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# RS Aggarwal Class 9 Chapter 11 Solutions (Areas of Parallelograms and Triangles)

RS Aggarwal Class 9 Maths Solutions Chapter 11 Areas of Parallelograms and Triangles is an essential chapter from the perspective of upcoming board exams next year. Chapter 11 of RS Aggarwal Class 9 Maths Solutions has one dedicated exercise, which contains 38 questions for your practice. Some of these questions have 4-5 sub-parts as well for a better revision. The main topics that are covered in this exercise are the Euclidean Area Axioms, Figures that are on a similar base and between the similar parallels lines, and also parallelograms lying on the same base and same parallels. The calculation of the base and altitude of a parallelogram is also covered in this exercise.

This chapter is prepared, keeping a steady pace for you throughout your preparation. We have presented you with the complete set of solutions for this chapter, which will clear your base and help you better understand the topic. This is a great reference book for your exam preparation. It will help you achieve good marks in your CBSE exams. We have made sure that the solution is in simple language for you. These solutions provided by Instasolv can be accessed free of cost.

## Important topics for RS Aggarwal Class 9 Maths Solutions Chapter 11 Areas of Parallelograms and Triangles

Area: It is the amount of planar surface that is being covered by a closed geometric figure in an even or uneven manner. Every closed figure imparts some areas that can always be calculated.

Identifying the Figures on the Common Base and Between the Same Parallels

It can only be said true for any two figures if:

a) They both have a common side.

b) The sides that are parallel to the common base and also the vertices opposite to that side lie on a similar straight line, which is parallel to the base.

Triangle: A plane figure that is bounded by three straight lines is called a triangle. It is the simplest form of a polygon. It is a closed figure that is formed by joining three line segments.

Area of a Parallelogram

Area of a parallelogram = b×h

where ‘b′ is known as the base, and ‘h′ is the corresponding altitude or height.

Area of a Triangle

Area of a triangle = (1/2)×b×h

where “b” is known as the base, and “h” is the corresponding altitude.

Many theorems need to be learned for comfortable solving of problems.

Theorem 1

Statement: Diagonals of a parallelogram divides it into two triangles of equal area.

Theorem 2

Statement: Parallelograms that are on a similar base and between the same parallel lines are equal in area.

Theorem 3

Statement: Triangles that share a similar base and are formed between the same parallels are equal in area.

Theorem 4

Statement: Triangles having equal areas and having one side of one of the triangles similar to one side of the other, always their corresponding altitudes similar.

### Exercise Discussion of RS Aggarwal Class 9 Maths Solutions Chapter 11 Areas of Parallelograms and Triangles

This chapter of RS Aggarwal Solutions for Class 9 Maths contains just one dedicated exercise which covers the whole range of questions:

• The first one is to identify the figures with a joint base and between the same parallel lines. It has six sub-parts.
• The next three items are about finding the area of parallelogram and length of its sides.
• The next three questions are based on trapezium. They are mainly focused on finding the area.
• The following many questions are based on proving the various theorems and their applications. These problems mainly focus on the triangle and its properties with variations of quadrilateral and rhombus.
• Few questions from 27 are based on finding the area of a parallelogram with intermediate difficulty.
• After this, all the questions till the end are about proving various applications over triangle and trapezium with a much higher level of difficulty.

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