Question

Which of the following statement(s) is correct regarding the given figure?

• A. MNOP is a rectangle
• B. $\text{Length of NO is}\sqrt{8}\text{unit}$
• C. All the above
• D. $\text{Length of MN is}\sqrt{13}\text{unit}$

Answer 1

The correct option is C All the above

Mark the outer points in the figure as A, B, C and D.

Plotting the coordinates of the points, we get:


Coordinate of point D = (3,6)
Coordinate of point M = (5,6)
Coordinate of point P = (3,4)

So, DM = 2 units and DP = 2 units

Consider △DMP by joining MP and applying the Pythagorean Theorem:

$M{P}^2=D{P}^2+D{M}^2$
$\Rightarrow M{P}^2=(2{)}^2+(2{)}^2$
$\Rightarrow M{P}^2=4+4=8$
$\Rightarrow MP=\sqrt{8}\text{unit}$

Again,
Coordinate of point C = (8,6)
Coordinate of point N = (8,4)

So, CM = 3 units and CN = 2 units

Consider △MCN by joining MN and applying the Pythagorean Theorem:

$M{N}^2=M{C}^2+C{N}^2$
$\Rightarrow M{N}^2=3^2+2^2$
$\Rightarrow M{N}^2=9+4$
$\Rightarrow MN=\sqrt{13}\text{unit}$

Again,
Coordinate of point B = (8,2)
Coordinate of point O = (6,2)

So, BO = 2 units and BN = 2 units

Consider △BON by joining ON and applying the Pythagorean Theorem:
$N{O}^2=N{B}^2+B{O}^2$
$\Rightarrow N{O}^2=2^2+2^2$
$\Rightarrow N{O}^2=4+4$
$\Rightarrow NO=\sqrt{8}\text{unit}$

Again,
Coordinate of point A = (3,2)
Coordinate of point P = (3,4)

So, AO = 3 units and PA = 2 units

Consider △PAO by joining PO and applying the Pythagorean Theorem:
$P{O}^2=P{A}^2+A{O}^2$
$\Rightarrow P{O}^2=2^2+3^2$
$\Rightarrow P{O}^2=4+9$
$\Rightarrow PO=\sqrt{13}\text{unit}$

In quadrilateral MNOP,
NO = MP = √8 unit.
MN = PO = √13 unit.

So, length of opposite sides are equal.

Therefore, quadrilateral MNOP is a rectangle.

So, all the options are correct.