ABCDis a rhombus in which AC=16cm , BC=10cm . Find the length of diagonal BD.

We know that the diagonals of a rhombus bisect each other at right angles

Therefore, CO = ¹/₂AC = (¹/₂ × 16) cm = 8 cm, BC = 10 cm and ∠COB = 90°.

From right ∆OBC, we have

BC² = CO² + BO²

⇒ BO² = (BC² – CO²) = {(10) ² - (8)²} cm²

                            = (100 - 64) cm²

                            = 36 cm²

     ⇒ BO = √36 cm = 6 cm.

Therefore, BD = 2 × BO = (2 × 6) cm = 12 cm.

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as diagonal of a rhombus bisect at 90^o 

so, name the point of intersetion as o

now co =ao & do = bo; so oc = 1/2AC = 8

they bisect at 90^o using pythagoras theorem 

OB² + OC² =  BC²

8² + OB² = 10²

OB = (100 - 64)^1/2

OB = (36)

OB = 6