Question

diagonals of a parallelogram intersect at point O .If AO =5, BO=12 and AB=13 then show that □ABCD is a rhombus.

Answer 1

Since 5, 12, 13 are Pythagorean triplets
So angle AOB = 90°
The diagonals bisect each other at right angles
So parallelogram ABCD is rhombus

Answer 2

Therefore ABCD is a RHOMBUS...the diagonal of a Rhombus bisect at perpendicular.

Step-by-step explanation:

Given:

[]ABCD is a Parallelogram

AO = 5

BO = 12

AB = 13

To Show:

[]ABCD is a Rhombus

Proof:

In the given Parallelogram diagonals AC and BD bisect at O we have

In  Δ AOB

$(AO)^{2}+(BO)^{2}=5^{2}+ 12^{2} =25+144\\\\(AO)^{2}+(BO)^{2}=169$...1

Also

$(AB)^{2}=13^{2}=169$....2

From 1 and 2

$(AO)^{2}+(BO)^{2}=(AB)^{2}$....Which is a Pythagoras Theorem

Therefore Δ AOB is a Right angle Triangle at ∠AOB=90°

i.e AO ⊥ BD

i.e Diagonals are Perpendicular to each other

    Diagonals Bisect each other

Therefore ABCD is a RHOMBUS...the diagonal of a Rhombus bisect at perpendicular.