prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonal.
Answers
Answer:
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°,
From right ∆AOB , we have
AB² = OA² + OB² [ by Pythagoras' theorem ]
==> 4AB² = ( AC² + BD² ) .
==> AB² + AB² + AB² + AB² = ( AC² + BD² ) .
•°• AB² + BC² + CD² + DA² = ( AC² + BD² ) .
[ In a rhombus , all sides are equal ] .
Step-by-step explanation:
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.