CBSE Class 9 Maths Chapter 9 Notes: Areas Of Parallelograms And Triangles

Key Concepts: Must Remember

• Area of a Figure: The measure of the region enclosed by a closed plane figure.
• Congruent Figures & Area: Congruent figures have equal areas. However, figures with equal areas are not necessarily congruent.
• Figures on the Same Base and Between the Same Parallels: This condition is central to all theorems in this chapter. It means two figures share one side as a common base, and their vertices opposite to the common base lie on a line parallel to the base.
• Theorem 1: Parallelograms on the same base and between the same parallels have equal areas.
• Theorem 2: A parallelogram and a triangle on the same base and between the same parallels have the area of the triangle equal to half the area of the parallelogram.
• Theorem 3: Triangles on the same base (or equal bases) and between the same parallels have equal areas.
• Converse of Theorem 3: If two triangles have the same base (or equal bases) and equal areas, then they lie between the same parallels.
• Median of a Triangle: A median divides a triangle into two triangles of equal area.

Key Definitions

Area

The magnitude or extent of a two-dimensional surface or shape, typically measured in square units.

Congruent Figures

Geometric figures that have the same shape and size. If two figures are congruent, their areas are equal.

Parallelogram

A quadrilateral with two pairs of parallel sides. Its area is given by Base × Height.

Triangle

A polygon with three edges and three vertices. Its area is given by ½ × Base × Height.

Altitude (Height)

The perpendicular distance from a vertex to the opposite side (or its extension) in a polygon, representing the height for area calculations.

Base

Any side of a polygon chosen to be the side to which an altitude is drawn.

Between the same parallels

A condition where two geometric figures share a common base, and the vertices opposite to this base lie on a straight line parallel to the common base.

Median of a Triangle

A line segment joining a vertex to the midpoint of the opposite side. It divides the triangle into two triangles of equal area.

Understanding Area Theorems

This chapter is primarily built upon a few critical theorems that relate the areas of different geometric figures, particularly parallelograms and triangles, when they share a specific configuration. The most important configuration to understand is "on the same base and between the same parallels". This means two figures have one common side (the base), and the line joining the vertices opposite to this base is parallel to the base itself. The perpendicular distance between these parallel lines is the height common to both figures relative to that base.

Theorem 1: Parallelograms on the same base and between the same parallels are equal in area.

Consider two parallelograms ABCD and EBCF, sharing the same base BC and lying between the same parallel lines BC and AE. Since the distance between parallel lines is constant, their heights corresponding to the base BC will be the same. The area of a parallelogram is calculated as Base × Height. As both parallelograms have the same base BC and the same height (distance between AE and BC), their areas must be equal. This theorem is crucial for comparing areas without direct measurement.

Theorem 2: If a parallelogram and a triangle are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.

Let parallelogram ABCD and triangle EBC be on the same base BC and between the same parallel lines BC and AE. Draw a diagonal AC in the parallelogram. The area of triangle ABC is half the area of parallelogram ABCD (since a diagonal divides a parallelogram into two congruent triangles of equal area). The height of triangle EBC from vertex E to base BC is the same as the height of parallelogram ABCD. The formula for the area of triangle EBC is ½ × Base × Height. Since the base (BC) and height are identical for both figures, it naturally follows that Area(EBC) = ½ × Area(ABCD). This theorem provides a direct relationship between the areas of these two fundamental shapes.

Theorem 3: Triangles on the same base (or equal bases) and between the same parallels are equal in area.

Suppose we have two triangles, ABC and DBC, sharing the same base BC and lying between the same parallel lines BC and AD. Just like with parallelograms, since they share the same base and lie between the same parallels, their heights corresponding to the base BC are equal. The area of a triangle is ½ × Base × Height. As both triangles have the same base BC and the same height, their areas, Area(ABC) and Area(DBC), must be equal. This theorem is frequently used in proofs involving area relations within complex figures. Its converse is also important: if two triangles have the same base (or equal bases) and equal areas, then they must lie between the same parallels. This helps prove that lines are parallel based on area information.

Area Relationships: Parallelograms vs. Triangles

Solved Examples

• {"title":"Example 1: Parallelogram and Triangle Area","bodyMarkdown":"Question: A parallelogram ABCD has an area of 50 cm². If a triangle ABE is drawn such that it shares the base AB with the parallelogram and lies between the same parallel lines AB and DC, what is the area of triangle ABE?\n\nSolution: According to Theorem 2, if a parallelogram and a triangle are on the same base and between the same parallels, the area of the triangle is half the area of the parallelogram.\nGiven, Area(ABCD) = 50 cm².\nArea(ABE) = ½ × Area(ABCD) = ½ × 50 cm² = 25 cm².\nAnswer: The area of triangle ABE is 25 cm²."}
• {"title":"Example 2: Median and Area","bodyMarkdown":"Question: In triangle PQR, PS is a median. If Area(PQR) = 60 cm², find Area(PQS).\n\nSolution: A median of a triangle divides it into two triangles of equal area.\nGiven, PS is the median to QR.\nTherefore, Area(PQS) = Area(PSR) = ½ × Area(PQR).\nArea(PQS) = ½ × 60 cm² = 30 cm².\nAnswer: The area of triangle PQS is 30 cm²."}

Exam Tips and Common Mistakes

1. Identify the Base and Parallels: The most crucial step in solving problems from this chapter is correctly identifying the common base and the pair of parallel lines. Misidentifying these can lead to incorrect applications of theorems.
2. "Equal Areas" vs. "Congruent": Remember that figures with equal areas are not necessarily congruent. Congruent figures always have equal areas, but the converse is false. Many students confuse these terms.
3. Proof-based Questions: For proofs, clearly state which theorem you are using. For example, "Since triangle ABC and triangle DBC are on the same base BC and between the same parallels BC and AD, Area(ABC) = Area(DBC) (Theorem 9.2)."
4. Auxiliary Constructions: Sometimes, drawing an auxiliary line (like a diagonal or a line parallel to a side) can help in applying the theorems. Practice identifying when such constructions are beneficial.
5. Area Calculation Formulas: Don't forget the basic area formulas for parallelograms (base × height) and triangles (½ × base × height). These are the foundation for the theorems.

Quick Revision Check

• Q: Can two triangles with equal areas be non-congruent? A: Yes, two triangles can have equal areas but different shapes, thus not being congruent (e.g., a tall thin triangle and a short wide one).
• Q: What is the relationship between the areas of two parallelograms that share the same base and are between the same parallel lines? A: Their areas are equal.
• Q: If a triangle and a parallelogram stand on the same base and between the same parallel lines, how is the area of the triangle related to the area of the parallelogram? A: The area of the triangle is half the area of the parallelogram.
• Q: What effect does a median have on the area of a triangle? A: A median divides the triangle into two triangles of equal area.

Frequently Asked Questions