Math, asked by Anonymous, 1 day ago

From a point in the interior of an equilateral triangle, perpendicular is drawn on the three sides. The lengths of the perpendicular are 14cm, 10cm and 6cm. Find the area of the triangle.

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Answers

Answered by Anonymous
8

\huge\underline{\underline{\mathfrak\red{Given::}}}

\small\text{From a point in the interior of an equilateral} \\  \small\text{triangle, perpendicular is drawn on the three sides.  } \\ \small\text{The lengths of the perpendicular are 14cm, 10cm and 6cm. } \\ \small\text{Find the area of the triangle.}

\huge\underline{\underline{\mathfrak\red{Solution::}}}

\text{Let each side of \: equilateral triangle be 'x' cm} \\  \\\bf{First\: Perpendicular} \\  \\  \bf{\implies \:  \frac{1}{2}  \times base \times height} \\  \\ \bf{\implies \:  \frac{1}{2}  \times x \times 14} =\red{ 7a \:  {cm}^{2} }

\bf{Second\: Perpendicular} \\  \\ \bf{\implies \:  \frac{1}{2}  \times base \times height} \\  \\ \bf{\implies \:  \frac{1}{2}  \times x \times 10} =\red{ 5x \:  {cm}^{2} }

\bf{Second\: Perpendicular} \\  \\ \bf{\implies \:  \frac{1}{2}  \times base \times height} \\  \\ \bf{\implies \:  \frac{1}{2}  \times x \times 6} =\red{ 3x \:  {cm}^{2} }

\bf{Third\: Perpendicular} \\  \\ \bf{\implies \:  \frac{1}{2}  \times base \times height} \\  \\ \bf{\implies \:  \frac{1}{2}  \times x \times 6} =\red{ 3x \:  {cm}^{2} }

\bf{Perpendiculars\: in\: total \:=7x+5x+3x=15x}

\bf{Area  \: of  \: equilateral  \: triangle  = \red{ \frac{ \sqrt{3} }{4}  \times  {x}^{2} }} \\ \bf{x =  \red{ side}} \\  \\  \bf{  \frac{ \sqrt{3} }{4} \times   {x}^{2} =   \red{15x}} \\  \\ \implies\bf{x = \red{ \frac{15 \times 4}{ \sqrt{3} } \times  \frac{ \sqrt{3} }{ \sqrt{3} }  }} \\  \\ \implies\bf{x = \red{ \frac{60 \sqrt{3} }{3} }} \\  \\ \implies\bf{x = \red{20 \sqrt{3}  \:  {cm}^{2} }} \\  \\ \\ \bf{ now \: by \: using \: formula}  \\  \\ \implies\bf{\red{ \frac{ \sqrt{3} }{4}  \times  {(20 \sqrt{3)} }^{2}  }}  \\  \\ \boxed{\implies\mathfrak{\pink{300 \sqrt{3 {cm}^{2} } }}} \\  \\ \boxed{\mathfrak{\green{hope \: it \: helps}}}

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