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Here we are providing Coordinate Geometry Class 9 Extra Questions Maths Chapter 3 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Coordinate Geometry with Answers Solutions

Extra Questions for Class 9 Maths Chapter 3 Coordinate Geometry with Solutions Answers

Coordinate Geometry Class 9 Extra Questions Very Short Answer Type

Coordinate Geometry Class 9 Extra Questions Question 1.
Write the signs convention of the coordinates of a point in the second quadrant.
Solution:
(-ve, +ve)

Class 9 Coordinate Geometry Extra Questions Question 2.
Write the value of ordinate of all the points lie on x-axis.
Solution:
0

Coordinate Geometry Extra Questions Class 9 Question 3.
Write the value of abscissa of all the points lie on y-axis.
Solution:
0

Class 9 Maths Chapter 3 Extra Questions Question 4.
If in coordinates of a point B(3, -2), signs of both coordinates are interchanged, then it will lie in which quadrant ?
Solution:
When signs of both coordinates of B(3, -2) are interchanged, then coordinates of new point are B'(-3, 2) and it will lie in second quadrant.

Extra Questions On Coordinate Geometry Class 9 Question 5.
Find distances of points C(-3, -2) and D(5, 2) from x-axis and y-axis.
Solution:
Distances of point C(-3, -2) from x-axis is 2 units in the negative direction and from y-axis is 3 units in the negative direction. Distances of point D(5, 2) from x-axis is 2 units and from y-axis is 5 units.

Extra Questions Of Coordinate Geometry Class 9 Question 6.
Find the values of x and y, if two ordered pairs (x – 3, – 6) and (4, x + y) are equal.
Solution:
Here, two ordered pairs are equal.
⇒ Their first components are equal and their second components are separately equal.
⇒ x – 3 = 4 and x + y = -6
⇒ x = 7 and 7 + y = -6 ⇒ y = – 13
Hence, x = 7 and y = – 13.

Questions On Coordinate Geometry Class 9 Question 7.
In which quadrant does the point (-1, 2) lie ?
Solution:
(-1, 2) lie in second quadrant.

Coordinate Geometry Class 9 Important Questions Question 8.
Find the distance of the point (0, -5) from the origin.
Solution:
5 units.

Class 9 Maths Coordinate Geometry Extra Questions Question 9.
Write the shape of the quadrilateral formed by joining (1, 1), (6, 1), (4, 5) and (3, 5) on graph paper.
Solution:
Trapezium.

Coordinate Geometry Class 9 Extra Questions Short Answer Type 1

Coordinate Geometry Class 9 Questions With Solutions Question 1.
In the given figure, ABCD is a rectangle with length 6 cm and breadth 3 cm. O is the mid-point of AB. Find the coordinates of A, B, C and D.

Solution:

We have taken 1 cm = 1 unit and origin O is the mid-point of AB
∴ OA = OB = 3 cm
and BC = AD = 3 cm
Thus, the coordinates of A are (-3, 0)
the coordinates of B are (3, 0)
the coordinates of C are (3, 3)
the coordinates of D are (-3, 3)

Ch 3 Maths Class 9 Extra Questions Question 2.
Write the coordinates of A, B, C and D from the figure given alongside.

Solution:
Coordinates of the point A are (5, 0)
Coordinates of the point B are (5, 3)
Coordinates of the point C are (-2, 4)
Coordinates of the point D are (0, -2)

Class 9 Maths Ch 3 Extra Questions Question 3.
A point lies on x-axis at a distance of 9 units from y-axis. What are its coordinates? What will be the coordinates of a point, if it lies on y-axis at a distance of -9 units from x-axis ?
Solution:
As shown in graph, the coordinates of a point which lies on x-axis at a distance of 9 units from y-axis are (9, 0) and the coordinates of a point which lies at a distance of -9 units from x-axis are (0, -9).

Coordinate Geometry Class 9 Hots Questions Question 4.
Plot the point P(2, -6) on a graph paper and from it draw PM and PN perpendiculars to x-axis and y-axis respectively. Write the coordinates of the points M and N.
Solution:

As shown in graph, coordinates of M are (2, 0) and coordinates of N are (0, -6).

