Question

what is the formula to find the bigger and smaller diagonal of rhombus from its side​

Answer 1

Answer:

Length of bigger diagonal = √{ 2ax }

Length of smaller diagonal = √{ 4a^2 - 2ax }

Step-by-step explanation:

From the given figure :

In ∆BCD' : Using Pythagoras Theorem.

= > BC^2 + D'C^2 = D'B^2

= > y^2 + ( x - a )^2 = a^2

= > y^2 + x^2 + a^2 - 2ax = a^2

= > x^2 + y^2 = 2ax

= > x^2 + y^2 = ( √2ax )^2 ... ( 1 )

In ∆BCD : Using Pythagoras Theorem.

= > BD^2 = BC^2 + DC^2

= > BD^2 = y^2 + x^2

= > BD^2 = ( √2ax )^2 { from ( 1 ) }

= > BD = √( 2ax )

= > Length of bigger diagonal of rhombus is √( 2ax ).

In ∆A'B'D' : Using Pythagoras Theorem.

= > ( A'D' )^2 = ( A'B' )^2 + ( B'D' )^2

= > ( A'D' )^2 = { a - ( x - a ) }^2 + ( y )^2

= > ( Length of smaller diagonal )^2 = ( a - x + a )^2 + y^2

= > ( Length of smaller diagonal )^2 = ( 2a - x )^2 + y^2

= > ( Length of smaller diagonal )^2 = 4a^2 + x^2 - 4ax + y^2

= > ( Length of smaller diagonal )^2 = 4a^2 + x^2 + y^2 - 4ax

= > ( Length of smaller diagonal )^2 = 4a^2 + 2ax - 4ax { from ( 1 ), x^2 + y^2 = 2ax }

= > ( Length of smaller diagonal )^2 = 4a^2 - 2ax

= > Length of smaller diagonal = √( 4a^2 - 2ax )

Hence the required formula to find the length of bigger diagonal is √{ 2ax } and to find the length of smaller side it is √{ 4a^2 - 2ax }, where a is the side of the rhombus and x is the perpendicular( attached with the side ) having the bigger diagonal as hypotenuse.

Answer 2

Step-by-step explanation:

this is for u may be its right