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Maths-

General

Easy

Question

The Diagonals of a square ABCD meet at O. Prove that $A B squared equals 2 A O squared$

hint Hint:

Hint :- using pythagoras theorem Find length of AC ,Find length of AO as the diagonals of square bisect each other.

The correct answer is: Hence proved

Aim :- Prove that $AB squared equals 2 AO squared$

Let length of side of square be a
Then ,applying using pythagoras theorem in triangle ADC
We get $A C squared equals A D squared plus D C squared not stretchy rightwards double arrow A C squared equals a squared plus a squared$
$not stretchy rightwards double arrow AC squared equals 2 straight a squared$
As diagonals bisect each other AC = 2OA
$20 straight A squared equals straight a squared$ where a =AB
We get $AB squared equals 2 AO squared$
Hence proved

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