question_answer

The area of a parallelogram, whose vertices are (-4, -2), (-2, 6), (10, 6) and (8, -2) is

A)  32                                         

B)  48

C)  72                         

D)         96  

Correct Answer: D

Solution :

 Draw a parallelogram which has the vertices \[(-4,-2),\,(-2,6),(10,6)\] and \[(8,-2)\].                                     As we can see from the figure above that the base of the given parallelogram is equal to the distance between the points \[(-4,-2)\] and \[(8,-2)\]. \[\therefore \]  Base of parallelogram                 \[=\sqrt{{{(-2-(-2))}^{2}}+{{(8-(-4))}^{2}}}\]                 \[=\sqrt{{{(0)}^{2}}+{{(8+4)}^{2}}}=\sqrt{{{(12)}^{2}}}=12\]and height h is the distance between the point   \[(-2,6)\] and \[(-2,-2)\]..                             \[\therefore \] Height of the parallelogram,  \[h=\sqrt{{{(-2-6)}^{2}}+{{(-2+2)}^{2}}}=8\] Now, Area of pa rallelogram = Base \[\times \]Height\[=12\times 8=96\]