CBSE Class 9

These NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry Ex 5.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry Exercise 5.2

Question 1.
How would you rewrite Euclid’s filth postulate so that it would be easier to understand?
Solution:
Two lines are said to be parallel if they are equidistant from one other and they do not have any point of intersection. eg.

Question 2.
Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.
Solution:
Yes,

According to Euclid’s 5th postulate when n line falls on 1 and m and if ∠1 + ∠2 < 180° and ∠3 + ∠4 > 180° and then producing line l & m further will meet in the side of ∠1 and ∠2 which is less than 180°. Which gave the clue about the condition when ∠1 + ∠2 = 180° and the line l and m will not meet at any point.

These NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry Exercise 5.1

Question 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.
(ii) There is an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both sides.
(iv) If two circles are equal then their radii are equal.
(v) In fig. 5.9, if AB = PQ and PQ = XY, then AB = XY

Solution:
(i) False, from one point an infinite number of lines can pass.
(ii) False, as from two distinct points only one line can pass through them.
(iii) True, as you can see in the figure.

(iv) True, if two circles are equal then their center and circumference will coincide. Therefore, radii will be equal.
(v) True. As AB = PQ and PQ = XY

Therefore, comparing figures of (a) and (b) we get AB = XY.

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
(ii) Perpendicular lines
(iii) Line Segment
(iv) Radius of a circle
(v) Square
Solution:
(i) Parallel lines: If two lines are at a constant distance from each other and don’t meet at any point are called parallel lines.

AB and CD are parallel to each other as there is a constant distance ‘d’ between them.
To define a parallel line, you should know the line, constant distance, Intersection points.

(ii) Perpendicular lines: If one line is standing on another line and makes 90° angle with the other is called perpendicular lines like in fig.
AB is perpendicular (1) on CD or vice versa.

(iii) Line Segment: A straight line drawn from anyone point to any other point.

(iv) Radius: Area bounded by a line whose one en 1 is fixed and completes a circle.

OA radius (O is fixed, a point called centre)

(v) Square: A square is a rectangle with a pair of consecutive sides that are equal.

Here ABCD is a square in which AB = BC = CD = AD and ∠A = ∠C = ∠D = 90°

Question 3.
Consider two ‘Postulates’ given below:
(i) Given any two distinct points A and B, there exist 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 Euclid’s Postulates? Explain?
Solution:
It does have an undefined term like line segment between two points, Third point is on the line or not.
Yes, they are consistent as these are two different situations in the
(i) the third point C lie on the line segment made by joining points A and B points.
(ii) The third point does not lie on the line segment made by joining the A and B points.
No, these postulates do not follow Eudid’s postulates. Actually, they are axiom 5.1.

Question 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.
Solution:

Given AC = BC
AC + AC = BC + AC (Equals are added to equals)
or, 2AC = AB (BC + AC coincides with AB)
AC = \(\frac { 1 }{ 2 }\) AB

Question 5.
In question 4, point C is called a midpoint of line segment AB, prove that every line segment has one and only one midpoint.
Solution:
Let there are two mid point C and D.
then AB = AC + CB = 2AC …..(i)
(∵ C is the mid point of AB)
and AB = AD + DB = 2AD ……(ii)
(∵ D is the mid point of AB)

From equation (i) and (ii)
AC = AD and CB = DB
But this will be possible only when D lies on point C.
So what we have assumed that there are two midpoints is wrong.

Question 6.
In fig. 5.10, If AC = BD, then prove that AB = CD

Solution:
AC = BD (Given)
AC = AB + BC ……(i)
(B lies between A and C)
BD = BC + CD ……(ii)
(C lies between B and D Points)
From the given equation (i) and (ii)
We get AB + BC = BC + CD
AB = CD (Substracting Equals from Equals)

Question 7.
Why is Axiom 5 in the list of Euclid’s axioms considered a ‘Universal Truth’?
Solution:
Axiom 5: The whole is greater than the part.
As we can see in this figure.

i.e., a = b + c
a > b and a > c,
a is whole and b and c are parts of a.
So, it is rightly said that the whole is greater than the part.

These NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.4

Question 1.
Give the geometric representation of y = 3 as an equation.
(i) in one variable.
(ii) in two variables.
Solution:
(i) We have given that y = 3
It is treated as an equation in one variable y only.
It has a unique solution y = 3, which is a point on the graph.

(ii) We have given that y = 3
It can be also expressed as
0 . x + y = 3
It is treated as an equation in two variables x and y.
It has infinitely many solutions which are in the form of (r, 3) where r is any real number.
The solutions of this equation represented by a line, we use the following table to draw the graph

Question 2.
Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable.
(ii) in two variables.
Solution:
(i) We have given that 2x + 9 = 0
or, x = \(\frac {-9}{2}\)
It is treated as an equation in one variable x only.
It has a unique solution x = \(\frac {-9}{2}\), which is a point on the graph.

