NCERT Solutions Class 9th Maths Chapter – 5 Introduction to Euclid Geometry Examples

NCERT Solutions Class 9th Maths Chapter – 5 Introduction to Euclid Geometry Examples
Last Doubt

NCERT Solutions Class 9th Maths Chapter – 5 Introduction to Euclid Geometry

TextbookNCERT
Class9th
SubjectMathematics
Chapter5th
Chapter NameIntroduction to Euclid Geometry
CategoryClass 9th Mathematics
MediumEnglish
SourceLast Doubt

NCERT Solutions Class 9th Maths Chapter – 5 Introduction to Euclid Geometry

Chapter -5

Introduction to Euclid Geometry

Examples

Example 1 : If A, B and C are three points on a line, and B lies between A and C (see Fig. 5.7), then prove that AB + BC = AC.

Solution – In the figure given above, AC coincides with AB + BC.
Also, Euclid’s Axiom (4) says that things which coincide with one another are equal to one another. So, it can be deduced that
AB + BC = AC
Note that in this solution, it has been assumed that there is a unique line passing
through two points.

Example 2 : Prove that an equilateral triangle can be constructed on any given line segment.


Solution – In the statement above, a line segment of any length is given, say AB [see Fig. 5.8(i)]. Fig. 5.8 Here, you need to do some construction. Using Euclid’s Postulate 3, you can draw a circle with point A as the centre and AB as the radius [see Fig. 5.8(ii)]. Similarly, draw another circle with point B as the centre and BA as the radius. The two circles meet at a point, say C. Now, draw the line segments AC and BC to form ∆ ABC
[see Fig. 5.8 (iii)].

So, you have to prove that this triangle is equilateral, i.e., AB = AC = BC.
Now, AB = AC, since they are the radii of the same circle (1)
Similarly, AB = BC (Radii of the same circle) (2)
From these two facts, and Euclid’s axiom that things which are equal to the same thing are equal to one another, you can conclude that AB = BC = AC.
So, ∆ ABC is an equilateral triangle.
Note that here Euclid has assumed, without mentioning anywhere, that the two circles drawn with centres A and B will meet each other at a point.
Now we prove a theorem, which is frequently used in different results:

You Can Join Our Social Account

YoutubeClick here
FacebookClick here
InstagramClick here
TwitterClick here
LinkedinClick here
TelegramClick here
WebsiteClick here