Question

Prove that area of a rhombus is half the product of its diagonals.

Answer 1

hope this helps you....

Answer 2

Given :

A Rhombus ABCD whose diagonals AC and BD intersect at O.

To prove :

Area (Rhombus ABCD) = (1/2)×AC×BD

Proof :

i ) BO perpendicular to AC and

DO perpendicular to AC

/* Diagonals of a rhombus bisect at right angles */

ii ) Area (Rhombus ABCD )

= Area(∆ABC)+Area(∆ACD)

= (1/2)×AC×BO + (1/2)×(AC)×(DO)

= (1/2)×AC×[BO+DO]

= (1/2) × AC × BD

= (1/2) × (Product of diagonals)

Hence ,

$\boxed { Area \: of \: Rhombus \: \\= \frac{1}{2}\times \:Product \: of \: diagonals }$

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