Coordinate Geometry Class 9 Extra Questions Short Answer Type 2

Chapter 3 Maths Class 9 Extra Questions Question 1.
Without plotting the points indicate the quadrant in which they lie, if :
(i) ordinate is 5 and abscissa is – 3
(ii) abscissa is -5 and ordinate is – 3
(iii) abscissa is – 5 and ordinate is 3
(iv) ordinate is 5 and abscissa is 3
Solution:
(i) Clearly, point (-3, 5) lies in 2nd quadrant.
(ii) Clearly, point (-5, – 3) lies in 3rd quadrant.
(ii) Clearly, point (-5, 3) lies in 2nd quadrant.
(iv) Clearly, point (3, 5) lies in 1st quadrant.

Coordinate Geometry Class 9 Extra Questions With Solutions Question 2.
Plot the points A(1, 4), B(-2, 1) and C(4, 1). Name the figure so obtained on joining them in order and also, find its area.
Solution:

Triangle.
Area of ∆ABC = \(\frac{1}{2}\) × BC × Height
= \(\frac{1}{2}\) × 6 × 3
= 9 sq. units

Coordinate Geometry Questions Class 9 Question 3.
Plot the following points, join them in order and identify the figure thus formed : A(1, 3), B(1, -1), C(7, -1) and D(7, 3)
Write the coordinates of the point of intersection of the diagonals.
Solution:

ABCD is a rectangle.
Point of intersection of the diagonals AC and BD is (4, 1).

Class 9 Chapter 3 Maths Extra Questions Question 4.
Plot the points A(2, 5), B(8,5) and C(5, -3) and join AB, BC and CA. What figure do you obtain ?
Solution:

We obtain an isosceles triangle in which AC = BC.

Extra Questions For Class 9 Maths Chapter 3 With Solution Question 5.
(i) Plot the points M(4, 3), N(4, 0), 0(0, 0), P(0, 3).
(ii) Name the figure obtained by joining MNOP.
(iii) Find the perimeter of the figure.
Solution:
(i)

(ii) As shown in graph, the figure obtained by joining MNOP is rectangle.
(iii) Perimeter of rectangle MNOP = 2 (ON + OP) = 2 (4 + 3) = 2 × 7 = 14 units.

Question 6.
Plot D(-2, -3) on the graph paper. Also, plot reflections of D in x-axis and y-axis.
Solution:

Reflection of D(-2, -3) in x-axis is D'(-2, 3) and reflection of D(-2, -3) in y-axis is D'(2, -3).

Question 7.
If the coordinates of a point M are (-2, 9) which can also be expressed as (1 + x, y) and y > 0, then find in which quadrant do the following points lie : Ply, x), Q(2, x), R(x, y − 1), S(2x, -3y).
Solution:
Here, M(-2, 9) can also be expressed as (1 + x, y²)
∴ (1 + x, y²) ⇔ (-2, 9) and y > 0
⇒ 1 + x = -2 and y² = 9
⇒ x = -3 and y = √9 = 3 (∵ y > 0)
P (y, x) = P(3, -3), it lies in IV quadrant
Q(2, x) = Q(2, -3), also lies in IV quadrant
R(x², y – 1) = R((-3)², 3 – 1) = R(9, 2), it lies in I quadrant
S(2x, -3y) = S(2 × (-3), -3 × 3) = S(-6, -9), it lies in III quadrant

Question 8.
In the given figure, PQR is an equilateral triangle with coordinates of Q and R as (-2, 0) and (2, 0) respectively. Find the coordinates of the vertex P.
Solution:

Here,
QR = OQ + OR.
= 2 + 2 = 4 units
∴ ∆PQR is an equilateral triangle.
∴ PQ = PR = QR = 4 units
In right-angled ∆OPQ, ∠POQ = 90°
∴ We have PQ² = OP² + OQ²
⇒ OP² = PQ² – OQ²
= 4² – 2² = 16 – 4 = 12
⇒ OP = √12 = 2√3 units
∴ The coordinates of P are (0, 2√3).

Coordinate Geometry Class 9 Extra Questions HOTS

Question 1.
Plot the points A(3, 2), B(-2, 2), C(-2, -2) and D(3, -2) in the cartesian plane. Join these points and name the figure so formed.
Solution:

Figure so formed is ABCD a rectangle.