(ii) We have given that 2x + 9 = 0
It can be also expressed as 2x + 0 . y + 9 = 0
It is treated as an equation in two variables x and y.
It has infinitely many solutions which are in the form of (r, 9) where r is any real number.
The solutions of this equation represented by a line.
To draw the graph we use the following table.

These NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.3

Question 1.
Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x – y = 2
(iii) y = 3x
(iv) 3 = 2x + y
Solution:
(i) We have given the equation
x + y = 4
To draw the graph, we need at least two solutions to the equation.
Put x = 0 then y = 4 and if x = 4 then y = 0 are solutions of the given equation.
So we use the following table to draw the graph:

(ii) We have given the equation x – y = 2
To draw the graph, we need at least two solutions to the equation.
Put x = 0 then y = -2 and if x = 2 then y = 0 are solutions of the given equation.
So we use the following table to draw the graph:

(iii) We have given the equation y = 3x
To draw the graph, we need at least two solutions to the equation.
Put x = 0 the y = 0, and if x = 1 then y = 3 are the solutions of the given equation.
So we use the following table to draw the graph:

(iv) We have given the equation 3 = 2x + y
To draw the graph, we need at least two solutions to the equation.
Put x = 0 then y = 3 and if x = 1 then y = 1 are the solutions of the given equation.
So we use the table to draw the graph:

Question 2.
Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Solution:
Here (2, 14) is the solution of a linear equation which is satisfied by the co-ordinate of the point (2,14).
For eg. 2x + 10 = y, 3x + 8 = y, 4x + 6 = y and so on.
It is because we know that from a given point there are infinitely many straight lines that pass.

Question 3.
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a?
Solution:
We have given that (3, 4) lies on the graph of the equation.
Therefore we put x = 3 and y = 4 in this equation.
3 × 4 = a × 3 + 7
or, 12 = 3a + 7
or, 12 – 7 = 3a
or, a = \(\frac {5}{3}\)
Therefore, the value of a = \(\frac {5}{3}\)

Question 4.
The taxi fare in a city is as follows:
For the first kilometer, the fare is Rs. 8 and for the subsequent distance, it is Rs. 5 per km. Taking the distance covered as x km and total fare is Rs. y, write a linear equation for this information and draw its graph.
Solution:
We have given that the total distance covered is x km and the total fare is Rs. y.
But, we have given that the rate of the first km is Rs. 8.
Therefore the fare of the remaining distance (x – 1) km is Rs. 5 per km.
Therefore, according to question, we can say
(x – 1) × 5 + 8 = y
or, 5x – 5 + 8 = y
or, 5x + 3 = y
Therefore, 5x + 3 = y is required equation for this information.
We use the following table to represent the graph.

Question 5.
From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.
(a) For Fig. 4.6
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x

(b) For Fig. 4.7
(i) y = x + 2
(ii) y = x – 2
(iii) y = -x + 2
(iv) x + 2y = 6

Solution:
(a) In Fig. 4.6 the points on the line are (-1, 1), (0, 0) and (1, -1). By inspection x + y = 0 is the equation corresponding to this graph. We find that the coordinate in each case is equal but an opposite sign of the x coordinate.

(b) In Fig. 4.7 the points on the line are (0, 2) and (2, 0). By inspection y = -x + 2 is the equation corresponding to this graph. We find that when we put x = 0 then the value of y = 2 and when we put x = 2 then the value of y = 0.

Question 6.
If the work done by a body on the application of a constant force is directly proportional to the distance travelled by the body. Express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also real from the graph the work done when the distance travelled by the body is
(i) 2 units
(ii) 0 units
Solution:
Let us assume,
The work done by a constant force is y-units and the distance travelled by the body is x-units. Therefore, According to the question.
Force y is directly proportional to distance travelled x
y ∝ x
y = kx (where k is constant)
Again according to the question.
We have given that constant force (k) = 5 units.
The required two-variable equation is y = 5x
(i) According to the graph,
When the distance travelled by the body is 2 units, then the work done is 10 units.
(ii) According to the graph,
When the distance travelled by the body is 0 units, the work done is 0 Units.
To draw the graph of y = 5x we use the following table.

