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Find the area of equilateral triangle inscribed in the circle $x^2+y^2+2gx+2fy+c=0$?

$\begin{array}{1 1}(a)\;\large\frac{3\sqrt 3}{2}\normalsize (g^2-f^2+c)sq.units\\(b)\;\large\frac{3\sqrt 3}{4}\normalsize (g^2+f^2-c)sq.units\\(c)\;\large\frac{\sqrt 3}{2}\normalsize (g^2-f^2+c)sq.units\\(d)\;\large\frac{3\sqrt 5}{2}\normalsize (g^2+f^2+c)sq.units\end{array}$

1 Answer

Given circle is $x^2+y^2+2gx+2fy+c=0$------(1)
Let O be the centre and ABC be equilateral triangle inscribed in the circle (1)
In $\Delta OBM$
$\sin 60^{\large\circ}=\large\frac{BM}{OB}$
$BM=OB\sin 60^{\large\circ}=OB\times \large\frac{\sqrt 3}{2}$
$BC=2BM=\sqrt 3OB$
Area of $\Delta ABC=\large\frac{\sqrt 3}{4}$$(BC)^2$
$\Rightarrow \large\frac{\sqrt 3}{4}$$\times 3(OB)^2$
$\Rightarrow \large\frac{3\sqrt 3}{4}\normalsize (g^2+f^2-c)sq.units$
Hence (b) is the correct answer.
answered Feb 11, 2014 by sreemathi.v

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