Math contest solution: the question asked an equilateral triangle and hexagon have the same area, what would?

Question

be the ratio of their perimeters

Answer 1

I find this an interesting question, and I suspect there are several possible methods for solving it. However, I recommend we deal with it straightforwardly by solving for, say, the base length, height ... any 'variables' found within these polygons' area formulas.

The understanding here is that both polygons are regular polygons (the triangle is therefore equilateral, the hexagon has 6 equal-length sides, etc.).

The most creative solutions (and least mathematical) would probably involve cutting up one of the shapes and "re-gluing" it to form the other, thereby gaining some hint at the relationship between their perimeters.

I normally prefer straight math, and would recall:

A(triangle) = (1/2) base * height
A(hexagon) = ([3sqrt(3)]/2) (side)^2

But let's "logic" our way through this one first. Think of the hexagon as being composed of 6 equilateral triangles, each containing 1/6 of the overall area. Because the original hexagon had an area in 1:1 ratio with the given triangle, we now can state:

Area (SmallTriangle) = (1/6) (BigTriangle)

Recalling a particular theorem concerning similar polygons ("The ratio of their areas will be the square of the ratio of their sides"), we can take this further:

Ratio of BigTri:SmallTri Areas = 6:1

So,

Ratio of BigTri:SmallTri Sides = sqrt(6):1

Picture now how the small triangles would be "glued together" to form their parent hexagon. Only the "outside side" would contribute to the hexagon's perimeter ... SIX such sides would be needed. In short:

Ratio of hexagon perimeter:BigTri perimeter = 6sqrt(6):1

Hope this (is correct and) helps!

Answer 2

Area of an equilateral triangle is 1/2 side times height, with height calculated as sin(60°) side. In the following observation, I will treat this as a constant factor which can be eliminated. A triangle area will be k*(side length) squared.

The equilateral triangle has 1 triangle area of side length t and a perimeter of 3 sidelengths t.

The equilateral hexagon has a smaller side length h, with an area of six triangle areas of side length h and a perimeter of six side lengths h.

So we have k * t^2 = 6* k * h^2. Eliminating k, we get

t^2 = 6 * h^2, or

t = sqrt(6) h.

comparing the perimeters, the ratio is

3*t / 6*h

or

3* sqrt(6) h / 6 h.

Eliminating h and 3 we get as the ratio

sqrt(6) / 2

or 1.22 : 1

Answer 3

The area of the hexagon is made up of 6 smaller equilateral triangles. let s = side of hexagon and S = side of triangle.
The height of an equilateral triangle is 1/2(S)(sqrt(3)) which is the long leg of a 30-60-90 right triangle where the short leg is half the side of the equilateral triangle.

Area of triangle = 1/2(b)(h) ==> 1/2(S)[(S)(sqrt(3)/2)] =
S^2sqrt(3)/4

Area of hexagon = 6(s^2)sqrt(3)/4 = 3s^2(sqrt(3))/2

Setting these equal and solving for the ratio of the side of the triangle to side of the hexagon yields S/s=sqrt(6)/1

The perimeter of the triangle P(t)=3S

The perimeter of the hexagon P(h)=6s

The ratio of P(t)/P(h) is then
3S/6s = S/2s = 1/2(S/s) ==> 1/2(sqrt(6)/1) = sqrt(6)/2

Answer 4

A hexagon can be considered to consist of six equilateral triangle. If the side = 1, then each triangle forms two right triangles with an area of 1/2bh.
The base would be 1/2 of a side or 1/2.
The height would be 1^2 = h^2 + (1/2)^2
h^2 = 3/4
h = (sqrt 3)/2
The area of the each equilateral triangle would be twice that or simply Sqrt 3.
Since there are six equ. triangles, the area of the hexagon is 6*Sqrt3.
The perimeter is 6.

The equ. triangle with an area of 6Sqrt3.
A=bh; h=sqrt3*b; A=b^2*sqrt3
b^2*sqrt3 = 6sqrt3
b^2=6
b=sqrt6

Perimeter of triangle is 3*sqrt6
Perimeter of hexagon is 6
ratio is 3sqrt6:6 = sqrt6:2
~1.22:1

Answer 5

A hexagon can be made out of 6 equilateral triangles. My intuition says 2:1. If you take the triangle and make a hex you'd get 6 times the area, and 3 times the perimeter, right?

Answer 6

Area of triangle of side x = (1/2) * base * height
= (1/2) * (x) * [x * sqrt(3) / 2]
= x^2 * sqrt(3) / 4

Area of hexagon of side y = 6 * Area of small triangle, formed by drawing lines from the centre to the vertices.
= 6 * (1/2) * base * height
= 6 * (1/2) * (y) * [y * sqrt(3) / 2]
= 6 * y^2 * sqrt(3) / 4

Equating these gives :
x^2 * sqrt(3) / 4 = 6 * y^2 * sqrt(3) / 4
Dividing through by sqrt(3) / 4 gives :
x^2 = 6* y^2
or, x = y * sqrt(6)

Perimeter of triangle = 3x = 3 * y * sqrt(6)
Perimeter of hexagon = 6y

Therefore, ratio = 3 * y * sqrt(6) / 6y
= sqrt(6) / 2

Answer 7

A(tri) = 1/2 h b; where h = height and b = base of the triangle with area A(tri)

A(hex) = 6 * A'(tri); where the area of a hexagon consists of 6 similar triangles each of equal area; so that A'(tri) = A(hex)/6 = 1/2 h' b', where h' and b' are the height and base for each of the included triangles for the hexagon.

Given A(hex) = A(tri), we have 6 A'(tri) = A(tri) = 6 1/2 h' b' = 1/2 h b; so that 6 h' b' = h b, which means the product of the single equilateral triangle parameters (height and base) is six times the product of the parameters for each included triangle of the hex.

Answer 8

A regular hexagon can be divided into 6 equilareral triangles. therefore the triangle must have sides √6 times larger than the hexagon
let s=side of hexagon
p(hex)=6s
p(tri)=3*√6 * s
p(tri)/p(hex)=3*√6 * s/(6s)=√6 / 2=1.225

Answer 9

1:2 ? lol idk