RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions is based on the concepts of the Constructions of different types of triangles. In this article, we will discuss- Constructions of Triangles, Equilateral Triangles, triangle when its base, total of other two edges, and one base angle are given, etc. The PDF attached below will help students know about several types of questions based on the RD Sharma exercise.

RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions clarifies the construction of various kinds of triangles. Going through these solutions and practicing the same, a student will score well in the exam. As the solutions are given in a stepwise manner, any student can understand them easily.

**Learn about RD Sharma Class 9 Chapter 17 (Constructions)**

Table of Contents

## Download RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions PDF

Solutions for Class 9 Maths Chapter 17 Construction Exercise 17.3

## What are the topics covered within the RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions?

Basically, it covers the following topics mentioned below-

- First of all, it covers the triangle construction part. Specifically, it covers the construction of the equilateral triangle, i.e., all sides are of equal length.
- It covers the construction of a triangle when its base, in which the sum of two different sides of the triangle is provided. At the same time, it also provides one base angle for the students to construct the triangle.
- It covers the construction of a triangle, in which one of its bases is given. Alongside this, here the difference between the lengths of the other two sides is provided. Also, it provides one base angle for the students to construct the triangle.
- It includes the topic meant for constructing a triangle where the perimeter of the triangle is provided. At the same time, the students are provided with two base angles as well.

## Important Definitions RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions

The concepts of the Constructions of different types of triangles are mentioned in the following points-

**Constructions of Triangles**

Triangle is basically a polygon having three sides and thus three angles. Triangles can be constructed of equal/unequal sides and angles. One can construct a triangle using rulers, compasses, and protractors.

**Construction of an Equilateral Triangle**

The equilateral triangle is one having equal sides and angles. It can be constructed using compass, scale, and rulers.

**Construction of a triangle when its base, total of the other two sides, and one base angle are given**

These triangles can be constructed using scale, compass, protractor, and ruler. However, one has to determine the other two angles upon calculating concerning given angle, and 180 degrees.

**Construction of a triangle when its base, difference of the other two sides, and one base angle are given**

These triangles can also be constructed using scale, compass, protractor, and ruler. Again, one needs to first determine the other two angles upon calculating concerning given angle, and 180 degrees.

**Construction of a triangle of given perimeter and two base angles**

Here one needs to draw a line of length as of given perimeter and then draw two base angles from both its terminal points, using compass. Then one needs to bisect both the angles and draw a perpendicular bisector of both. The point where these perpendicular bisectors meet the perimeter length becomes the base of the desired triangle, and the point where the bisectors meet become the other point.

**Examples of RD Sharma Chapter 17 Class 9 Maths Exercise 17.3 Solutions**

**Ques- Construct a △ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.**

**Solution- **Steps of Construction of triangle is mentioned below-

- Draw a line section BC = 3.6 cm.
- At spot B, draw ∠XBC = 60°.
- Draw an arc that crosses XB at point ‘D’ from point B’’ and with a radius of 4.8 cm.
- Join DC.
- Draw a perpendicular bisector of DC, which crosses DB at ‘A.’
- Join AC.

Hence, △ABC is the required triangle.

**Ques- Using rulers and compasses (protractor) only, construct a △ABC, given base BC = 7cm, ∠ABC = 60°, and AB + AC = 12cm.**

**Solution- **Steps of Construction of a triangle is mentioned below-

- Draw a line segment BC = 7cm.
- Draw an arc from point ‘B’ cutting BC at ‘N.’ (Choose any radius.)
- Keep compass (protractor) at point ‘N’ with the same radius (selected in step 2), make the previous arc at ‘M.’
- Join the line segment BM.
- Produce BM to any point ‘P.’
- Mark BR = 12cm, from BP.
- Join CR.
- Draw a perpendicular bisector of RC which crosses BR at ‘A.’
- Join the line segment AC.

Hence, △ABC is the required triangle.