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Maths-

General

Easy

Question

Prove that the sum of squares of sides of a rhombus is equal to twice the sum of the squares of its diagonals.

The correct answer is: Hence proved

Solution :-
Aim :- Prove that $A B squared plus B C squared plus C D squared plus D A squared equals A C squared plus B D squared$
Hint :- Applying pythagoras theorem to both triangles AIB. find the AB² and multiply by 4 and substitute proper conditions to prove the condition .
Explanation(proof ) :-
Given ABCD is a rhombus
The diagonals in the rhombus are perpendicular bisectors.
$BI equals 1 divided by 2 BD minus Ea 1$
$AI equals 1 divided by 2 AC minus Eg 2$
All sides of the rhombus are equal
$AB equals BC equals CD equals DA minus Eg 3$

Applying pythagoras theorem in ΔAIB,
$A B squared equals A I squared plus B I squared$
Multiplying with 4 on both sides
$4 A B squared equals 4 A I squared plus 4 B I squared not stretchy rightwards double arrow A B squared plus A B squared plus A B squared plus A B squared equals left parenthesis 2 A I right parenthesis squared plus left parenthesis 2 B I right parenthesis squared$
Applying Eq 1 ,2 and 3
We get $A B squared plus B C squared plus C D squared plus D A squared equals A C squared plus B D squared$

Hence proved .

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