NCERT Solutions Class 9th Maths Chapter – 5 Introduction to Euclid Geometry Exercise – 5.1

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

TextbookNCERT
Class 9th
Subject Mathematics
Chapter5th
Chapter NameIntroduction to Euclid Geometry
CategoryClass 9th Mathematics
Medium English
SourceLast Doubt

NCERT Solutions Class 9th Maths Chapter – 5 Introduction to Euclid Geometry Ex 5.1 In This Chapter We Will Learn About Linear Equations in Two Variables, Linear equation, Linear equation in two variable, Graph  with Class 9th Maths Chapter – 5 Introduction to Euclid Geometry Ex 5.1.

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

Chapter -5

Introduction to Euclid Geometry

Exercise – 5.1

Question 1. Which of the following statements are true and which are false? Give reasons for your answers.
Ch 5 5.1
(i) Only one line can pass through a single point.
Solution – False. This can be seen usually.

(ii) There are an infinite number of lines which pass through two distinct points.
Solution – False. This contradicts Axiom 5.1.

(iii) A terminated line can be produced indefinitely on both the sides.
Solution – True. Postulate 2.

(iv) If two circles are equal, then their radii are equal.
Solution – True. If we superimpose the region bounded by one circle on the other, then they coincide. So, their centres and boundaries coincide therefore, their radii will coincide.

(v) In figure, if AB – PQ and PQ = XY, then AB = XY.
Solution – True. The first Axiom of Euclid.

Question 2. Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they and how might you define them?
(i) Parallel lines
Solution – Parallel Lines: Two lines l and m in a plane are said to be parallel, if they have no common point and we write them as l ॥ m.
Ch 5 5.1
(ii) Perpendicular lines
Solution – Perpendicular Lines: Two lines p and q lying in the same plane are said to be perpendicular if they form a right angle and we write them as p ⊥ q.
Ch 5 5.1
Solution – Line Segment: A line segment is a part of line and having a definite length. It has two end-points. In the figure, a line segment is shown having end points A and B. It is written as AB¯¯¯¯¯¯¯¯ or BA¯¯¯¯¯¯¯¯.
Ch 5 5.1
(iv) Radius of a circle
Solution – Radius of a circle: The distance from the centre to a point on the circle is called the radius of the circle. In the figure, P is centre and Q is a point on the circle, then PQ is the radius.
Ch 5 5.1
(v) Square
Solution – Square: A quadrilateral in which all the four angles are right angles and all the four sides are equal is called a square. Given figure, PQRS is a square.
Ch 5 5.1

Question 3. Consider two ‘postulates’ given below:
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain.

Solution – Yes, these postulates contain undefined terms such as ‘Point and Line’. Also, these postulates are consistent because they deal with two different situations as
(i) says that given two points A and B, there is a point C lying on the line in between them. Whereas
(ii) says that, given points A and B, you can take point C not lying on the line through A and B.
No, these postulates do not follow from Euclid’s postulates, however they follow from the axiom, “Given two distinct points, there is a unique line that passes through them.”

Question 4. If a point C lies between two points A and B such that AC = BC, then prove that AC = 1/2 AB. Explain by drawing the figure.

Solution – We have,
Ch 5 5.1
AC = BC [Given]
∴ AC + AC = BC + AC [If equals added to equals then wholes are equal]
or 2AC = AB [∵ AC + BC = AB]
or AC = 1/2AB

Question 5. In question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.

Solution – Let the given line AB is having two mid points ‘C’ and ‘D’.
Ch 5 5.1
AC = 1/2AB ……(i)
and AD = 1/2AB ……(ii)
Subtracting (i) from (ii), we have
AD – AC = 1/2A− 1/2AB
or AD – AC = 0 or CD = 0
∴ C and D coincide.
Thus, every line segment has one and only one mid-point.

Question 6. In figure, if AC = BD, then prove that AB = CD.

Solution – From the figure, it can be observed that
AC = AB + BC
BD = BC + CD
It is given that AC = BD
AB + BC = BC + CD ……(1)
According to Euclid’s axiom, when equals are subtracted from equals, the remainders are also equal.
Subtracting BC from equation (1), we obtain
AB + BC − BC = BC + CD − BC
AB = CD

Question 7. Why is axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that, the question is not about the fifth postulate.)

Solution – As statement is true in all the situations. Hence, it is considered a ‘universal truth.’

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