question_answer

Area of the rectangle whose vertices are (-2 5) (8, 5), (8,-2) and (-2,-2) is               

A)  45                         

B)         50

C)  55                         

D)         70  

Correct Answer: D

Solution :

 Let us draw a rectangle which has given vertices. Here, length of the rectangle will be the distance. By the distance formula distance between \[(-2,-2)\] and \[(8,-25)\]                 \[=\sqrt{{{(-2-(-2))}^{2}}+{{(8-(-2))}^{2}}}\]                 \[=\sqrt{{{(0)}^{2}}+{{(8+2)}^{2}}}\]                 \[=\sqrt{{{(10)}^{2}}}\]                 \[=10.\] So, length of the rectangle is 10 and the distance between the points \[(8,-2)\] and \[(8,5)\] will be the width of the rectangle. \[\Rightarrow \]Width of rectangle \[=\sqrt{{{(5-(2))}^{2}}+{{(8-8)}^{2}}}\]                                                 \[=\sqrt{{{(5+2)}^{2}}}\]                                                 \[=\sqrt{{{(7)}^{2}}}\]                                                 \[=7.\] Now, area of rectangle is given by the product of its length and width between points \[(-2,-2)\] and \[(8,-2)\] \[\therefore \] Area of rectangle with given vertices\[=Length\times Width=10\times 7=70\]