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

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

Answers (1)

[300\sqrt{3}cm^{2} ]

Here \DeltaABC is equilateral triangle i.e.

AB = BC = CA = x (Let)

                     

Now we can see here three perpendicular

OX \perp AB, OZ \perp BC, OY \perp AC

OX = 10 cm, OY = 14 cm\; and\; OZ = 6 cm

So\; that\; area\; of\; \Delta ABC = Ar(\Delta AOC) + Ar(\Delta BOC) + Ar(\Delta AOB)

\frac{\sqrt{3}}{4} side^{2}=\frac{1}{2} \times base \times height + \frac{1}{2} \times base \times height + \frac{1}{2} \times base \times height

[\Theta \Delta ABC\; is\; equilateral\; \Delta ,so\; area\; of\; equilateral\; \Delta = \frac{\sqrt{3}}{4}side^{2} ]

\Rightarrow \frac{^{\sqrt{3}}}{4}\times x^{2}=\frac{1}{2}\times AC\times OY+\frac{1}{2}\times BC\times OZ+\frac{1}{2}\times AB\times OX

Taking\; \frac{1}{2} \; common

\frac{\sqrt{3}}{4}\times x^{2}=\frac{1}{2}\times \left [ AC\times OY+BC\times OZ+AB\times OX \right ]

\frac{\sqrt{3}}{4}\times x^{2}=\frac{1}{2}\times \left [ AC\times OY+AC\times OZ+AC\times OX \right ]

[\Theta  AB = BC = AC, \DeltaABC equilateral triangle]

\frac{\sqrt{3}}{4}\times x^{2}=\frac{1}{2}AC\times \left [ OY+OZ+OX \right ]

\frac{\sqrt{3}}{4}\times x^{2}=\frac{1}{2}AC\times \left [ 14+6+10 \right ]

\frac{\sqrt{3}}{4}\times x^{2}=\frac{1}{2}AC\times \left [ 30 \right ]

\frac{\sqrt{3}}{4}\times x^{2}=15\times AC

\sqrt{3}\times x^{2}=15\times x\times 4\; \; \; \; \; \; [\because AC = BC = AC = x]

\frac{x^{2}}{x}=\frac{15\times 4}{\sqrt{3}}

x=\frac{60}{\sqrt{3}}cm

Hence \; the\; length\; of\; the\; sides\; of\; \Delta ABC\; is\; \frac{60}{\sqrt{3}}cm

Area\; of\; \Delta ABC =\frac{\sqrt{3}}{4}\times \left ( \frac{60}{\sqrt{3}} \right )^{2}=\frac{\sqrt{3}}{4}\times \frac{60}{\sqrt{3}}\times \frac{60}{\sqrt{3}}

=\frac{15\times 60}{\sqrt{3}}=\frac{900}{\sqrt{3}}

On rationalisation

\frac{900}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{900\times \sqrt{3}}{3}=300\sqrt{3}cm^{2}

 

Posted by

infoexpert21

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