The parallelogram has the angle measures shown can you conclude that it is a rhombus

Solution:

Properties of Parallelogram

(a) Opposite sides are equal and parallel.

(b) Opposite angles are equal.

(3) Diagonals bisect each other.

A rhombus is a parallelogram apart from the above properties,  if it's all the sides are equal and diagonals bisect each other at right angles.

Name the vertices of Parallelogram as A B CD, Join AC and B D.

The point of intersection of diagonal AC and B D is point O.

Now, ∠A+∠B +∠C +∠D= 360°→→The sum of angles of any quadrilateral is 360°.

2 ∠A +144° +144°= 360°

2 ∠A= 360°-288°

2 ∠A= 72°

∠A=36°=∠C

As, AO bisects BD.

In ΔAOD and ΔAOB

BO=DO→→→Diagonals of a parallelogram bisect each other.

AO=AO [Common side]

∠ADO=∠ABO =72°

But with the given information ,we can't prove the two triangles ΔAOD and ΔAOB are congruent, so that's why we can't prove that, ∠AOD=∠AOB=90°, neither AD=AB.

So,we can't prove that the , given Parallelogram is Rhombus.

the correct answer is the second option