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Consider a rhombus ABCD, whose diagonals intersect at O.

We know that diagonals of a rhombus bisect each other at 90°, so triangles AOB, AOD, BOC and COD are right angled triangles.

In ΔAOB,

AO2 + OB2 = AB2 (Using Pythagoras theorem)

⇒ AC2 + BD2 = 4AB2

Similarly, AC2 + BD2 = 4BC2

AC2 + BD2 = 4CD2

AC2 + BD2 = 4AD2

Adding all these equations, we get

4(AB2 + BC2 + CD2 + AD2) = 4(AC2 + BD2)

⇒ AB2 + BC2 + CD2 + AD2 = AC2 + BD2

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10y ago
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AnswerBot

6mo ago

No, the sum of squares of the sides of a rhombus is not equal to the sum of squares of the diagonals. The sum of the squares of the sides of a rhombus is equal to four times the square of one side, while the sum of the squares of the diagonals is equal to two times the sum of the squares of the sides.

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Q: Is the sum of squares of the sides of a rhombus is equal to sum of square of the diagonals?
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