(Q10) Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Solution :

OABCD

Given :□ABCD is a rhombus.

Let its diagonals AC and BD bisect each other at 'O'.We know that “the diagonals in a rhombus are perpendicular to each other”.

In ∆AOD; AD2 = ( )2 + OD2 ..... (1)

[Pythagoras theorem]

In ∆COD; CD2 = ( )2 + OD2 ..... (2)

[Pythagoras theorem]

In ∆AOB; AB2 = ( )2 + OB2 ..... (3)

[Pythagoras theorem]

In ∆BOC; BC2 = ( )2 + OC2 ..... (4)

[Pythagoras theorem]

Adding the above equations we get AD2 + CD2 + AB2 + BC2 = 2 (( )2 + ( )2 + ( )2 + ( )2)

= 2[(
1
2
 AC)2 + (
1
2
 BD)2 + (
1
2
 AC)2 + (
1
2
 BD)2 ]
[∵ AO = OC =
1
2
 AC
BO = OD =
1
2
 BD ]
= 2[
2
4
 +
2
4
 +
2
4
 +
2
4
 ] `
= 2[
2 + 2
4
 +
2 + 2
4
 ]
= 2[
2( )2 + 2( )2
4
 ]
=
4( )2
4
 +
4( )2
4

= ( )2 + ( )2