Question

QUESTION:-

The area of a regular hexagon is 54√3 cm². Find the length of each side.
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Show the steps, don't write the answer only in a few sentence.

(Use Heron's Fomula / Question from RD Sharma - IX)

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Answer 1

Area of regular hexagon = 54√3 cm^2

=> 3√3a^2 /2 = 54√3

=> 3√3 × a^2 = 108√3

=> a^2 = 108/3 = 36

=> a = √36 = 6cm

∴ The side of the hexagon = 6cm

Answer 2

Solution :

First of all divide the regular hexagon into 6 equilateral triangles.

Let the side of the triangle be a

Now,

Semi-Perimeter of Triangle = Side + Side + Side/2

→ Semi-Perimeter = a + a + a/2

→ Semi-Perimeter = 3a/2

• Given area = 54√3 cm²

Using Heron's Formula :

$\bigstar \: \sf \triangle \: = \sqrt{s(s - a)(s - b)(s - c)}$

$\longrightarrow \sf 54 \sqrt{3} \: = 6[\sqrt{ \frac{3a}{2} ( \frac{3a}{2} - \frac{a}{2} )( \frac{3a}{2} - \frac{a}{2} )( \frac{3a}{2} - \frac{a}{2}}]$

$\longrightarrow \sf 54 \sqrt{3} \: = 6(\sqrt{ \frac{3a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}})$

$\longrightarrow \sf 54 \sqrt{3} \: = 6(\sqrt{ 3 \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2} })$

$\longrightarrow \sf 54 \sqrt{3} \: = 6(\sqrt{ 3 \times \frac{ {a}^{2} }{4} \times \frac{ {a}^{2} }{4}})$

$\longrightarrow \sf 54 \sqrt{3} \: = 6(\sqrt{ 3 \times \frac{a}{2} \times \frac{a}{2} })$

$\longrightarrow \sf 54 \sqrt{3} \: = \cancel6 \times \sqrt{3} \times \frac{ {a}^{2} }{ \cancel4} }$

$\longrightarrow \sf 54 \sqrt{3} \: = \dfrac{3 \sqrt{3} {a}^{2} }{2}$

$\longrightarrow \sf 108 \sqrt{3} \: = 3 \sqrt{3} {a}^{2}$

$\longrightarrow \sf \dfrac{108 \sqrt{3} }{3 \sqrt{3} } = {a}^{2}$

$\longrightarrow \sf \cancel \dfrac{108}{3} = {a}^{2}$

$\longrightarrow \sf 36 = {a}^{2}$

$\longrightarrow \sf a = \sqrt{36}$

$\longrightarrow \sf a = \sqrt{6 \times 6}$

$\longrightarrow \sf a = 6 \: cm$

$\therefore$ Length of each side of regular hexagon is 6 cm.