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.

Open in App

Solution

In $△$ AOB,
AO = 5 cm
OB = 12 cm
AB = 13 cm
5, 12 and 13 form a Pythagorean triplet.
Thus, $△$ AOB is a right angle triangle, right angled at O.
ABCD is a parallelogram.
So, diagonals bisect each other.
$\Rightarrow$ AO = OC = 5 cm
Also, OB = OD = 12 cm.
Thus, in parallelogram ABCD, diagonals bisect at right angles.
Hence, ABCD is a rhombus.

Suggest Corrections

Similar questions

A B C D

A B | | D C

^{'} O^{'}

\frac{A O}{B O} = \frac{C O}{D O}

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that

A B C D

O

\frac{A O}{B O} = \frac{C O}{D O}

A B C D

Join BYJU'S Learning Program