Question

Diagonal of a parallelogram intersect each other at point O.if AO=5 ,BO=12and AB=13then show that quadrilateral ABCD is a rhombus
​

Answer 1

Knowledge required

We should know that

●if In a triangle the square of the largest side is equal to the sum of the squares of other two sides then the angle opposite to the largest side will be 90 degree. this theorem is known as converse of Pythagorean theorem.

● A parralelogram whose diagonals intersect each other perpendicularly is known as rhombus.

REFER TO THE ATTACHMENT FOR SOLUTION

Answer 2

QUESTION :-

Diagonal of a parallelogram intersect each other at point O.if AO=5 ,BO=12and AB=13then show that quadrilateral ABCD is a rhombus

SOLUTION :-

=> Given : ABCD is a parallelogram.

=> to prove : ABCD is a rhombus.

=> proof : AO²+ BO² = 5²+12² = 25+144

=> .°. AO²+BO² = 169 --------(1)

=> AB² = 13² = 169 ------------(2)

=> In, ΔAOB , AB² = AO² +BO²

[From (1) & (2) ]

=> .°. ∠AOB = 90° -----[Converse of Pythagoras theorem.]

=> Diagonal of ABCD are perpendicular to each other at point O.

=> also, ABCD is a parallelogram. ---(given)

=> but, we know that a parallelogram in which diagonal are perpendicular is a rhombus.

=> .°. ABCD IS A RHOMBUS.