Area of Polygons - Formulas

Area of Polygons

A parallelogram can be rearranged into a rectangle with the same area.

Animation for the area formula $K=bh$.

All of the area formulas for general convex quadrilaterals apply to parallelograms. Further formulas are specific to parallelograms:

A parallelogram with base b and height h can be divided into a trapezoid and a right triangle, and rearranged into a rectangle, as shown in the figure to the left. This means that the area of a parallelogram is the same as that of a rectangle with the same base and height:

$K=bh.$

The area of the parallelogram is the area of the blue region, which is the interior of the parallelogram

The base × height area formula can also be derived using the figure to the right. The area K of the parallelogram to the right (the blue area) is the total area of the rectangle less the area of the two orange triangles. The area of the rectangle is

$K_{\text{rect}}=(B+A)\times H\,$

and the area of a single orange triangle is

$K_{\text{tri}}={\frac {A}{2}}\times H.\,$

Therefore, the area of the parallelogram is

$K=K_{\text{rect}}-2\times K_{\text{tri}}=((B+A)\times H)-(A\times H)=B\times H.$

Another area formula, for two sides B and C and angle θ, is

$K=B\cdot C\cdot \sin \theta .\,$

The area of a parallelogram with sides B and C (B ≠ C) and angle $\gamma$ at the intersection of the diagonals is given by

$K={\frac {|\tan \gamma |}{2}}\cdot \left|B^{2}-C^{2}\right|.$

When the parallelogram is specified from the lengths B and C of two adjacent sides together with the length D₁ of either diagonal, then the area can be found from Heron's formula. Specifically it is

$K=2{\sqrt {S(S-B)(S-C)(S-D_{1})}}$

where $S=(B+C+D_{1})/2$ and the leading factor 2 comes from the fact that the chosen diagonal divides the parallelogram into two congruent triangles.

Area of Trapezoid

The area K of a trapezoid is given by

$K={\frac {a+b}{2}}\cdot h=mh$

where a and b are the lengths of the parallel sides, h is the height (the perpendicular distance between these sides), and m is the arithmetic mean of the lengths of the two parallel sides. In 499 AD Aryabhata, a great mathematician-astronomer from the classical age of Indian mathematics and Indian astronomy, used this method in the Aryabhatiya (section 2.8). This yields as a special case the well-known formula for the area of a triangle, by considering a triangle as a degenerate trapezoid in which one of the parallel sides has shrunk to a point.

The 7th-century Indian mathematician Bhāskara I derived the following formula for the area of a trapezoid with consecutive sides a, c, b, d:

$K={\frac {1}{2}}(a+b){\sqrt {c^{2}-{\frac {1}{4}}\left((b-a)+{\frac {c^{2}-d^{2}}{b-a}}\right)^{2}}}$

where a and b are parallel and b > a. This formula can be factored into a more symmetric version

$K={\frac {a+b}{4|b-a|}}{\sqrt {(-a+b+c+d)(a-b+c+d)(a-b+c-d)(a-b-c+d)}}.$

When one of the parallel sides has shrunk to a point (say a = 0), this formula reduces to Heron's formula for the area of a triangle.

Another equivalent formula for the area, which more closely resembles Heron's formula, is

$K={\frac {a+b}{|b-a|}}{\sqrt {(s-b)(s-a)(s-b-c)(s-b-d)}},$

where $s={\tfrac {1}{2}}(a+b+c+d)$ is the semiperimeter of the trapezoid. (This formula is similar to Brahmagupta's formula, but it differs from it, in that a trapezoid might not be cyclic (inscribed in a circle). The formula is also a special case of Bretschneider's formula for a general quadrilateral).

From Bretschneider's formula, it follows that

$K={\sqrt {{\frac {(ab^{2}-a^{2}b-ad^{2}+bc^{2})(ab^{2}-a^{2}b-ac^{2}+bd^{2})}{(2(b-a))^{2}}}-\left({\frac {b^{2}+d^{2}-a^{2}-c^{2}}{4}}\right)^{2}}}.$

The line that joins the midpoints of the parallel sides, bisects the area.

Licensing

Content obtained and/or adapted from:

• Parallelogram, Wikipedia under a CC BY-SA license
• Triangle, Wikipedia under a CC BY-SA license
• Trapezoid, Wikipedia under a CC BY-SA license