In the given figure, ABCD is a rhombus. Then,

Options

A. `AC^(2)+BD^(2)=AB^(2)`
B. `AC^(2)+BD^(2)=2AB^(2)`
C. `AC^(2)+BD^(2)=4AB^(2)`
D. `2(AC^(2)+BD^(2))=3AB^(2)`

Answer

C

Explanation

We known that the diagonals of a rhombus bisect each other at right angles.
`therefore OA=(1)/(2)AC, OB=(1)/(2)BD and angle AOB=90^(@)`
`therefore AB^(2)=OA^(2)+OB^(2)=(1)/(4)AC^(2)+(1)/(4)BD^(2)`
`rAArr 4AB^(2)=(AC^(2)+BD^(2)).`