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RD Sharma Class 9 Solutions Chapter 15 - Areas of Parallelograms and Triangles (Ex 15.2) Exercise 15.2

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Last updated date: 26th May 2024
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RD Sharma Class 9 Solutions Chapter 15 - Areas of Parallelograms and Triangles (Ex 15.2) Exercise 15.2 - Free PDF

Free PDF of RD Sharma Class 9 Solutions Chapter 15 – Areas of Parallelograms and Triangles Exercise 15.2 solved by expert teachers is available for download on Vedantu. The fundamental theorems of areas of triangles and parallelograms are explained briefly in this chapter to ensure that students can have a clear understanding of the advanced topics. All Chapter 15 – Areas of Parallelograms and Triangles Ex 15.2 Questions with Solutions for RD Sharma are provided in this PDF so that students can revise the sums  thoroughly during their exam preparation. 

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About the Chapter

In the RD Sharma book, Areas of Parallelograms and Triangles is the 15th chapter. Students will learn about Parallelograms and Triangles, and how to find their areas.  They will get to practice all concepts of this chapter with the extensive collection of questions in RD Sharma. Hence, it will boost their confidence for the exam.


Quadrilateral is a 2D figure having four sides. A parallelogram is a quadrilateral in which the opposite sides are equal and the diagonals bisect each other. A triangle is a three-sided polygon. It has  three edges and three vertices. Based on the measure of sides and angles, triangles can be classified as equilateral, isosceles, and scalene triangles. In an equilateral triangle, three sides have the same length. In an isosceles triangle, any two sides are of equal length whereas the third side has unequal length. In a scalene triangle, three sides have unequal lengths.


The following three exercises are there in RD Sharma Class 9 Chapter 15 Areas of Parallelogram and Triangles

  1. Exercise 15.1

  2. Exercise 15.2

  3. Exercise 15.3


Areas of Parallelogram and Triangles: Important Formulas

Area of an equilateral triangle= \[ \frac{\surd3}{4}a^{2}\]

(a = side of the given equilateral triangle)


Area of an isosceles triangle= \[\frac{1}{2}bh\]

( b = base; h= height of the given isosceles triangle) 


Area of a scalene triangle=\[\frac{1}{2}bh\]

( b = base; h= height of the given isosceles triangle)


Area of a scalene triangle= √s(s–a)(s–b)(s–c) 

(a, b, and c are the sides of the triangle)

\[s=\frac{a+b+c}{2}\]

(s = the semi perimeter of the triangle)


The theorems and axioms of triangles and parallelogram and the sums on areas of parallelograms and triangles with same base are covered in this chapter.


Key Benefits of Solving Class 9 RD Sharma Areas of Parallelograms and Triangles

  1. RD Sharma Class 9 Chapter 15 provides a comprehensive understanding of the fundamental concepts of areas enclosed by parallelogram and triangles.

  2. The detailed explanation of theorems and axioms of triangles and parallelograms given in this book undoubtedly guide students to identify their applications in the sums.

  3. Students can refer to the solved examples given in Class 9 RD Sharma Chapter 15 to learn the step-wise problem-solving approach for all types of sums covered in this chapter.

  4. Also, they can check the answer-key given at the back of the chapters to verify if they can solve the sums correctly.


The Class 9 RD Sharma Chapter 15 ex.-15.2 Solutions are provided in a PDF format on Vedantu app and website. So students can refer to these solutions online as well as offline. They can download the PDF for free and refer to the solutions for their practice purpose. After practising the sums in Ex.-15.2, they can begin with the next exercise right away, without wasting any time. Class 9 RD Sharma Ex.-15.3 Solutions PDF is also available on Vedantu for free download.

FAQs on RD Sharma Class 9 Solutions Chapter 15 - Areas of Parallelograms and Triangles (Ex 15.2) Exercise 15.2

1. How to find the areas of a triangle and a parallelogram formed on same base?

When a triangle and parallelogram are constructed on the same base, they are said to be formed between the same pair of parallel lines. Therefore, the height of the parallelogram and the triangle is the same. Now, let the length of the base be and the height be h.


Area of triangle= \[\frac{1}{2}bh\]


Area of parallelogram= bh


Therefore, the area of the triangle is half the area of the parallelogram constructed on the same base.

2. Give two properties of parallelogram and triangles.

Two properties of a parallelogram are as follows.

  • There are two diagonals in a parallelogram, which bisect each other.

  • Every two opposite angles in a parallelogram are equal.

Two properties of a triangle are as follows.

  • The sum of all angles in any given  triangle is 180 degrees.

  • The sum of any two sides of a given triangle is always greater than the measure of the third side.