## Area of parallelogram

The following quadrilaterals can be split into rectangles and triangles in order to calculate their area.

### Method 1

$Area\,of\,triangle\,1 = \frac{1}{2}bh$

$= \frac{1}{2} \times 4 \times 5$

$= \frac{1}{2} \times 20$

$= 10c{m^2}$

Area of rectangle (2) $$= l \times b$$

$= 6 \times 5$

$= 30c{m^2}$

Area of triangle (3) = same as area of triangle (1)

$= 10c{m^2}$

$Total\,area = 10 + 30 + 10 = 50c{m^2}$

### Method 2

Split the parallelogram into two congruent triangles along one of the diagonals.

$Area of triangle = \frac {1}{2}bh$

$= \frac {1}{2} \times 10 \times 5$

$= 25cm^{2}$

$Area of parallelogram = 2 \times 25 = 50 cm^{2}$

### Method 3

Split the parallelogram into two congruent triangles along one of the diagonals.

$Area\,of\,triangle\ = \frac{1}{2}bh$

$= \frac{1}{2} \times 10 \times 5$

$= 25{cm^2}$

$Area\ of\ parallelogram\ = 2 \times 25 = 50 {cm^2}$

## Area of a kite

Since a has a vertical line of symmetry, then triangles 1 and 2 will have the same area. This will also be the same for triangles 3 and 4.

$Area\,of\,triangle\,1 = \frac{1}{2}bh$

$= \frac{1}{2} \times 6 \times 10$

$= \frac{1}{2} \times 60$

$= 30c{m^2}$

Area of triangle 2 $$= 30c{m^2}$$

$Area\,of\,triangle\,3 = \frac{1}{2}bh$

$= \frac{1}{2} \times 6 \times 18$

$= \frac{1}{2} \times 108$

$= 54c{m^2}$

Area of triangle 4 $$= 54c{m^2}$$

$Total\,area = 30 + 30 + 54 + 54 = 168c{m^2}$