Question

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

Answer 1

Step-by-step explanation:

If ABCD is a Rhombus,than it's diagonals meet at 90°.

Or we can say that ∆AOB is a right triangle ,right angle at O.

So,according to the converse of Pythagoras Theorem,if

${AB}^{2} = {AO}^{2} + {BO}^{2}$

here AB= 13 cm

AO=5 cm

BO = 12 cm

${(13)}^{2} = {(5)}^{2} + {(12)}^{2} \\ \\ 169 = 25 + 144 \\ \\ 169 = 169$

Hence according to the converse of Pythagoras theorem AOB is a right triangle,right angle at O.

Thus ABCD is a rhombus.

Hope it helps you.