Question

prove that the sum of squares of a rhombus is equal to the sum of the square of diagonal
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Answer 1

Answer:

Given ;

Rhombus ABCD

With diagonal AC and BD intersecting at D .

To prove ;

sum of square of all side = sum of square of it's diagonals .

= AC^2 + BC^2 + CD^2 + AD^2 = AC^2 +BD^2

Proof ;

since sides of a rhombus are equal

AB = BC =CD = AD

we know that,

diagonals of a rhombus bisect each other at right angle .

Therefore ,

ANGLE [ AOB = BOC = COD =DOA ] = 90°

Also , AO =CO= 1/ 2 AC . ........( 1 )

& BO = DO = 1/2 BD . .......... (2)

Now AOB is a right triangle .

Using Pythagoras theorem ,,

( Hypotenuse ) ^2 = (Height)^2 + (Base)^2

(AB)^2 = (OA)^2 + ( OB ) ^2 .

( AB ) ^2 = ( 1/2 AC^2 ) + (1/2 BD^2 ) . ....( 1 ) &( 2 )

(AB)^2 = AC^2 /4 + BD^2 /4 .

(AB)^2 = AC^2 + BD^2 / 4 .

(AB)^2 + (AB)^2 + (AB)^2 + (AB)^2 = AC^2 + BD^2.

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

Hence proved ...