Question

Consider an equilateral triangle with side 'x' . Drive a formula by using herons formula to calculate area . Use the formula and calculate area for an equilateral triangle with side 4cm​

Answer 1

Solution

☞︎︎︎ Given :

• Equilateral triangle

(All sides are equal)

• Side = x

$\large \sf \underline{ \underline{\bold{ \star \: \: finding \: \: area : }}}$

➪ S = semi - Perimeter

➪ a , b , c = 3 sides

$\large \rm \underline{ \underline{{ \star \: \: formula: }}}$

$\\ \large \longrightarrow \: \underline{ \boxed{\tt{s = \frac{a + b + c}{2} }}}$

• Since , All sides = x

$\\ \large \sf : \implies \: s = \dfrac{x + x + x}{2} \\ \\ \large \sf \therefore \: s = \frac{3x}{2}$

$\large \tt \star \: \underline{ \brack{\bold{ \large \: herons \: \: \: formula \: }}}$

Area = a

$\\ \mapsto\large \sf \:a = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ \\ \tt \implies \: a = \sqrt{ \frac{3x}{2} \left( \frac{3x}{2} - x\right) \left( \frac{3x}{2} - x\right) \left( \frac{3x}{2} - x\right)} \\ \\ \tt \implies \: a = \sqrt{ \frac{3x}{2} \left( \frac{3x - 2x}{2} \right) \left( \frac{3x - 2x}{2} \right) \left( \frac{3x - 2x}{2} \right)} \\ \\ \tt \implies \: a = \sqrt{ \frac{3x}{2} \left( \frac{x}{2} \right) \left( \frac{x}{2} \right) \left( \frac{x}{2} \right)} \\ \\ \tt \implies \: a = \sqrt{ \frac{3 {x}^{4} }{16} } \\ \\ \tt \implies \: a = { \frac{ \sqrt{ 3 } \sqrt{ {x}^{4} }}{ \sqrt{ 16} } }$

$\large \sf \leadsto \: \underline{ \boxed{ \sf \: a = \dfrac{ {x}^{2} \sqrt{3} }{4} }}$

☆ Hence, Formula Drived

Applying Formula for :

• equilateral triangle of side 4cm

$\large \sf \implies{{ \sf \: a = \dfrac{ {x}^{2} \sqrt{3} }{4} }} \\ \\ \large \sf \implies{{ \sf \: a = \dfrac{ {4}^{2} \sqrt{3} }{4} }} \\ \\ \large \sf \implies{{ \sf \: a = \dfrac{ {16}^{} \sqrt{3} }{4} }} \\ \\ \large \sf \implies{{ \sf \: a = \dfrac{ { \cancel{16} \: \: 4}^{} \sqrt{3} }{ \cancel4} }} \\ \\ \large \sf \implies{{ \sf \: a = { {4}^{} \sqrt{3} }{} }} \\ \\ \large \implies \sf \: a = 4 \times 1.73 \: (approx) \\ \\ \large \sf \: \therefore \: \: \underline{ \underline{area = 6.92c {m}^{2} }}$

Area = 6.92cm² (approx)

☞︎︎︎ Check :

• Side = 4

$\\ \large \longrightarrow \: \underline{ \boxed{\tt{s = \frac{a + b + c}{2} }}}$

$\\ \sf : \implies \: s = \dfrac{4 + 4 + 4}{2} \\ \\ \sf : \implies \: s = \dfrac{12}{2} \\ \\ \sf \therefore \: \underline{ s = 6cm}$

• Applying Herons Formula

$\\ \mapsto\tt \:a = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ \implies \tt \:a = \sqrt{6(6 - 4)(6 - 4)(6 - 4)} \\ \\ \implies \tt \:a = \sqrt{6 \times 2 \times \bar2 \times \bar2} \\ \\ \implies \tt \:a = 2 \sqrt{3 \times \bar 2 \times \bar2} \\ \\\implies \tt \:a = 2 \times 2\sqrt{3 } \\ \\\implies \tt \:a = 4 \times 1.73 \\ \\ \sf \underline{\boxed{ \sf\therefore \: \: area = 6.29 {cm}^{2} }}$