A quadrilateral with congruent sides and perpendicular diagonals bisect each other

https://hi-static.z-dn.net/files/d51/6725cef4693e911aec525b6d1eb00637.jpeg

Let the center be O,

Now,

p^2 =√h^2 - b^2 [ By using Pythagoras Theorem]

or, x= √10^2 - 6^2

=8cm

So, OB = OD = 8 cm

Therefore, BD =OB +OD =( 8 +8 )cm

=16 cm

And,

AC = AO + OC = (6 +6 )cm

=12cm

In BOC :

BC = √ BO^2+OC^2[ By using Pythagoras Theorem]

= √ 6^2 + 8^2

= 10cm

The given figure is rhombus so it's all sides are equal i.e.

AB= BC = CD = AD = 10 cm.

Hence ; It's solved !