NCERT Class 9 Maths · Chapter 9

NCERT Solutions Class 9 Maths Chapter 9Areas of Parallelograms & Triangles

Step-by-step solutions for all exercises in NCERT Class 9 Maths Areas of Parallelograms & Triangles.

Chapter Overview

Prove theorems relating areas of figures on the same base and between the same parallels.

This chapter is part of the NCERT Mathematics textbook for Class 9 and is important for CBSE school examinations. The concepts covered here build the foundation for more advanced topics in higher classes.

Below you will find solved examples from this chapter. Each solution includes detailed step-by-step working so you can understand the method, not just the answer.

Solved Examples from Areas of Parallelograms & Triangles

1For two planar figures to be considered 'between the same parallels', what is the essential condition they must satisfy?

A.Their bases must lie on one parallel line, and their vertices opposite to the base must lie on the other parallel line.
B.They must have at least one common side.
C.Their areas must be equal.
D.They must be congruent.

Answer: A

Solution:

Step 1: The definition states that if two figures have a common base (or equal bases) and the vertices opposite to the base of each figure lie on a line parallel to the base, then they are said to be on the same base and between the same parallels.

Step 2: Option A directly captures this definition, explaining the geometric arrangement required.

2In a figure, parallelogram ABCD and triangle EBC are drawn such that point E lies on the side AD. For these two figures to be on the same base and between the same parallels, which condition is necessary?

A.E must be the midpoint of AD.
B.The line AD must be parallel to the line BC.
C.The line AB must be parallel to the line CD.
D.Area(ABCD) must be equal to Area(EBC).

Answer: B

Solution:

Step 1: The common base for both parallelogram ABCD and triangle EBC is BC.

Step 2: For them to be between the same parallels, the line containing the vertex opposite to the base (AD for the parallelogram, and E for the triangle, which lies on AD) must be parallel to the base BC.

Step 3: Therefore, the line AD must be parallel to BC.

3Parallelogram PQRS and parallelogram MNRS are on the same base SR and between the same parallels SR and PM. If Area(PQRS) = 45 cm², what is Area(MNRS)?

A.90 cm²
B.45 cm²
C.22.5 cm²
D.Cannot be determined without side lengths.

Answer: B

Solution:

Step 1: According to Theorem 9.1, parallelograms on the same base and between the same parallels are equal in area.

Step 2: Given that PQRS and MNRS are on the same base SR and between the same parallels SR and PM, their areas must be equal.

Step 3: Therefore, Area(MNRS) = Area(PQRS).

Step 4: Area(MNRS) = 45 cm².

4Which of the following conditions is *not necessarily true* for two triangles on the same base and between the same parallels?

A.Their heights corresponding to the common base are equal.
B.Their areas are equal.
C.They have the same perimeter.
D.Their vertices opposite to the common base lie on the same parallel line.

Answer: C

Solution:

Step 1: If two triangles are on the same base and between the same parallels, it implies they have the same base length and the same height corresponding to that base.

Step 2: Based on the formula Area = 1/2 × base × height, their areas will be equal (Option B is true). Their heights are indeed equal (Option A is true). And by definition, their opposite vertices lie on the same parallel line (Option D is true).

Step 3: However, having the same base and height does not guarantee that the other two sides of the triangles are equal. Therefore, their perimeters are not necessarily equal.

5A triangle and a parallelogram are on the same base and between the same parallels. If the area of the parallelogram is 60 cm², what is the area of the triangle?

A.30 cm²
B.60 cm²
C.120 cm²
D.45 cm²

Answer: A

Solution:

Step 1: According to a key theorem, if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.

Step 2: Given Area(parallelogram) = 60 cm².

Step 3: Area(triangle) = 1/2 × Area(parallelogram).

Step 4: Area(triangle) = 1/2 × 60 cm² = 30 cm².

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Frequently Asked Questions

Where can I find NCERT Solutions for Class 9 Maths Chapter 9?+
You can find complete NCERT Solutions for Class 9 Maths Chapter 9 (Areas of Parallelograms & Triangles) on this page with step-by-step explanations for all exercises.
Are these NCERT Solutions for Class 9 Areas of Parallelograms & Triangles updated for 2025-26?+
Yes, these solutions follow the latest NCERT textbook for the 2025-26 academic session and cover all exercise questions.
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Is Areas of Parallelograms & Triangles important for Class 9 exams?+
Yes, Areas of Parallelograms & Triangles is an important chapter in Class 9 CBSE Maths. Questions from this chapter regularly appear in school exams and board assessments.
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