If an equilateral triangle and a hexagon have the same perimeter, which area is greater and by how much? Please show work.

Answer 1

Let's say the perimeter of the equilateral triangle is #2 + 2 + 2 = 6#, while the perimeter of the regular hexagon is #1 + 1 + 1 + 1 + 1 + 1 = 6#. Then:

#A_"triangle" = (bh)/2#

From the diagram you can see that a 30-60-90 triangle has height #b/2*sqrt3#, so we have, with #b = 2#:

#A_"triangle" = (b*b/2*sqrt3)/2 = (b^2sqrt3)/4 = sqrt3 ~~ color(blue)(1.732)#

For the hexagon, you can think of it as six equilateral triangles of perimeter #3# (draw straight lines connecting corners across the hexagon).

With each of those triangles, #b = 1#, and we get:

#A_"hexagontriangle" = (b^2sqrt3)/4 = sqrt3/4 ~~ 0.43#

And with six triangles in the hexagon, we get:

#= 6 * sqrt3/4 = (3sqrt3)/2 ~~ color(blue)(2.60)#

And so we get:

#((3sqrt3) / 2 - sqrt3)/(sqrt3) * 100% = color(highlight)"50% larger area"#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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