Question

The area of the parallelogram with vertices $(0,0),(7,2),(5,9)$ and $(12,11)$ is

• A. 50
• B. 54
• C. 51
• D. 52
• E. 53

Answer 1

Answer: (E)

Let $A(0,0),B(7,2),C(5,9),D(12,11)$
$\therefore$ Area of $\Delta ABC=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 7& 2& 1\\ 5& 9& 1\end{array}\right|$
$=\frac{1}{2}\cdot 1(63-10)=\frac{53}{2}$ sq unit
$\therefore$ Area of $\Delta ACD=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ 5& 9& 1\\ 12& 11& 1\end{array}\right|$
$=\frac{1}{2}\cdot 1(55-108)$
$=\frac{53}{2}$ sq unit
$\therefore$ Area of parallelogram $ABCD$
$=$ Area of $△ABC+$ Area of $\Delta ACD$
$=\frac{53}{2}+\frac{53}{2}$
$=53$ sq unit