Question 7.
Yamini and Fatima, two students of class IX of a school, together contributed ₹ 100 towards Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data, (you may take their contributions as ₹ x and ₹ y). Draw the graph of the same.
Solution:
Let Yamini contributed towards Prime Minister Relief Fund is ₹ x and Fatima Contributed ₹ y.
Therefore, According to question,
x + y = 100
The linear equation which satisfies this data is x + y = 100.
To draw the graph we use the following table:

Question 8.
In countries like the USA and Canada, the temperature is measured in Fahrenheit, whereas in countries like India it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:
F = (\(\frac {9}{5}\)) C + 32
(i) Draw the graph of the linear equation above using Celsius for the x-axis and Fahrenheit for the y-axis.
(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95° F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes find it.
Solution:
(i) We have given that the linear equation.
F = (\(\frac {9}{5}\)) C + 32
To draw the graph, we need at least two solutions to the equation.
Put C = 0 then F = 32 and if C = 5 then F = 41 are the solutions of the given equation.
So we use the following table to draw the graph.

Taking C on the x-axis and F on the y-axis.

(ii) We have given that C = 30°
Then by the linear equation F = (\(\frac {9}{5}\)) C + 32
Put C = 30
F = \(\frac {9}{5}\) × 30 + 32
F = 54 + 32
∴ F = 86
Therefore, If the temperature is 30°C then the temperature in Fahrenheit is 86°.

(iii) We have given that F = 95°
Then by the linear equation
F = (\(\frac {9}{5}\)) C + 32
Put F = 95°
or, 95 = \(\frac {9}{5}\) C + 32
or, 95 – 32 = (\(\frac {9}{5}\)) C
63 × \(\frac {5}{9}\) = C
∴ C = 35°
The temperature of 95°F is 35° Celsius.

(iv) We have given that C = 0°
Then by the linear equation
F = (\(\frac {9}{5}\)) C + 32
Put C = 0°
F= (\(\frac {5}{9}\)) × 0 + 32
or, F = 0 + 32
F = 32
The temperature of 0°C is 32 Fahrenheit.
Again in case II,
We have given that F = 0°
Then by liner equation
F = (\(\frac {9}{5}\)) C + 32
put F = 0
0 = (\(\frac {9}{5}\)) C + 32
or, 0 – 32 = (\(\frac {9}{5}\)) C
or, -32 × \(\frac {5}{9}\) = C
or, \(\frac {-160}{9}\) = C
C = -17.77
or, C = 17.7
∴ The temperature of 0°F is -17.7 Celsius.

(v) Yes, there is a temperature which is numerically the same in both Fahrenheit and Celsius.
This is -40°C
At -40°C the value of the Fahrenheit scale is also -40°F.

These NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.2

Question 1.
Which one of the following statements is true and why?
y = 3x + 5 has
(i) a unique solution.
(ii) only two solutions.
(iii) infinitely many solutions.
Solution:
We have given the equation
y = 3x + 5
Put x = 0, then y = 5
Therefore, (0, 5) is the solution of this equation. Again, put x = 1 then y = 8.
Therefore, (1, 8) is also the solution to this equation. Again, put x = 2, then y = 11.
Therefore, (2, 11) is also the solution of this equation. Again x = 3, x = 4 and so on.
Now, it is clear that this equation has infinitely many solutions.
Note: A linear equation in two variables has infinitely many solutions.

Question 2.
Write four solutions for each of the following equations.
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y
Solution:
(i) We have 2x + y = 7
Put x = 0 then y = 7
Therefore, (0, 7) is the solution of this equation.
Again put x = 1, then y = 5.
Therefore (1, 5) is also the solution of this equation. Again put x = 2, then y = 3.
Therefore (2, 3) is also the solution of this equation. Again put x = 3 then y = 1.
Therefore (3, 1) is also the solution of this equation.
Therefore, (0, 7), (1, 5), (2, 3) and (3, 1) are the four solutions of the equation 2x + y = 7.

(ii) We have given πx + y = 9
or, \(\frac {22}{7}\) x + y = 9 (∵ π = \(\frac {22}{7}\))
or, y = 9 – \(\frac {22}{7}\) x
Put x = 0 then y = 9
Therefore (0, 9) is the solution of this equation.
Again, put x = 1 then y = \(\frac {41}{7}\)
Therefore (\(\frac {22}{7}\)) is also the solution of this equation.
Again put x = 7, then y = -13
Therefore (7, – 13) is also the solution of this equation.
Again put x = 14, then y = -3
Therefore (14, -35) is also the solution of this equation.
Hence (0, 9) (\(\frac {22}{7}\)) (7, 13) and (14, -35) are the four solutions of equation πx + y = 9

(iii) We have given that x = 4y
Put y = 0, then x = 0
Therefore, (0, 0) is the solution of this equation.
Again put y = 1, then x = 4 Therefore, (4, 1) is also the solution of this equation.
Again, put y = 2 then x = 8 Therefore, (8, 2) is also the solution of this equation.
Again put y = 3, then x = 12 Therefore, (12, 3) is also the solution of this equation.
Hence, (0, 0), (4, 1) (8, 2), and (12, 3) are the four solution of equation x = 4y.

