Question

prove that the sum of squares of the sides of rhombus is equal to the sum of the sum of squares of its diagonal

Answer 1

Answer:

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.

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Answer 2

Step-by-step explanation:

Poof: ABCD is rhombus and let diagonals AC and BD bisect each other at O.

∴ ∠AOB =∠BOC =∠COD =∠DOA = 90°

In ∆AOB

AB2 = AC2 + BO2

AB2 = (1/2AC2) + (1/2BD2)

AB2 = 1/4(AC2 + BD2)

4AB2 = AC2 + BD2 -(1)

Similarly, in ∆BOC

4BC2 = AC2 + BD2 -(2)

4CD2 = AC2 + BD2 -(3)

4AD2 = AC2 + BD2 -(4)

Adding 1, 2, 3, and 4

4AB2 + 4BC2 + 4CD2 + 4DA2 = 4(AC2 + BD2)

4(AB2 + BC2 + CD2 + DA2) = 4(AC2 + BD2)

∴ AB2 + BC2 + CD2 + DA2 = AC2 + BD2