Euclid’s Geometry
NCERT Solutions • Class 9 Maths • Chapter 5Exercise 5.1
1. Which of the following statements are true and which are false? Give reasons for your answers.
(i) Only one line can pass through a single point.
False Reason: There can be infinite number of lines passing through a single point.
False Reason: There can be infinite number of lines passing through a single point.
(ii) There are an infinite number of lines which pass through two distinct points.
False Reason: Axiom 5.1 states that given two distinct points, there is a unique line that passes through them.
False Reason: Axiom 5.1 states that given two distinct points, there is a unique line that passes through them.
(iii) A terminated line can be produced indefinitely on both the sides.
True Reason: This is Euclid’s Postulate 2.
True Reason: This is Euclid’s Postulate 2.
(iv) If two circles are equal, then their radii are equal.
True Reason: If two circles are equal, their regions coincide perfectly (Axiom 4), implying their centers and boundaries coincide, thus radii are equal.
True Reason: If two circles are equal, their regions coincide perfectly (Axiom 4), implying their centers and boundaries coincide, thus radii are equal.
(v) In Fig. 5.9, if $AB = PQ$ and $PQ = XY$, then $AB = XY$.
True Reason: Axiom 1 states “Things which are equal to the same thing are equal to one another”.
True Reason: Axiom 1 states “Things which are equal to the same thing are equal to one another”.
2. Give a definition for each of the following terms. Are there other terms that need to be defined first?
Prerequisite Terms:
Before defining these, we need to have an idea of point, line, plane, ray, and angle.
(i) Parallel Lines:
Two straight lines in a plane that never intersect each other are called parallel lines.
(ii) Perpendicular Lines:
Two lines are said to be perpendicular if they intersect each other at a right angle ($90^\circ$).
(iii) Line Segment:
A part of a line with two definite end points is called a line segment.
(iv) Radius of a Circle:
The distance from the center of the circle to any point on its boundary (circumference).
(v) Square:
A quadrilateral in which all four sides are equal and all four interior angles are right angles.
3. Consider two ‘postulates’ given below… Do these contain undefined terms? Are they consistent? Do they follow from Euclid’s postulates?
Undefined Terms: Yes, these postulates contain undefined terms such as ‘point’, ‘line’, and ‘in between’.
Consistency: Yes, these postulates are consistent because they deal with two different situations:
1. Given two points, a third point exists between them.
2. Given two points, there is a third point not on the line connecting them.
1. Given two points, a third point exists between them.
2. Given two points, there is a third point not on the line connecting them.
Relation to Euclid: No, these do not follow directly from Euclid’s postulates. However, they follow from Axiom 5.1 (Given two distinct points, there is a unique line that passes through them).
4. If a point C lies between two points A and B such that $AC = BC$, then prove that $AC = \frac{1}{2} AB$. Explain by drawing the figure.
Given: $AC = BC$
To Prove: $AC = \frac{1}{2} AB$
Proof:
Since $AC = BC$, add $AC$ to both sides (Euclid’s Axiom 2: If equals are added to equals, the wholes are equal).
$\Rightarrow AC + AC = BC + AC$
$\Rightarrow 2AC = BC + AC$
From the figure, $BC + AC$ coincides with $AB$ (Axiom 4).
$\Rightarrow 2AC = AB$
$\Rightarrow AC = \frac{1}{2} AB$
Given: $AC = BC$
To Prove: $AC = \frac{1}{2} AB$
Proof:
Since $AC = BC$, add $AC$ to both sides (Euclid’s Axiom 2: If equals are added to equals, the wholes are equal).
$\Rightarrow AC + AC = BC + AC$
$\Rightarrow 2AC = BC + AC$
From the figure, $BC + AC$ coincides with $AB$ (Axiom 4).
$\Rightarrow 2AC = AB$
$\Rightarrow AC = \frac{1}{2} AB$
5. Prove that every line segment has one and only one mid-point.
Let us assume there are two mid-points, $C$ and $D$, for the line segment $AB$.
If $C$ is the mid-point: $AC = \frac{1}{2} AB$ … (1)
If $D$ is the mid-point: $AD = \frac{1}{2} AB$ … (2)
From (1) and (2):
$AC = AD$
This means point $C$ and point $D$ coincide. Therefore, every line segment has one and only one mid-point.
If $C$ is the mid-point: $AC = \frac{1}{2} AB$ … (1)
If $D$ is the mid-point: $AD = \frac{1}{2} AB$ … (2)
From (1) and (2):
$AC = AD$
This means point $C$ and point $D$ coincide. Therefore, every line segment has one and only one mid-point.
6. In Fig. 5.10, if $AC = BD$, then prove that $AB = CD$.
Given: $AC = BD$
From the figure:
$AC = AB + BC$
$BD = BC + CD$
Substituting these into the given equation:
$AB + BC = BC + CD$
Subtracting $BC$ from both sides (Euclid’s Axiom 3: If equals are subtracted from equals, the remainders are equal):
$AB = CD$
From the figure:
$AC = AB + BC$
$BD = BC + CD$
Substituting these into the given equation:
$AB + BC = BC + CD$
Subtracting $BC$ from both sides (Euclid’s Axiom 3: If equals are subtracted from equals, the remainders are equal):
$AB = CD$
7. Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’?
Axiom 5: “The whole is greater than the part.”
Reason: This statement is true for everything in the universe, not just in geometry. Whether we talk about numbers, physical objects, or quantities, a part is always contained within the whole, making the whole greater than the part. Hence, it is considered a universal truth.
Reason: This statement is true for everything in the universe, not just in geometry. Whether we talk about numbers, physical objects, or quantities, a part is always contained within the whole, making the whole greater than the part. Hence, it is considered a universal truth.