Question

A rhombus, prove that 4 times the square of any side is equal to the sum of squares of its diagonals

Answer 1

Let ABCD be a rhombus.
since,
all sides of rhombus are equal.
therefore,
AB=BC=CD=AD=x (say).--------------(1)

Also since,
diagonals of rhombus are bisect each other and are perpendicular.
therefore,
AO=CO=AC/2 and BO=DO=BD/2--------------(2)
Also, angle COD = 90 Degrees.
[After this pls see the second image]

Hope it will help you.........

Answer 2

Proved that in rhombus 4 times the square of any side =  sum of squares of its diagonals

Step-by-step explanation:

Property of rhombus

Diagonal of rhombus bisects each other perpendicularly

and all sides of rhombus are equal

Let say Say side of Rhombus = a

and diagonal are d₁  & d₂

d₁  & d₂ bisect perpendicularly at O

Applying pythagorus theorem

(d₁/2)² + (d₂/2)²   = a²

=> d₁²/4  + d₂²/4   = a²

Multiplying by 4 both sides

=>  d₁²  + d₂²   = 4a²

4 times the square of any side =  sum of squares of its diagonals

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