Open Question: How to figure out this hard area and perimeter problem?

19:51 Publicado por Flechado16

So an equilateral has 3 sides of length s. So P = s + s + s. or, P = 3s.
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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