What is the numerical value of the product of the lengths of all diagonals of a regular hexagon of side length 1?
+592 We can start by looking at the angles, and we see that each interior angle is 120$^\circ$. this means that we can find so 30-60-90 triangles.
we can start by looking at the top half(notice we are cutting the hexagon from one of its diagonals), and if we draw two straight lines as heights, we can two 30-60-90 triangles on the side, and we get 1/2 as the base of the triangle. we can do the same for the other triagle, which also has a base of 1/2, and the length between these two triangles is 1, so 1/2+1/2+1=2, which is the length of one diagonal.
there are 6 diagonals in total, so 2 $\cdot$ 2 $\cdot$ 2 $\cdot$ 2$\cdot$ 2 $\cdot$ 2 = 2$^6$=$\boxed{64}$
+1641 Important: The number of diagonals in a polygon that can be drawn from any vertex in a polygon is three less than the number of sides. To find the total number of diagonals in a polygon, multiply the number of diagonals per vertex (n - 3) by the number of vertices, n, and divide by 2.
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The number of diagonals in a hexagon is => [(6 - 3) * 6]/2 = 9
If the side of a hexagon is 1 unit, then 3 diagonals have the lengths of 2 units, and 6 diagonals have the lengths of √3.
So, the product of all 9 diagonals is => 8 ( 81√3 )
