Question

The perimeters of a square and a regular hexagon are equal.Find the ratio of the area of the hexagon to the area of a square.

Answer 1

Answer:

perimeter of square-4a

a=4a

Parimeter of r.hexagon=6x

4a=6x

a=6/4x

area of square-a2

= (6/4x) 2 = 36/16xX2 = 9/4 x2

area of R.hexagon=3v3x2/2

Ratio:9x2/4:.3v3x2/2

18x^2:12 v3x^2

18:12 v3

3:2v 3. Ans

Answer 2

The ratio of the area of hexagon to the area of a square is $2\sqrt{3} :3$

Step-by-step explanation:

Consider the provided information.

Let a is the length of the side of square and x is the length of the side of hexagon.

Then, the perimeters of a square is:

$P=4\times side\\P=4a$

The perimeter of Hexagon is:

$P=6\times side\\P=6x$

The perimeters of a square and a regular hexagon are equal.

$4a=6x$

$a=\frac{3x}{2}$

We need to find the ratio of the area of the hexagon to the area of a square.

The area of hexagon is: $A=\frac{3\sqrt{3}}{2}(side)^2$

$A=\frac{3\sqrt{3}}{2}(x)^2$

The area of square is: $A=(side)^2$

$A=(a)^2$

Thus, the ratio will be:

$\frac{\text{Area of hexagon}}{\text{Area of squre}} =\frac{\frac{3\sqrt{3}}{2}(x)^2}{a^2}$

Substitute the value of a in above.

$\frac{\text{Area of hexagon}}{\text{Area of squre}} =\frac{\frac{3\sqrt{3}}{2}(x)^2}{(\frac{3x}{2})^2}$

$\frac{\text{Area of hexagon}}{\text{Area of squre}} =\frac{\frac{3\sqrt{3}}{2}(x)^2}{\frac{9x^2}{4}}$

$\frac{\text{Area of hexagon}}{\text{Area of squre}} =\frac{3\sqrt{3}\times4}{2\times9}$

$\frac{\text{Area of hexagon}}{\text{Area of squre}} =\frac{\sqrt{3}\times2}{3}$

$\frac{\text{Area of hexagon}}{\text{Area of squre}} =\frac{2\sqrt{3}}{3}$

Hence, the ratio of the area of hexagon to the area of a square is $2\sqrt{3} :3$

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