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

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
                      \small = \frac{1}{2} \times EO \times HF + \frac{1}{2} \times HF \times OG

                      \small =\frac{1}{2} \times (\frac{d2}{2}) \times d1+\frac{1}{2} \times (\frac{d2}{2}) \times d1

                      \small =\frac{(d1\times d2)}{4}+\frac{(d1\times d2)}{4}
                      \small =\frac{2(d1 \times d2))}{4}

                      \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.

 \small \boldsymbol{A}\left ( PQRS \right ) = \frac{1}{2}\times d1 \times d2

\small = \frac{1}{2}\times 24 \times 18

\small = 216

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

Cite this Simulator: