Question

a Side of an equilateral triangle is 6 cm
find the height of the triangle and area​

Answer 1

⋆ Given:-

• Sides of equilateral triangle = 6 cm

⋆ To find:-

• Height of Triangle

• Area of Triangle

⋆ Solution:-

★ Area of equilateral ∆ = $\sf{ \dfrac{ \sqrt{3}}{4} \times a^2}$

$\implies \sf{ \dfrac{ \sqrt{3}}{4} \times 6 \times 6}$

$\implies \sf{ \dfrac{ \sqrt{3}}{4} \times 36}$

$\implies \sf{ \dfrac{ \sqrt{3}}{ \cancel{4} }\times \cancel{36}}$

$\implies \sf{9 \sqrt{3} \; cm^2}$

Also,

★ Area of ∆ = $\sf{ \dfrac{1}{2} \times base \times height}$

$\implies \sf{9 \sqrt{3} = \dfrac{1}{ \cancel{2}} \times \cancel{6} \times H}$

$\implies \sf{\dfrac{9 \sqrt{3}}{3} = H}$

$\implies \sf{H = 3 \sqrt{3}}$

Hence,

★ Area of equilateral ∆ = $\sf{9 \sqrt{3} \; cm^2}$

★ And, Height of equilateral ∆ = $\sf{3 \sqrt{3} \; cm}$

$\rule{200}{2}$

Answer 2

Given that ,

• Side of equilateral triangle (a) = 6 cm

We know that , the area of equilateral triangle is given by

$\large \sf \underline{ \fbox{Area \: of \: Δ = \frac{ \sqrt{3} \times {(a)}^{2} }{4} }}$

Thus ,

$\sf \Rightarrow Area = \frac{ \sqrt{3} \times {(6)}^{2} }{4} \\ \\\sf \Rightarrow Area = \frac{ \sqrt{3} \times 36 }{4} \\ \\\sf \Rightarrow Area = 9 \sqrt{3} \: \: {m}^{2}$

$\therefore \underline \bold{ \sf{The \: area \: of \: equilateral \: triangle \: is \: 9 \sqrt{ 3} \: \: {m}^{2} }}$

Now ,

$\sf \underline{\fbox{Area \: of \: Δ = \frac{1}{2} \times base \times height }}$

Thus ,

$\sf \Rightarrow 9 \sqrt{3} = \frac{1}{2} \times 6 \times height \\ \\\sf \Rightarrow height = 3 \sqrt{3} \: \: m$

$\therefore \sf{\underline{ The\: height \: of \: triangle \: is \: 3 \sqrt{3} \: \: m}}$