Question

The height of an equilateral triangle is √6 cm. find Its area​

Answer 1

Let side of equilateral TRAINGLE = a cm

$\begin{gathered}\color{red}h = \frac{ \sqrt{3} }{2} a \\ \\ </p><p>\color{blue} \sqrt{6} = \frac{ \sqrt{3} }{2} a \\ \\\color{red}\frac{ \sqrt{6} \times 2 }{ \sqrt{3} } = a \\ \\ \color{hotpink} \boxed{ a = > 2 \sqrt{2} cm } \\ \\ \color{red}area = \frac{ \sqrt{3} }{4} {a}^{2} \\ \\ \color{skyblue} area = \frac{ \sqrt{3} }{4} \times ( 2\sqrt{2} {)}^{2} \\ \\ \color{pink}area = > \frac{ \sqrt{3} }{4} \times 8cm \\ \\ \color{red} \boxed{area = > 2 \sqrt{3} c {m}^{2} }\end{gathered}$

$\huge\underline\bold\orange{Thanks}$

Answer 2

Answer :

›»› The area of an equilateral traingle = 2√3 cm²

Given :

• Height of an equilateral traingle = √6 cm

To Find :

• Area of an equilateral traingle = ?

Required Solution :

To find the area of an equilateral traingle, at first we need to find the side of an equilateral traingle, after that we will find Area of an equilateral traingle on the basis of conditions given above.

In right angle ∆ADB

From Pythagoras theorem

$\tt{:\implies {\bigg(\dfrac{a}{2} \bigg)}^{2} + { (\sqrt{6} )}^{2} = {a}^{2} }$

$\tt{:\implies \bigg( \dfrac{a}{2} \times \dfrac{a}{2} \bigg) + \big(\sqrt{6} \times \sqrt{6} \big) = {a}^{2} }$

$\tt{:\implies \dfrac{ {a}^{2} }{4} + 6 = {a}^{2} }$

$\tt{:\implies \dfrac{3 {a}^{2} }{4} = 6}$

$\tt{:\implies {a}^{2} = 8}$

$\bf{:\implies \underline{ \: \: \underline{ \orange{ \: \: a = 2\sqrt{2}\: cm \: \: }} \: \: }}$

Now, we have one elements that used in formula Side of an equilateral traingle.

• Side of an equilateral traingle (a) = 2√3 cm

And we need to find Area of an equilateral traingle.

As we know that

$\tt{:\implies Area = \dfrac{ \sqrt{3} }{2} {a}^{2} }$

$\tt{:\implies Area = \dfrac{ \sqrt{3} }{4} \times \big( {{2 \sqrt{2} } \big)}^{2} }$

$\tt{:\implies Area = \dfrac{ \sqrt{3} }{4} \times 2 \sqrt{2} \times 2 \sqrt{2} }$

$\tt{:\implies Area = \dfrac{ \sqrt{3} }{ \cancel{4}} \times \cancel{8}}$

$\tt{:\implies Area = \sqrt{3} \times 2 }$

$\bf{:\implies \underline{ \: \: \underline{ \red{ \: \: Area = 2 \sqrt{3} \: {cm}^{2} \: \: }} \: \: }}$

║Hence, the Area of an equilateral traingle is 2√3 cm².║