Prove that sum of the squares on the diagonals of a rhombus is equal to four times the square on the side of the rhombus?

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Prove that sum of the squares on the diagonals of a rhombus is equal to four times the square on the side of the rhombus?

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Answer 1 · 2016-07-26T16:54:33

Please see below.

Explanation:

In the absence of any figure I have drawn it as given below.
enter image source here
In a rhombus diagonals intersect each other at right angles. In the given rhombus #ABCD# diagonal bisect each other at #O# hence

#BC^2+AD^2=(2BO)^2+(2AO)^2=4BO^2+4AO^2#

= #4(BO^2+AO^2)=4AB^2#

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Prove that sum of the squares on the diagonals of a rhombus is equal to four times the square on the side of the rhombus?

1 Answer

Explanation:

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