Question 2.
Write the coordinates of two points on X-axis and two points on Y-axis which are at equal distances from the origin. Connect all these points and make them as vertices of quadrilateral. Name the quadrilateral thus formed.
Solution:
Let a be the equal distance from origin on both axes. Now, the coordinates of two points on equal distance ‘a’on x-axis are Pla, 0) and R(-a, 0). Also, the coordinates of two points on equal distance ‘a’ on Y-axis are Q(0, a) and S(0, -a). Join all the four points on the graph. Now, PQRS, thus formed is a square.

Coordinate Geometry Class 9 Extra Questions Value Based (VBQs)

Question 1.
On environment day, class-9 students got five plants of mango, silver oak, orange, banyan and amla from soil department. Students planted the plants and noted their locations as (x, y).

Plot the points (x, y) in the graph and join them in the given order. Name the figure you get. Which social act is being done by students of class-9 ?
Solution:

The given trees (points) are Mango (2, 0), Silver Oak (3, 4), Orange (0,7), Banyan (-3, 4) and Amla (-2, 0). The location of these trees are Orange (0,7) shown in the graph.
On joining the points of mango, silver oak, orange, banyan and amla in order, the figure so formed is a regular pentagon.
Planting more trees helpful in reducing pollution and make the environment clean and green for the coming generations.

Here we are providing Linear Equations for Two Variables Class 9 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Linear Equations for Two Variables with Answers Solutions

Extra Questions for Class 9 Maths Chapter 4 Linear Equations for Two Variables with Solutions Answers

Linear Equations for Two Variables Class 9 Extra Questions Very Short Answer Type

Linear Equations In Two Variables Class 9 Extra Questions Question 1.
Linear equation x – 2 = 0 is parallel to which axis ?
Solution:
Here, linear equation is x – 2 ⇒ 0 x = 2
Thus, it is parallel to the y-axis.

Linear Equations Class 9 Extra Questions Question 2.
Express x in term of y: \(\frac{x}{7}\) + 2y = 6
Solution:
Given equation is
\(\frac{x}{7}\) + 2y = 6
⇒ \(\frac{x}{7}\) = 6 – 2y
Thus, x = 7(6 – 2y).

Linear Equations In Two Variables Extra Questions Class 9 Question 3.
If we multiply or divide both sides of a linear equation with a non-zero number, then what will happen to the solution of the linear equation ?
Solution:
Solution remains the same.

Class 9 Maths Chapter 4 Extra Questions Question 4.
Find the value of k for which x = 0, y = 8 is a solution of 3x – 6y = k.
Solution:
Since x = 0 and y = 8 is a solution of given equation
3x – 6y = k
3(0) – 6(8) = k
⇒ k = – 48

Class 9 Linear Equations In Two Variables Extra Questions Question 5.
Write the equation of a line which is parallel to x-axis and is at a distance of 2 units from the origin.
Solution:
Here, required line is parallel to x-axis and at a distance of 2 units from the origin.
∴ Its equation is
y + 2 = 0
or y – 2 = 0

Extra Questions For Class 9 Maths Chapter 4 With Solution Question 6.
Find ‘a’, if linear equation 3x – ay = 6 has one solution as (4, 3).
Solution:
Since (4, 3) is a solution of given equation.
∴ 3(4) – a(3) = 6
⇒ 12 – 3a = 6
⇒ a = \(\frac{-6}{-3}\)
Hence, a = 2

Linear Equations In Two Variables Class 9 Extra Questions With Solutions Question 7.
Cost of a pen is two and half times the cost of a pencil. Express this situation as a linear equation in two variables.
Solution:
Let cost of a pen be ₹ x and cost of a pencil be ₹ y.
According to statement of the question, we have
x = 2\(\frac{1}{2}\) y
⇒ 2x = 5y or 2x – 5y = 0

Extra Questions On Linear Equations In Two Variables Class 9 Question 8.
In an one day international cricket match, Raina and Dhoni together scored 198 runs. Express the statement as a linear equation in two variables.
Solution:
Let runs scored by Raina be x and runs scored by Dhoni be y.
According to statement of the question, we have
x + y = 198
x + y – 198 = 0

Extra Questions For Class 9 Maths Linear Equations In Two Variables Question 9.
The cost of a table is 100 more than half the cost of a chair. Write this statement as a linear equation in two variables.
Solution:
Let the cost price of a table be ₹ x and that of a chair be ₹ y.
Since the cost price of a table is 100 more than half the cost price of a chair.
∴x = \(\frac{1}{2}\)y + 100
⇒ 2x = y + 200 or 2x – y – 200 = 0.

