Question

If the coordinates of quadrilateral ABCD are (–3, 5), (3, 5), (3, –5) and (–3, –5), then the area of the quadrilateral ABCD is equal to

• A.
  50 sq. units
• B.
  60 sq. units
• C.
  100 sq. units
• D.
  40 sq. units

Answer 1

The correct option is B

60 sq. units
Let the points (–3, 5), (3, 5), (3, –5) and (–3, –5) be denoted by A, B, C and D respectively. By distance formula, we have



$AB=\sqrt{(3+3{)}^2+(5-5{)}^2}=6unitsBC=\sqrt{(3-3{)}^2+(-5-5{)}^2}=10unitsCD=\sqrt{(-3-3{)}^2+(-5+5{)}^2}=6unitsAD=\sqrt{(-3+3{)}^2+(-5-5{)}^2=10units}$

$\therefore$ AB = CD = 6 units and BC = AD = 10 units

Thus, ABCD is a rectangle.

$\therefore$ Area of rectangle $ABCD=AB\times BC=6\times 10=60sq.units$

Hence, the correct answer is option (b).