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The maximum area of the triangle formed by the circle $x^2+y^2=a^2$ with two of its vertices are diametrically opposite points is?

$\begin{array}{1 1}(a)\;a^2\\(b)\;\large\frac{a^2}{2}\\(c)\;\large\frac{a^2}{4}\\(d)\;\text{none of these}\end{array}$

1 Answer

The maximum area of triangle $ABC$ will be when A is (0,a),B(-a,0),C(a,0)
Hence area of $\Delta ABC$ is $\large\frac{1}{2}$$\begin{vmatrix}0 & a& 1\\-a & 0 & 1\\a &0&1\end{vmatrix}$
$\Rightarrow \large\frac{1}{2}$$(|a(-a-a)|)$
$\Rightarrow \large\frac{1}{2}$$\times 2a^2$
$\Rightarrow a^2$
Hence (a) is the correct option.
answered Feb 11, 2014 by sreemathi.v
edited Mar 19, 2014 by rvidyagovindarajan_1

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