# 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:**

As we know, in a rhombus, diagonals bisect each other perpendicularly.

In rhombus ABCD

AC ⊥ BD and OA = OC;

And OB = OD

In ΔAOB, ∠AOB = 90°

⇒ AB^{2} = OA^{2} + OB^{2}..... (1)

Similarly, we can prove

BC^{2} = OB^{2} + OC^{2}.................................... (2)

CD^{2} = OC^{2} + OD^{2}.................................. (3)

AD^{2} = OD^{2} + OA^{2}.................................. (4)

Adding (1), (2), (3) and (4)

AB^{2} + BC^{2} + CD^{2} + AD^{2} = OA^{2} + OB^{2} + OB^{2} + OC^{2} + OC^{2} + OD^{2} + OD^{2} + OA^{2}

AB^{2} + BC^{2} + CD^{2} + AD^{2} = 2OA^{2} + 2OB^{2} + 2OC^{2} + 2OD^{2}

AB^{2} + BC^{2} + CD^{2} + AD^{2} = 2[OA^{2} + OB^{2} + OC^{2} + OD^{2}]

AB^{2} + BC^{2} + CD^{2} + AD^{2} = 2 [(AC/2)^{2} + (BD/2)^{2 } + (AC/2)^{2} + (BD/2)^{2}] [Since, OA = AC = AC/2 and OB = OD = BD/2]

BC + CD + AD = 2 [(AC^{2} + BD^{2} + AC^{2} + BD^{2})/4]

AB + BC + CD + AD = 2[(2 AC^{2} + 2BD^{2})/4]

AB + BC + CD + AD = 4[(AC^{2} + BD^{2})/4]

AB^{2} + BC^{2} + CD^{2} + AD^{2} = AC^{2} + BD^{2}

**Video Solution:**

## Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.5 Question 7:

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals

Hence it is proved that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals