Question

Area of a triangle whose vertices are $(a\cos \theta ,b\sin \theta ),(-a\sin \theta ,b\cos \theta )$ and $(-a\cos \theta ,-b\sin \theta )$ is

• A. $ab\sin \theta \cos \theta$
• B. $a\cos \theta \sin \theta$
• C. $\frac{1}{2}ab$
• D. $ab$

Answer 1

Answer: (D)

$ab$

Area of triangle is
$\Delta =\frac{1}{2}\left|\begin{array}{ccc}a\cos \theta & b\sin \theta & 1\\ -a\sin \theta & b\cos \theta & 1\\ -a\cos \theta & -b\sin \theta & 1\end{array}\right|$
Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$
Then
$\Delta =\frac{1}{2}\left|\begin{array}{ccc}a\cos \theta & b\sin \theta & 1\\ -a\sin \theta -a\cos \theta & b\cos \theta -b\sin \theta & 0\\ -2a\cos \theta & -2b\sin \theta & 0\end{array}\right|$
$=\frac{1}{2}[-2b\sin \theta (-a\sin \theta -a\cos \theta )+2a\cos \theta (b\cos \theta -b\sin \theta )]$
$=\frac{1}{2}(2ab{\sin }^2\theta +2ab\sin \theta \cos \theta +2ab{\cos }^2\theta -2ab\sin \theta \cos \theta )$
$=\frac{1}{2}\times 2ab=ab$