Question

Find the area of an equilateral triangle with side 2√3cm

Answer 1

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Answer 2

Answer:

$\underline\mathfrak{Given:-}$

$\: \: \: \: \: An \: \: equilateral \: \: triangle \: \: of \: \: side \: \: = \: \: {2}\sqrt{3} \: cm$

$\underline\mathfrak{To \: \: Find:-}$

$\: \: \: \: \: find \: \: it's \: \: area.?$

$\underline\mathfrak{Solutions:-}$

$\: \: \: \: \: Let \: \: the \: \: area \: \: of \: \: triangle \: \: be \: \: x \: cm.$

$\: \: \: \: \: Area \: \: of \: \: equilateral \: \: triangle \: \: = \: \: \frac{\sqrt{3}}{4} \: \times \: {(sides)}^{2}$

$\: \: \: \: \: \frac{\sqrt{3}}{4} \: \times \: {(sides)}^{2} \: \: = \: \: x$

$\: \: \: \: \: \leadsto \frac{\sqrt{3}}{4} \: \times \: {2}\sqrt{3} \: \: = \: \: x$

$\: \: \: \: \: \leadsto \frac{{2} \: \times \: {3}}{4} \: \: = \: \: x$

$\: \: \: \: \: \leadsto \frac{6}{4} \: \: = \: \: x$

$\: \: \: \: \: \leadsto \frac{3}{2} \: \: = \: \: x$

$\: \: \: \: \: Hence, \: \: the \: \: area \: \: of \: \: equilateral \: \: is \: \: \frac{3}{3}.$

$\underline\mathfrak{Important \: \: formula:-}$

$\: \: \: \: \: base \: \: and \: \: height \: \leadsto \: A \: \: = \: \: \frac{1}{2} \: bh \: \: \: \: {(where \: \: b \: \: = \: \: base, \: \: h \: \: = \: \: height)}$

$\: \: \: \: \: three \: \: sides \: \leadsto \: A \: \: = \: \: \sqrt{s \: (s \: - \: a) \: (s \: - \: b) \: (s \: - \: c)} \: \: \: {(where \: \: a, \: \: b, \: \: and \: \: c \: \: are \: \: the \: \: length \: \: of \: \: the \: \: side.)}$

$\: \: \: \: \: two \: \: sides \: \: and \: \: including \: \: angles \: \: \leadsto \: \: A \: \: = \: \: \frac{1}{2} \: Sin \: c \: \: \: {(where \: \: a, \: \: b, \: \: are \: \: two \: \: sides \: \: and \: \: c \: \: is \: \: the \: \: angle \: \: between \: \: them.)}$

$\: \: \: \: \: equilateral \: \: triangle \: \: \leadsto \: \: A \: \: = \: \: \frac{{s}^{2} \: \sqrt{3}}{4} \: \: \: \: {(where \: \: s \: \: = \: \: side.)}$