Spliting 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 A kite has two pairs of adjacent sides with equal lengths, like two isosceles triangles with their bases joined. 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\( = 30 + 30 + 54 + 54 = 168c{m^2}\)
