the middle rectangle is $5\sqrt{37}$. The area of the two triangles is found by $\frac{1}{2}ab\sin(C)$, so here it is $3\sqrt3$. There are two of them, so combined with the rectangle, the total area is $$6\sqrt{3}+5\sqrt{37}$$
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