Open Question: How to figure out this hard area and perimeter problem?
A regular has 6 sides of length t. so P = t + t + t + t + t + t. or, P = 6t.
We also know that they have equal perimeters... so 3s = 6t => s = 2t
Now, A = 10 = 1/2 bh. Or, 20 = bh. Well what's bh? b is obviously s. h is trickier.
consider the triangle with legs length h and s/2 and with hypotenuse length s.
by pythagorean theorem we have h = sqrt( s^2 - s^2 / 4 ) and so h = sqrt(3)*s / 2
so returning to our area...
20 = s * sqrt(3)/2 * s
40 / sqrt(3) = s^2
by substituting....
40 / sqrt(3) = 4t^2
and so t = sqrt( 10 / sqrt(3) ).
remember t is the length of each side of the regular hexagon.
I'm pretty sure you can find a formula for the area of a regular hexagon given that you have the side length.
Hope this helps!
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