Area Axioms

Area Axiom: Every polygonal region has an area and is measured in square units.

Congruent Area Axiom: If $\triangle ABC$ and $\triangle DEF$ are two congruent triangles, then $ar(\triangle ABC) = ar(\triangle DEF)$. ie. two congruent regions have equal area.

Rectangle Area Axiom: If $ABCD$ is a rectangular  region such that $AB = a$ units and $BC = b$ units, then $ar(ABCD) = ab$ square units.

Parallelograms on the same base and between the same parallels Theorem 1: A diagonal of a parallelogram divides it into two triangles of same area

In this case $ar (\triangle ABC) = ar (\triangle ADC)$

Also $ar (\triangle ABD) = ar (\triangle BCD)$

Theorem 2: Parallelograms on the same base and between the same parallels are equal  in area.  $ar (ABCD) = AB \times h$ $ar (ABFE) = AB \times h$ $\therefore ar (ABCD) = ar (ABFE)$

Theorem 3: Area of a parallelogram is the product of its base and the corresponding altitude. Let the two adjacent sides of the parallelogram be $a$ and $b$.  Area $= Base \times Height$

Triangle Area Axiom

Theorem 4: The area of a triangle is half the product of any  of its sides and the corresponding altitude.  $a, b, c$ denotes the sides of the Triangle. Then: $\displaystyle \text{Area } = \frac{1}{2} \times Base \times Height = \frac{1}{2} bh$

Area $= \sqrt{s(s-a)(s-b)(s-c)}$. This is known as Heron’s Theorem.

Theorem 5: If a triangle and parallelogram are on the same base and between the same parallels, the area of the triangle is equal to the half of the parallelogram.  $ar (ABCD) = AB \times h$ $\displaystyle ar ( \triangle ABE) = \frac{1}{2} \times AB \times h$ $ar ( \triangle ABE) = ar (ABCD)$

Trapezium Area Axiom

Theorem 6: The area of a trapezium is half the product of its heights and the sum of parallel sides. A trapezium is a quadrilateral two of whose sides are parallel. A trapezium whose non-parallel sides are equal is known as an isosceles trapezium.

Let $a$ and $b$ be the parallel sides and $h$ be the distance between the parallel sides. $\displaystyle \text{Then Area } = \frac{1}{2} (a+b) \times h$

Triangles on the same base and between the same parallels

Theorem 7: Triangles on the same base and the same parallels have the same area.  $\displaystyle ar ( \triangle ABD) = \frac{1}{2} \times AB \times h$ $\displaystyle ar ( \triangle ABC) = \frac{1}{2} \times AB \times h$ $\displaystyle \therefore ar ( \triangle ABD) = ar ( \triangle ABC)$

Theorem 8: Triangles having equal areas and having one side of one of the triangles equal to one side of the other triangle, have their corresponding altitudes the same.

Theorem 9: Two triangles having the same bases (or equal bases) and equal area lie between the same parallels.