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Maharashtra State BoardSSC (English Medium) 10th Standard

Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. - Geometry Mathematics 2

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Question

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

Sum
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Solution

Diagonals of a parallelogram bisect each other, i.e., O is the midpoint of AC and BD.

In ∆ABD, point O is the midpoint of side BD.

BO = OD = `1/2`BD

∴ by Apollonius theorem,

AB2 + AD2 = 2AO2 + 2BO2   ...(1)

In ∆CBD, point O is the midpoint of side BD.

BO = OD = `1/2`BD

∴ by Apollonius theorem,

CB2 + CD2 = 2CO2 + 2BO2   ...(2)

Adding (1) and (2), we get,

AB2 + AD2 + CB2 + CD2 = 2AO2 + 2BO2 + 2CO2 + 2BO2

⇒ AB2 + AD2 + CB2 + CD2 = 2AO2 + 4BO2 + 2AO2   ...(∵ OC = OA)

⇒ AB2 + AD2 + CB2 + CD2 = 4AO2 + 4BO2

⇒ AB2 + AD2 + CB2 + CD2 = (2AO)2 + (2BO)2

⇒ AB2 + AD2 + CB2 + CD2 = (AC)2 + (BD)2

⇒ AB2 + AD2 + CB2 + CD2 = AC2 + BD2

∴ the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

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Chapter 2: Pythagoras Theorem - Problem Set 2 [Page 45]

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Balbharati Mathematics 2 [English] Standard 10 Maharashtra State Board
Chapter 2 Pythagoras Theorem
Problem Set 2 | Q 9 | Page 45

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