Question

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of it's diagonals plzzz ans

Answer 1

Heya user!!!
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Here's the answer u r lukng for :-
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Let ABCD be the rhombus with diagonals AC and BD....

Here we have to keep two properties of rhombus in mind. They r

1. A rhombus has its all sides equal.

2. Diagonals of a rhombus bisect each other at right angles....

So,
Since we know that diagonals bisect each other at right angles.

Therefore,

We have :-

AB²=OB²+OA²

Also we know that they bisect...

Therefore we can write the above as :-

AB²=(1/2BD)²+(1/2AC)²

=>AB²=1/4BD²+1/4 AC² ........(i)

Now multiplying by 4 on either side in equation i we get :-

4AB²=(4)*1/4(AC²+BD²)

=>AB²+BC²+CD²+AD²=AC²+BD²

(Since all sides are equal, therefore 4AB²=AB²+BC²+CD²+AD²)

Thus proved....

Hope this helps!!!!!!