Linear Equations for Two Variables Class 9 Extra Questions Short Answer Type 1

Class 9 Maths Ch 4 Extra Questions Question 1.
Write linear equation representing a line which is parallel to y-axis and is at a distance of 2 units on the left side of y-axis.
Solution:
Here, required equation is parallel to y-axis at a distance of 2 units on the left side of y-axis.
x = -2 or x + 2 = 0

Chapter 4 Maths Class 9 Extra Questions Question 2.
In some countries 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 = \(\left(\frac{9}{5}\right)\)C + 32°
If the temperature is – 40°C, then what is the temperature in Fahrenheit?
Solution:
Given linear equation is
F = \(\left(\frac{9}{5}\right)\)C + 32°
Put C = -40°, we have
F = \(\frac{9}{5}\)(-40°) + 32°
F = – 72° + 32°
F= -40°

Ch 4 Maths Class 9 Extra Questions Question 3.
Give equation of two lines on same plane which are intersecting at the point (2, 3).
Solution:
Since there are infinite lines passing through the point (2, 3).
Let, first equation is x + y = 5 and second equation is 2x + 3y = 13.
Clearly, the lines represented by both equations intersect at the point (2, 3).

Class 9 Maths Linear Equations In Two Variables Extra Questions Question 4.
If ax + 3y = 25, write y in terms of x and also, find the two solutions of this equation.
Solution:

Linear Equations In 2 Variables Class 9 Extra Questions Question 5.
Find the value of k, if (1, -1) is a solution of the equation 3x – ky = 8. Also, find the coordinates of another point lying on its graph.
Solution:
Since (1, -1) is a solution of the equation 3x – ky = 8
∴ 3(1) – k(-1) = 8
⇒ k = 8 – 3 = 5
Thus, the given equation is
3x – 5y = 8

Hence, the coordinates of another point lying on the graph of 3x – 5y = 8 is (6, 2).

Linear Equations In Two Variables Class 9 Important Questions Question 6.
Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is the value of y, when x = 5 ?
Solution:
Given y varies directly as x implies y = kx
But y = 12 for x = 4
⇒ 4k = 12 = k = 3
Put k = 3 in y = kx, we have
y = 3x
Now, when x = 5, y = 3 x 5 = y = 15 …(i)

Linear Equations for Two Variables Class 9 Extra Questions Short Answer Type 2

Linear Equations In Two Variables Class 9 Extra Questions Pdf Question 1.
A fraction becomes , when 2 is subtracted from the numerator and 3 is added to the denominator. Represent this situation as a linear equation in two variables. Also, find two solutions for this.
Solution:
Let numerator and denominator of the given fraction be respectively x and y. According to the statement, we obtain
\(\frac{x-2}{y+3}\) = \(\frac{1}{4}\)
⇒ 4x – 8 = y + 3
⇒ 4x – y – 11 = 0
Which is the required linear equation. When y = 1, then x = 3. When y = 5, then x = 4. Hence, the two solutions are (3, 1) and (4, 5).

Extra Questions For Class 9 Maths Chapter 4 Question 2.
The path of an aeroplane is given by the equation 3x – 4y = 1 2: Represent the path graphically. Also, show that the point (-4,-6) lies on the graph.
Solution:
Given equation is 3x – 4y = 12
∴ y = \(\frac{3 x-12}{4}\).

When x = 0, then y = -3
When x = 4, then y = 0
Table of solutions is

Plot the points (0, – 3), (4, 0) on the graph and join them to get the required graph. From the graph, we see, when
x = -4, then y = -6.
Therefore, (-4,- 6) lies on the graph of given equation.

Class 9 Chapter 4 Maths Extra Questions Question 3.
Express y in terms of x for the equation 3x – 4y + 7 = 0. Check whether the points (23, 4) and \(\left(0, \frac{7}{4}\right)\) lie on the graph of this equation or not.
Solution:
Given equation is 3x – 4y + 7 = 0
∴y = \(\frac{3 x+7}{4}\)
When x = -1, then y = 1
When x = 3, then y = 4
When x = 7, then y = 7
Table of solutions is :


Plot the points (-1, 1), (3, 4), (7, 7). and join them to get the given graph. From the graph, we see that, when x = 0, then y \(\frac{7}{4}\) = 5 and when y = 4,
then x = 3. Thus, (0, \(\frac{7}{4}\)) lies on the graph, whereas (23, 4) does not lie on the graph.

