Question

Find the area of a triangle whose vertices are $(a\cos \theta ,b\sin \theta ),( - a\sin \theta ,b\cos \theta )$ and $( - a\cos \theta , - b\sin \theta )$.
A. $ab\sin \theta \cos \theta $
B. $a\cos \theta \sin \theta $
C. $\dfrac{1}{2}ab$
D. $ab$

Answer 1

Hint: First write the formula of the area of a triangle, then substitute the given coordinates in the formula to obtain the area of the triangle.

Formula Used:
Area=$\dfrac{1}{2}\left[ {{x_1} \left( {{y_{2 - }} {y_3} } \right) + {x_2} \left( {{y_3} - {y_1} } \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$ , where $({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})$ are the vertices of the triangle.
And, ${\cos ^2}\theta + {\sin ^2}\theta = 1$

Complete step by step solution:
Substitute $({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})$ by $(a\cos \theta ,b\sin \theta ),( - a\sin \theta ,b\cos \theta )$ ,$( - a\cos \theta , - b\sin \theta )$
 in the formula $\dfrac{1}{2}\left[ {{x_1} \left( {{y_{2 - }} {y_3} } \right) + {x_2} \left( {{y_3} - {y_1} } \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$and calculate to obtain the required area.
$\dfrac{1}{2}\left[ {a\cos \theta \left( {b\cos \theta - ( - b\sin \theta )} \right) - a\sin \theta \left( { - b\sin \theta - b\sin \theta } \right) + \left( { - a\cos \theta } \right)\left( {b\sin \theta - b\cos \theta } \right)} \right]$
$ = \dfrac{1}{2}\left[ {a\cos \theta \left( {b\cos \theta + b\sin \theta } \right) - a\sin \theta \left( { - 2b\sin \theta } \right) - a\cos \theta \left( {b\sin \theta - b\cos \theta } \right)} \right]$
$ = \dfrac{1}{2}\left[ {ab{{\cos }^2}\theta + ab\sin \theta \cos \theta + 2ab{{\sin }^2}\theta - ab\sin \theta \cos \theta + ab{{\cos }^2}\theta } \right]$
$ = \dfrac{1}{2}\left[ {2ab{{\cos }^2}\theta + 2ab{{\sin }^2}\theta } \right]$
$ = \dfrac{1}{2} \times 2ab\left[ {{{\cos }^2}\theta + {{\sin }^2}\theta } \right]$
$ = \dfrac{1}{2} \times 2ab\left[ 1 \right] = ab$

Option ‘D’ is correct

Note: The another formula can be used to obtain the area of a triangle is $\begin{vmatrix}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{vmatrix}$ , where $({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})$ are the vertices of the triangle.