Comprehensive theory, key formulas, diagrams, and memory aids for Areas of Parallelograms and Triangles.
In previous classes, you learned how to calculate the areas of various geometric figures using standard formulae. In this chapter, we will look at area from a different perspective. We will study the relationship between the areas of figures that lie on the same base and between the same parallels. This provides a powerful geometric method for comparing areas without needing to know specific measurements.
The part of the plane enclosed by a simple closed figure is called a planar region corresponding to that figure. The magnitude or measure of this planar region is called its area. Area is always expressed as a positive real number (in some unit like $cm^2, m^2,$ etc.).
We often abbreviate $Area(A)$ as $ar(A)$.
Two figures are said to be on the same base and between the same parallels if they have a common base (side) and the vertices (or vertex) opposite to the common base of each figure lie on a line parallel to the base.
For example, if parallelogram $ABCD$ and parallelogram $ABEF$ share the base $AB$, and points $C, D, E, F$ all lie on a straight line parallel to $AB$, then they are on the same base and between the same parallels.
Theorem 1: Parallelograms on the same base and between the same parallels are equal in area.
Proof Concept: Let parallelograms $ABCD$ and $ABEF$ be on the same base $AB$ and between the same parallels $AB$ and $CF$. We can prove that $\triangle ADF \cong \triangle BCE$ (using ASA or AAS). Since congruent triangles have equal area, $ar(\triangle ADF) = ar(\triangle BCE)$. Now, $ar(ABCD) = ar(ABED) + ar(\triangle BCE)$ Substitute $ar(\triangle BCE)$ with $ar(\triangle ADF)$: $ar(ABCD) = ar(ABED) + ar(\triangle ADF)$ $ar(ABCD) = ar(ABEF)$. Thus, their areas are equal.
A direct consequence of this theorem is the formula for the area of a parallelogram. A parallelogram and a rectangle on the same base and between the same parallels are equal in area. Since the area of a rectangle is length $\times$ breadth, the area of the parallelogram is base $\times$ corresponding altitude (height). Area of Parallelogram = base $\times$ height
The relationship between parallelograms also extends to triangles.
Theorem 2: Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Proof Concept: Let $\triangle ABC$ and $\triangle ABD$ be on the same base $AB$ and between parallels $AB$ and $CD$. If we draw a line through $B$ parallel to $CA$ and a line through $A$ parallel to $CB$ to form parallelograms, we can use the property that a diagonal divides a parallelogram into two triangles of equal area. Since the parallelograms on the same base and between the same parallels are equal in area, their halves (the triangles) must also be equal in area.
A triangle can be viewed as exactly half of a parallelogram on the same base and between the same parallels. Therefore: Area of Triangle = $\frac{1}{2} \times$ base $\times$ height
Theorem 3: If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram. $ar(\triangle) = \frac{1}{2} ar(\text{Parallelogram})$
A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side.
Theorem: A median of a triangle divides it into two triangles of equal areas.
Proof: Let $AD$ be the median of $\triangle ABC$. Therefore, $BD = DC$. Draw altitude $AL \perp BC$. $ar(\triangle ABD) = \frac{1}{2} \times BD \times AL$ $ar(\triangle ADC) = \frac{1}{2} \times DC \times AL$ Since $BD = DC$, the areas are equal.
This is a very useful property. It means that while the two triangles created by a median might not be congruent (they might look completely different), they contain the exact same amount of planar space.
The geometric theorems in this chapter allow us to manipulate and compare the areas of shapes based solely on their structural relationships (sharing a base and lying between parallel lines), rather than relying entirely on numeric calculations. Understanding that a median bisects the area of a triangle is a critical tool for solving complex area problems.