Question

show that in rhombus sum of the square of its diagonal is equals to sum of the square of its side​

Answer 1

Answer:

Step-by-step explanation:

Let ABCD is a rhombus  in which AC and BD are the diagonals which intersect at O.

To prove: $AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}$

We know, diagonals of a rhombus bisect at right angles.

∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

Therefore, from triangle AOB,

From right ∆AOB , we have

By Pythagoras' theorem, we know that

AB² = OA² + OB²

OA = $\frac{1}{2} AC$

OB = $\frac{1}{2} BD$

$AB^{2} = (\frac{1}{2}AC)^{2} + (\frac{1}{2}BD)^{2}$

= $\frac{1}{4} AC^{2} + \frac{1}{4} BD^{2}$

$4AB^{2} = AC^{2} + BD^{2}$

AB² + AB² + AB² + AB² = ( AC² + BD² )

We know that In a rhombus , all sides are equal

AB² + BC² + CD² + DA² = ( AC² + BD² )

Hence its Proved