Question

Compare the areas of an equilateral triangle, a square and a regular hexagon of equal perimeter

Answer 1

Answer:

Formula for the area of an equilateral triangle with side length a is At=√34⋅a2

Let 2x be the side length of the equilateral triangle,

given that area of the equilateral At=2 units2

⇒At=√34⋅(2x)2=√34⋅4x2=2

⇒x2=2√3 units2

A regular hexagon can be divided into 6 congruent equilateral triangles, as shown in the figure.

given that the equilateral triangle and the regular hexagon have equal perimeter,

⇒ side length of the hexagon =3⋅2x6=x units

⇒ area of the regular hexagon =Ah=6⋅√34⋅x2

⇒Ah=6⋅√34⋅2√3=3 units2

Answer 2

The area of the given three shapes with equal perimeters will be in the following proportions -

Equilateral Triangle: Square: Hexagon:: $\frac{\sqrt{3}}{9} : \frac{1}{4} : \frac{3\sqrt{3} }{18}$

Given,

Perimeters of an equilateral triangle, a square and a regular hexagon are equal

To Find,

Areas of these 3 shapes

Solution,

Let the perimeter of the equilateral triangle, the square and the regular hexagon be 'x'

If 'x' is the perimeter of the equilateral triangle then,

Side length = $\frac{x}{3}$

Area = $\frac{ (side)^2\sqrt{3}}{4}$

Area = $\frac{x^2\sqrt{3} }{36}$

If 'x' is the perimeter of the square then,

Side length = $\frac{x}{4}$

Area = $(side)^2$

Area = $\frac{x^2}{16}$

If 'x' is the perimeter of the hexagon then,

Side length = $\frac{x}{6}$

Area = $\frac{3\sqrt{3}}{2} (side)^2$

Area = $\frac{3\sqrt{3} x^2}{72}$

Comparing we get the proportion as -

Equilateral Triangle: Square: Hexagon:: $\frac{\sqrt{3}}{36} : \frac{1}{16} : \frac{3\sqrt{3} }{72}$

Simplifying we get,

Equilateral Triangle: Square: Hexagon:: $\frac{\sqrt{3}}{9} : \frac{1}{4} : \frac{3\sqrt{3} }{18}$

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