CBSE Class 9 Maths Exemplar Chapter 9: Areas of Parallelograms and Triangles NCERT Solutions

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NCERT Solutions PDF Class 9 PDF

CBSE Class 9 Maths Exemplar Chapter 9, 'Areas of Parallelograms and Triangles,' NCERT Solutions, offers a detailed exploration of geometric area calculations. This resource covers key concepts such as the properties of medians within triangles and the crucial relationships between polygons that share the same base and lie between the same parallel lines. Students will find clear explanations for calculating the area of figures derived from joining the midpoints of a rectangle, as well as a thorough understanding of parallelogram area computation. These solutions are meticulously crafted to support students in their academic journey, providing step-by-step guidance to solidify their grasp of the subject matter and enhance their problem-solving abilities for examinations.

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Quick info

BoardCBSE
ClassClass 9
SubjectMaths Exemplar
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 9

Chapter summary

Chapter 9 of the CBSE Class 9 Maths Exemplar focuses on the areas of parallelograms and triangles. This NCERT Solutions set clarifies key theorems and their applications, such as how a median divides a triangle into two equal areas and the properties of figures situated on the same base and between the same parallels. It also includes problems on calculating areas of derived figures and parallelograms, providing a solid foundation for geometry.

Learning outcomes

  • Understand that a median divides a triangle into two triangles of equal area.
  • Identify figures on the same base and between the same parallels.
  • Calculate the area of a figure formed by joining the mid-points of a rectangle.
  • Determine the area of a parallelogram using its base and height.

Topics covered

Paper topics

  • Areas of Parallelograms and Triangles
  • Median of a Triangle
  • Triangles of Equal Area
  • Polygons on the Same Base
  • Polygons Between the Same Parallels
  • Figure formed by joining mid-points of a rectangle
  • Area of a Rhombus
  • Area of a Parallelogram

Important topics

  • Median and its area property
  • Figures on the same base and between same parallels
  • Area calculation for figures formed by mid-points
  • Area of parallelogram formula

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Questions and Solutions

Question 1

The median of a triangle divides it into two:
  1. triangles of equal area
  2. congruent triangles
  3. right triangles
  4. isosceles triangles
Solution:

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. A key geometric property states that a median divides the triangle into two smaller triangles that have equal areas. Therefore, the correct option is (A).

Question 2

In which of the following figures, you find two polygons on the same base and between the same parallels?

(A)

Figure A

(B)

Figure B

(C)

Figure C

(D)

Figure D
Solution:

The property of figures lying on the same base and between the same parallels is a fundamental concept in geometry. Observing the provided figures, figure (d) clearly depicts two polygons (specifically, parallelograms PQRA and BQRS) that share the same base (QR) and are situated between the same pair of parallel lines (PS and QR). This configuration is crucial for proving theorems related to equal areas. Hence, the correct option is (D).

Question 3

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is:
  1. a rectangle of area 24 cm<sup>2</sup>.
  2. a square of area 25 cm<sup>2</sup>.
  3. a trapezium of area 24 cm<sup>2</sup>.
  4. a rhombus of area 24 cm<sup>2</sup>.
Solution:

Let the rectangle be ABCD with sides AB = CD = 8 cm and BC = DA = 6 cm. Let E, F, G, and H be the mid-points of sides AB, BC, CD, and DA, respectively. Joining these mid-points forms the quadrilateral EFGH. According to the mid-point theorem applied to the triangles formed within the rectangle, EF is parallel to AC and EF = 1/2 AC, and HG is parallel to AC and HG = 1/2 AC. Similarly, EH is parallel to BD and EH = 1/2 BD, and FG is parallel to BD and FG = 1/2 BD. Since the diagonals of a rectangle are equal (AC = BD), all sides of EFGH are equal (EF = FG = GH = HE). Thus, EFGH is a rhombus. The diagonals of this rhombus are EG and FH. EG is parallel to BC and DA, and its length is equal to BC = 6 cm. FH is parallel to AB and CD, and its length is equal to AB = 8 cm. The area of a rhombus is given by half the product of its diagonals. Therefore, the area of rhombus EFGH = \frac{1}{2} \times EG \times FH = \frac{1}{2} \times 6 \text{ cm} \times 8 \text{ cm} = 24 \text{ cm}^2. Hence, the correct option is (D).

Question 4

In the given figure, the area of parallelogram ABCD is:
Parallelogram ABCD with height BM

The options provided are:

(A) AB \times BM

Solution:

The area of a parallelogram is defined as the product of its base and its corresponding perpendicular height. In the given parallelogram ABCD, if we consider AB as the base, then BM is the perpendicular height drawn from vertex B to the base AB (or its extension). Therefore, the area of parallelogram ABCD is given by the formula: Area = base × height = AB \times BM. Hence, the correct option is (A).

Common mistakes

  • Confusing congruent triangles with triangles of equal area.
  • Incorrectly identifying figures on the same base and between the same parallels.
  • Errors in calculating the area of figures derived from rectangles.
  • Misapplying the formula for the area of a parallelogram.

Revision tips

  • Review the properties of a median and its effect on a triangle's area.
  • Practice identifying parallelograms and triangles that share the same base and lie between the same parallel lines.
  • Work through the examples involving mid-points of rectangle sides to solidify area calculations.
  • Ensure you can correctly state and apply the formula for the area of a parallelogram.

Practice MCQs

Q1. What is the effect of a median on the area of a triangle?

Q2. Which geometric property is illustrated by two polygons on the same base and between the same parallels?

Q3. When the mid-points of the adjacent sides of a rectangle are joined, what figure is formed?

Q4. If a rectangle has sides 8 cm and 6 cm, what is the area of the rhombus formed by joining its mid-points?

Q5. The area of a parallelogram is given by the formula:

Frequently asked questions

What is the main concept covered in CBSE Class 9 Maths Exemplar Chapter 9?

Chapter 9 focuses on the areas of parallelograms and triangles, including theorems about medians, figures on the same base and between the same parallels, and area calculations for specific geometric shapes.

How does a median affect the area of a triangle?

A median divides a triangle into two triangles of equal area. This is a fundamental property explained in the solutions.

What is the area of the figure formed by joining the mid-points of a rectangle?

When the mid-points of the adjacent sides of a rectangle are joined, a rhombus is formed. Its area is half the product of the diagonals of the rectangle (which are equal to the sides of the rectangle).

How can these NCERT Solutions help with exam preparation?

These solutions provide clear, step-by-step explanations for each question, reinforcing understanding of key concepts and formulas, which is crucial for exam revision and problem-solving.

What is the formula for the area of a parallelogram?

The area of a parallelogram is calculated by multiplying its base by its perpendicular height (Area = base × height).

Important links

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