Math, asked by dianabisht, 8 months ago



Prove that the sum of the squares of the diagonals of a parallelogram is equal to its sides ​

Answers

Answered by rohitkumar200688
1

Answer:

Draw perpendiculars from C and D on AB as shown. Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answered by ppadmadevi12
0

Answer:

ANSWER

In parallelogram ABCD,

AB=CD and BC=AD

Draw perpendiculars from C and D on AB as shown.

In right angled △AEC,

⇒ (AC)

2

=(AE)

2

+(CE)

2

[ By Pythagoras theorem ]

⇒ (AC)

2

=(AB+BE)

2

+(CE)

2

⇒ (AC)

2

=(AB)

2

+(BE)

2

+2×AB×BE+(CE)

2

----- ( 1 )

From the figure CD=EF [ Since CDFE is a rectangle ]

But CD=AB

⇒ AB=CD=EF

Also CE=DF [ Distance between two parallel lines

⇒ △AFD≅△BEC [ RHS congruence rule ]

⇒ AF=BE [ CPCT ]

Considering right angled △DFB

⇒ (BD)

2

=(BF)

2

+(DF)

2

[ By Pythagoras theorem ]

⇒ (BD)

2

=(EF−BE)

2

+(CE)

2

[ Since DF=CE ]

⇒ (BD)

2

=(AB−BE)

2

+(CE)

2

[ Since EF=AB ]

⇒ (BD)

2

=(AB)

2

+(BE)

2

−2×AB×BE+(CE)

2

----- ( 2 )

Adding ( 1 ) and ( 2 ), we get

(AC)

2

+(BD)

2

=(AB)

2

+(BE)

2

+2×AB×BE+(CE)

2

+(AB)

2

+(BE)

2

−2×AB×BE+(CE)

2

⇒ (AC)

2

+(BD)

2

=2(AB)

2

+2(BE)

2

+2(CE)

2

⇒ (AC)

2

+(BD)

2

=2(AB)

2

+2[(BE)

2

+(CE)

2

] ---- ( 3 )

In right angled △BEC,

⇒ (BC)

2

=(BE)

2

+(CE)

2

[ By Pythagoras theorem ]

Hence equation ( 3 ) becomes,

⇒ (AC)

2

+(BD)

2

=2(AB)

2

+2(BC)

2

⇒ (AC)

2

+(BD)

2

=(AB)

2

+(AB)

2

+(BC)

2

+(BC)

2

∴ (AC)

2

+(BD)

2

=(AB)

2

+(BC)

2

+(CD)

2

+(AD)

2

∴ The sum of the squares of the diagonals of a parallelogram is equal to the sum of squares of its sides.

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