Question

Prove that the sum of the squares of the diagonals of a rhombus is equal to the sum of the squares of all its sides

Answer 1

here is ur answer....

Answer 2

Answer:

In a rhombus all sides are equal and diagonals bisect and perpendicular

figure: draw a quadrilateral ABCD with diagonals AC and BD intersecting at O

In triangle DOC,

∠DOC = 90°

applying pythagoras theorem,

DC²=DO²+OC²

DC²=(DB/2)²+(AC/2)²

DC²=DB²/4 + AC²/4                              (OPENING THE BRACKET)

4DC²=DB²+AC²     ( ALL SIDES R EQUAL)

⇒AB²+BC²+CD²+AD²=DB²+AC²

HENCE PROVED!

Step-by-step explanation: