Question

prove that the sum of squares of the sides of rhombus is equal to the square of the it's
diagonals​

Answer 1

Step-by-step explanation:

given : In rhombus ABCD AB=BC=CD=DA

AC and BD are diagonals are meet at point o

R.T.P. : AB2 +BC2+ CD2+DA2 = AC2 + BD2

Proof :from triangle i.e AOB angle AOB = 90

by phythagoras theorem

AB2 = OA2 +OB2

AB2 =[AC/2]2 + [BD/2]2

AB2 =AC2/4 + BD2/ 4

AB2 = AC2 + BD2/4

4AB2 = AC2 + BD2

AB2 + AB2+AB2+ AB2=AC2 + BD2

AB2 +BC2+CD2+ DA2 = AC2 +BD2

hence proved