prove 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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Given:
Rhombus ABCD
AC and BD are Diagonals intersecting at O.
Figure is in the attachment provided below.
To Prove that:
Sum of squares of the sides of a rhombus = Sum of squares of its diagonal.
Proof:
AB = BC = CD = AD (All sides of a rhombus are equal)
Diagonals of a rhombus bisect each other at 90 degree or right angles.
∴ ∠ AOB = 90 Degree
AO = 1 / 2 AC
BD = 1 / 2 BO
Consider ΔAOB,
∠ AOB = 90 Degree
Applying Pythagoras theorem in this triangle we get,
This means that,
(∵All sides are equal)
Hence Proved
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Answer:
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Step-by-step explanation:
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