Question

find the area of an equilateral triangle with side 2√3 cm for class 9

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Answer 1

GIVEN :-

• Side of an equilateral ∆ = 2√3 cm.

TO FIND :-

• The area of an equilateral ∆.

SOLUTION :-

As we know that the area of ∆ id given by,

$\\ : \implies \: \displaystyle \sf \: Area _{( \triangle)} = \sqrt{s(s - a)(s - b)(s - c)}$

• a = 2√3 cm.
• b = 2√3 cm.
• c = 2√3 cm.
• s = ( a + b + c )/2 = (2√3 + 2√3 + 2√3)/2 = 3√3.

$\\ \\ : \implies \: \displaystyle \sf \: Area _{( \triangle)} = \sqrt{3 \sqrt{3} \bigg(3 \sqrt{3} - 2 \sqrt{3}\bigg)\bigg(3 \sqrt{3} - 2 \sqrt{3} \bigg)\bigg(3 \sqrt{3} - 2 \sqrt{3} \bigg)}$

$: \implies \: \displaystyle \sf \: Area _{( \triangle)} = \sqrt{3 \sqrt{3} \times \sqrt{3} \times \sqrt{3} \times \sqrt{3} }$

$: \implies \: \displaystyle \sf \: Area _{( \triangle)} = \sqrt{3 \sqrt{9} \times \sqrt{9} }$

$: \implies \: \displaystyle \sf \: Area _{( \triangle)} = \sqrt{3 \times 3 \times 3}$

$: \implies \: \displaystyle \sf \: Area _{( \triangle)} = \sqrt{27}$

$: \implies \: \displaystyle \sf \: Area _{( \triangle)} = \sqrt{9 \times 3}$

$: \implies \: \underline{ \boxed{ \displaystyle \sf \: Area _{( \triangle)} = 3 \sqrt{3} }}$

Answer 2

Hope this helps.......