Math, asked by prinjaldubey49, 1 year ago

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

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Answered by ckroyofficial
6

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


ckroyofficial: thik h
prinjaldubey49: ok thanks yarr
prinjaldubey49: for the help
Answered by althaf97531
0

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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