Area of quadrilaterals 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 kite 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 \(= 54c{m^2}\)

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