Question

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❣️❣️prove that the sum of the square of the sides of a rhombus is equal to the sum of the square of its diagonal.❣️❣️

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Answer 1

▶ Answer :-

▶ Step-by-step explanation :-

Given :- A rhombus ABCD whose diagonals AC and BD intersect at O.

To Prove :- ( AB² + BC² + CD² + DA² ) = ( AC² + BD² ).

Proof :-

➡ We know that the diagonals of a rhombus bisect each other at right angles.

==>∠AOB = ∠BOC = ∠COD = ∠DOA = 90°,

\bf { OA = \frac{1}{2} AC \: and \: OB = \frac{1}{2} BD . }OA=

2

1

ACandOB=

2

1

BD.

From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]

\bf { \implies {AB}^{2} = (\frac{1}{2} AC}^{2}) + ( { \frac{1}{2}BD) }^{2} .⟹AB

2

=(

2

1

AC

2

)+(

2

1

BD)

2

.

\bf { \implies {AB}^{2} = \frac{1}{4} ( {AC}^{2} + {BD}^{2} ) }⟹AB

2

=

4

1

(AC

2

+BD

2

)

==> 4AB² = ( AC² + BD² ) .

==> AB² + AB² + AB² + AB² = ( AC² + BD² ) .

•°• AB² + BC² + CD² + DA² = ( AC² + BD² ) .

[ In a rhombus , all sides are equal ] .

Hence, it is proved. mark as brain list plz

Answer 2

$\huge\bf\pink{BONJOUR\: JIJU}$

$\huge\boxed{ANSWER}$

$<b> <i>$

Let the diagonals AC and BD of rhombus ABCD intersect each other at O. since the diagonals of a rhombus bisect each other at right Angles.

Angle AOB= Angle BOC= Angle COD = Angle DOA= 90°.

and, AO= CO, BO= OD.

Since, AOB is a Right ∆ ,right angled at O.

$AB^2= OA^2 + OB^2$

=>$AB^2= (1/2AC)^2+ (1/2BD)^2$ [ •.• OA=CO and OB= OD.

=> $4AB^2 = AC^2+ BD^2.$........①

Similarly,.We Have

$4BC^2= AC^2+ BD^2$........②

$4CD^2= AC^2+ BD^2$..........③

$4AD^2= AC^2+ BD^2$..........④

Adding all these results, we get

$4(AB^2+BC^2+ CD^2+ AD^2) = 4 (AC^2+ BD^2)$

=> $AB^2 + BC^2 + CD^2+ DA^2 =AC^2+ BD^2$

$\huge\bf\orange{THANKS}$