Question

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

Answer 1

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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.

Answer 2

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$<u>Answer :</u>$

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

ABCD is a rhombus in which diagonals AC and BD intersect at point O.

We need to prove AB2 + BC2 + CD2 + DA2 = AC2 + DB2

⇒ In Δ AOB; AB2 = AO2 + BO2

⇒ In Δ BOC; BC2 = CO2 + BO2

⇒ In Δ COD; CD2 = DO2 + CO2

⇒ In Δ AOD; AD2 = DO2 + AO2

⇒ Adding the above 4 equations we get

⇒ AB2 + BC2 + CD2 + DA2 = AO2 + BO2

+ CO2 + BO2 + DO2 + CO2 + DO2 + AO2

⇒ = 2(AO2 + BO2 + CO2 + DO2)

Since, AO2 = CO2 and BO2 = DO2

= 2(2 AO2 + 2 BO2).

= 4(AO2 + BO2) ……eq(1)

Now, let us take the sum of squares of diagonals

⇒ AC2 + DB2 = (AO + CO)2 + (DO+ BO)2

= (2AO)2 + (2DO)2

= 4 AO2 + 4 BO2 ……eq(2)

From eq(1) and eq(2) we get

⇒ AB2 + BC2 + CD2 + DA2 = AC2 + DB2

Hence, proved