Question 3.
Check which of the following are solutions of the equation x – 2y = 4 and which are not.
(i) (0, 2)
(ii) (2, 0)
(iii) (4, 0)
(iv) (√2, 4√2)
(v) (1, 1)
Solution:
(i) we have given that x – 2y = 4
Put x = 0 and y = 2
Then, 0 – 2 × 2 = 4
-4 = 4
Here, L.H.S. ≠ R.H.S.
Therefore, (0, 2) is not the solution of equation x – 2y = 4

(ii) Put x = 2 and y = 0
Then, 2 – 2 × 0 = 4
2 = 4
Here, L.H.S. ≠ R.H.S.
Therefore (2, 0) is not the solution of equation x – 2y = 4

(iii) Put x = 4 and y = 0 in equation x – 2y = 4
Then 4 – 2 × 0 = 4
4 = 4
Here, L.H.S. = R.H.S.
Therefore, (4, 0) is the solution of equation x – 2y = 4

(iv) Put x = √2 and y = 4√2 in equation x – 2y = 4
Then √2 – 2 × 4√2 = 4
√2 – 8√2 = 4
-7√2 = 4
Here, L.H.S. ≠ R.H.S.
Therefore, (√2, 4√2) is not the solution of equation x – 2y = 4.

(v) (1, 1)
Put x = 1 and y = 1 in equation x – 2y = 4
Then 1 – 2 × 1 = 4
1 – 2 = 4
-1 = 4
L.H.S. ≠ R.H.S.
Therefore (1, 1) is not the solution of equation x – 2y = 4.

Question 4.
Find the value of k if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
We have given that x = 2 and y = 1 is the solution of the equation 2x + 3y = k.
Put x = 2 and y = 1 in equation 2x + 3y = k
2 × 2 + 3 × 1 = k
4 + 3 = k
k = 7
Hence, the value of k in equation 2x + 3y = k is 7.

These NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.1

Question 1.
The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
Solution:
Let the cost of a notebook is Rs. x and the cost of a pen is Rs. y
According to question
x = 2y or, x – 2y = 0
Therefore the linear equation in two variables to represent this statement is x – 2y = 0

Question 2.
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 9.\(\bar{35}\)
(ii) x – \(\frac {y}{5}\) – 10 = 0
(iii) -2x + 3y = 6
(iv) x = 3y
(v) -2x = -5y
(vi) 3x + 2 = 0
(vii) y – 2 = 0
(viii) 5 = 2x
Solution:
(i) We have given that
2x + 3y = 9.\(\bar{35}\)
2x + 3y – 9.\(\bar{35}\) = 0
Now, it is the form of ax + by + c = 0
Therefore the values of a, b and c are 2, 3 and -9.\(\bar{35}\) respectively.

(ii) We have given that
x – \(\frac {y}{5}\) – 10 = 0
We can also write
1x + (\(\frac {-y}{5}\)) y + (-10) = 0
Now, it is in the form of ax + by + c = 0
Therefore values of a, b and c are 1, \(\frac {-y}{5}\), and -10 respectively.

(iii) We have given that
-2x + 3y = 6
or, -2x + 3y – 6 = 0
We can also write
(-2)x + 3y + (-6) = 0
Now, it is in the form of ax + by + c = 0
Therefore, the values of a, b and c are -2, 3 and -6 respectively.

(iv) We have given that,
x = 3y
or, x – 3y = 0
We can also write
1x + (-3)y + 0 = 0
Now, it is in the form of ax + by + c = 0 and the values of a, b and c are 1, -3 and 0 respectively.

(v) We have given that
2x = -5y
or, 2x + 5y = 0
We can also write 2x + 5y + 0 = 0
Now, it is in the form of ax + by + c = 0.
Therefore, the values of a, b and c are 2, 5 and 0 respectively.

(vi) We have given that
3x + 2 = 0
We can also write 3x + 0y + 2 = 0
Now, it is in the form of ax + by + c = 0
Therefore, the values of a, b and c are 3, 0 and 2 respectively.

(vii) y – 2 = 0
We can also write,
0x + 1y + (-2) = 0
Now, it is in the form of ax + by + c = 0.
Therefore, the values of a, b and c are 0, 1 and -2 respectively.

(viii) we have given that
5 = 2x
or, 2x – 5 = 0
We can also write
2x + 0y + (-5) = 0
Now it is in the form of ax + by + c = 0.
Therefore, the values of a, b and c are 2, 0 and (-5) respectively.