Question

prove that the sum of the square of the sides of a rhombus is equal to the sum of the squares of its diagonal.​

Answer 1

Given:

• ABCD is a rhombus.
• AC & BD are the diagonals.

To proof:

• Sum of square of all sides = Sum of the squares of it's diagonals.

→ AB^2 + BC^2 + CD^2 + AD^2 = AC^2 + BD^2

Solution:-

Since sides of rhombus are equal.

OA = OC = OB = OD

• In ∆AOB, ∠AOB = 90°

→ AB^2 = AO^2 + BO^2 --(1)

• In ∆BOC, ∠BOC = 90°

→ BC^2 = OB^2 + OC^2 --(2)

• In ∆COD, ∠COD = 90°

→ CD^2 = OC^2 + OD^2 --(3)

• In ∆ AOD, ∠AOD = 90°

→ AD^2 = OA^2 + OD ^2 --(4)

Adding (1), (2), (3), & (4).

→ AB^2 + BC^2 + CD^2 + AD^2 = AO^2 + BO^2 OB^2 + OC^2 + OD^2 + OA^2 + OD^2.

→ 2OA^2 + 2OB^2 + 2OC^2 + 2OD^2

→ 2OA ^2 + 2OB^2 + 2OA^2 + 2OB^2

→ 4OA^2 + 4OB^2

→ (2OA)^2 + (2OB)^2

→ AC^2 + BD^2

Hence,

• Sum of square of all sides = Sum of the squares of it's diagonals.

Answer 2

Answer:

In rhombus ABCD, AB = BC = CD = DA

We know that diagonals of a rhombus bisect each other perpendicularly.

That is AC ⊥ BD, ∠AOB=∠BOC=∠COD=∠AOD=90° and

Consider right angled triangle AOB

AB2 = OA2 + OB2   [By Pythagoras theorem]

⇒  4AB2 = AC2+ BD2

⇒  AB2 + AB2 + AB2 + AB2 = AC2+ BD2

∴ AB2 + BC2 + CD2 + DA2 = AC2+ BD2

Thus the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.