An equilateral triangle and a regular hexagon have equal perimeters. if the area of the triangle is 2, what is the area of the hexagon?

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An equilateral triangle and a regular hexagon have equal perimeters. if the area of the triangle is 2, what is the area of the hexagon?

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Answer 1 · 2018-03-29T08:00:57

$A_{h} = 3 {units}^{2}$ $A_h=3 " units"^2$

Explanation:

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Formula for the area of an equilateral triangle with side length $a$ $a$ is $A_{t} = \frac{\sqrt{3}}{4} \cdot a^{2}$ $A_t=sqrt3/4*a^2$
Let $2 x$ $2x$ be the side length of the equilateral triangle,
given that area of the equilateral $A_{t} = 2 {units}^{2}$ $A_t=2 " units"^2$
$\Rightarrow A_{t} = \frac{\sqrt{3}}{4} \cdot {(2 x)}^{2} = \frac{\sqrt{3}}{4} \cdot 4 x^{2} = 2$ $=> A_t=sqrt3/4*(2x)^2=sqrt3/4*4x^2=2$
$\Rightarrow x^{2} = \frac{2}{\sqrt{3}} {units}^{2}$ $=> x^2=2/sqrt3 " units"^2$

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,
$\Rightarrow$ $=>$ side length of the hexagon $= \frac{3 \cdot 2 x}{6} = x$ $= (3*2x)/6=x$ units
$\Rightarrow$ $=>$ area of the regular hexagon $= A_{h} = 6 \cdot \frac{\sqrt{3}}{4} \cdot x^{2}$ $=A_h=6*sqrt3/4*x^2$
$\Rightarrow A_{h} = 6 \cdot \frac{\sqrt{3}}{4} \cdot \frac{2}{\sqrt{3}} = 3 {units}^{2}$ $=> A_h=6*sqrt3/4*2/sqrt3=3 " units"^2$

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An equilateral triangle and a regular hexagon have equal perimeters. if the area of the triangle is 2, what is the area of the hexagon?

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