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Area of rhombus

## Objective

To show that the area of rhombus is half the product of its diagonals.

### Theory

1. A rhombus is a simple (non-self-intersecting) quadrilateral whose all four sides are of same length.
2. If a parallelogram has two consecutive sides congruent, it is a rhombus.
3. If two triangles are congruent then their areas are equal.
4. Area of a triangle = 1/2 X base X height
5. Area of a rectangle = Length X Breadth
6. Diagonals of rhombus are perpendicular to each other.

## Proof

In above figure EHGF is rhombus with diagonal HF (length d1) and diagonal EG (length d2)

Area of rhombus EHGF = Area of triangle EFH + Area of triangle FHG
$\dpi{80} \fn_phv \small = \frac{1}{2} \times EO \times HF + \frac{1}{2} \times HF \times OG$

$\dpi{80} \fn_phv \small =\frac{1}{2} \times (\frac{d2}{2}) \times d1+\frac{1}{2} \times (\frac{d2}{2}) \times d1$

$\dpi{80} \fn_phv \small =\frac{(d1\times d2)}{4}+\frac{(d1\times d2)}{4}$
$\dpi{80} \fn_phv \small =\frac{2(d1 \times d2))}{4}$

$\dpi{80} \fn_phv \small =\frac{d1 \times d2}{2}$
= half of the product of the diagonals

### Example

Find the area of the following rhombus.

### Solution:

In the given figure,

PR = d1= 24 cm.

SQ = d2 = 18 cm.

So, the area of the rhombus PQRS is 216 cm2.