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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:

We know that, in a rhombus, diagonals bisect each other at 90 degrees.

In rhombus ABCD as shown below, AC and BD are the diagonals.

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

AC ⊥ BD and OA = OC;

And OB = OD

In ΔAOB, ∠AOB = 90°

Thus, by using Pythagoras theorem,

AB2 = OA2 + OB2............................ (1)

Similarly, we can prove

BC2 = OB2 + OC2.................................... (2)

CD2 = OC2 + OD2.................................. (3)

AD2 = OD2 + OA2.................................. (4)

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

AB2 + BC2 + CD2 + AD2 = OA2 + OB2 + OB2 + OC2 + OC2 + OD2 + OD2 + OA2

AB2 + BC2 + CD2 + AD2 = 2OA2 + 2OB2 + 2OC2 + 2OD2

AB2 + BC2 + CD2 + AD2 = 2[OA2 + OB2 + OC2 + OD2]

AB2 + BC2 + CD2 + AD2 = 2 [(AC/2)2 + (BD/2)2  + (AC/2)2 + (BD/2)2] [Since, OA = AC = AC/2 and OB = OD = BD/2]

AB2 + BC2 + CD2 + AD2 = 2 [(AC2 + BD2 + AC2 + BD2)/4]

AB2 + BC2 + CD2 + AD2 = 2[(2AC2 + 2BD2)/4]

AB2 + BC2 + CD2 + AD2 = 4[(AC2 + BD2)/4]

AB2 + BC2 + CD2 + AD2 = AC2 + BD2

☛ Check: NCERT Solutions for Class 10 Maths Chapter 6

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

Summary:

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.

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