Question

Let the four vertices of a quadrilateral be $A(-3,5),$ $B(3,7),$ $C(6,-2)$ and $D(0,-4)$. Which of the following statements is/are true?

• A. Length of one of the diagonals is $\sqrt{130}$ units
• B. Area of the quadrilateral is $60$ sq. units
• C. The quadrilateral is a rectangle.
• D. Both the diagonals of the quadrilateral are equal.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Length of one of the diagonals is $\sqrt{130}$ units
B Area of the quadrilateral is $60$ sq. units
C The quadrilateral is a rectangle.
D Both the diagonals of the quadrilateral are equal.

$A(-3,5),B(3,7),C(6,-2)$ and $D(0,-4)$
Now, finding the distances,
$AB=\sqrt{(-3-3{)}^2+(5-7{)}^2}=\sqrt{40}\text{units}BC=\sqrt{(3-6{)}^2+(7+2{)}^2}=\sqrt{90}\text{units}CD=\sqrt{(0-6{)}^2+(-4+2{)}^2}=\sqrt{40}\text{units}AD=\sqrt{(-3-0{)}^2+(5+4{)}^2}=\sqrt{90}\text{units}$

As $AB=CD$ and $BC=AD$
Finding the diagonal length,
$AC=\sqrt{(-3-6{)}^2+(5+2{)}^2}=\sqrt{130}\text{units}BD=\sqrt{(0-3{)}^2+(-4-7{)}^2}=\sqrt{130}\text{units}$
Diagonal length is also same.
Therefore it is a rectangle.
Area of rectangle $=\sqrt{90}\times \sqrt{40}=60\text{sq. units}$