Extra Questions Of Chapter 4 Class 9 Maths Question 4.
The following observed values of x and y are thought to satisfy a linear equation. Write the linear equation :

Draw the graph using the values of x, y as given in the above table. At what points the graph of the linear equation :
(i) cuts the x-axis (ii) cuts the y-axis
Solution:
Let ax + by + c = 0 ……(i)
be the linear equation in two variables. From the table, we have two points A(6, -2) and B(-6, 6) which lie on the graph of the linear equation
⇒ 6a – 2b + c = 0
and -6a + 6b + c = 0
Adding the above two equations, we obtain

Which is the required linear equation in two variables. Plots the points A(6, -2) and B(-6, 6) on the graph. Join them to get line AB.
From the graph, we see that the graph cuts the x-axis at (3, 0) and the y-axis at (0, 2).

Linear Equations for Two Variables Class 9 Extra Questions Long Answer Type

Question 1.
Write the equations of the lines drawn in following graph :

Also, find the area enclosed between these lines.
Solution:
Equations of the lines drawn in the graph are as :
x = -1 or x + 1 = 0,
x = 2 or x – 2 = 0,
y = 1 or y – 1 = 0 and
y = 3 or y – 3 = 0
Figure formed by these lines is a rectangle of dimensions 3 units by 2 units.
Hence, the area enclosed between given lines = 6 sq. units.

Question 2.
If (2, 3) and (4, 0) lie on the graph of equation ax + by = 1. Find the value of a and b. Plot the graph of equation obtained.
Solution:
(2, 3) and (4, 0) lie on the graph of equation
ax + by = 1 …(i)
∴ We have 2a + 3b = 1 … (ii)
and 4a + 0 = 1

Which is required linear equation.
Put x= 0 in eq. (iii), we have
⇒ 3(0) + 2y = 12
⇒ 2y = 12
⇒ y = 6
Put x = 2 in eq. (iii), we have
⇒ 3(2) + 2y = 12
⇒ 2y = 6
⇒ y = 3
Put x = 4 in eq. (iii), we have
⇒ 3(4) + 2y = 12
⇒ 2y = 0
⇒ y = 0
We have the following table :

By plotting the points (0, 6), (2, 3) and (4, 0). Joining them, we obtained the graph of 3x + 2y = 12.

Question 3.
Draw the graphs of the following equations on the same graph sheet :
x = 4, x = 2, y = 1 and y – 3 = 0.
Solution:
Graphs of the given equations are drawn on graph sheet.

Question 4.
Cost of 1 pen is ₹ x and that of 1 pencil is ₹ y. Cost of 2 pens and 3 pencils together is ₹ 18. Write a linear equation which satisfies this data. Draw the graph for the same.
Solution:
Here, cost of 1 pen is ₹x and that
of 1 pencil is ₹ y According to the statement of the question, we have
2x + 3y = 18
⇒ x = \(\frac{18-3 y}{2}\)
When y = 0, x = 9
When y = 4, x= 3.
When y = 6, x = 0
Table of solutions is :

Plot the points (0, 6), (3, 4) and (9, 0). Join them in pairs to get the required line.

Question 5.
Sum of two numbers is 8. Write this in the form of a linear equation in two variables. Also, draw the line given by this equation. Find graphically the numbers, if difference between them is 2.
Solution:
Let the two numbers be x and y.
It is given that sum of two numbers is 8.
∴ x + y = 8
y = 8 – x
When x = 0,
When x = 4, y = 4
When x = 8, y = 0

Table of solutions is :

Plot the points (0, 8), (4, 4), (8, 0) and join them in pairs, we get the required graph.
When difference between two number is 2, then
x – y = 2, x > y
⇒ x = y + 2

When x = 0, y = -2
When x = 2, y = 0
When x = 4, y = 2
Table of solutions is :

Plot these poitns (0, -2), (2, 0), (4, 2) and join them to get the required line.
Graphically, the numbers are : (-2; 4), (-1, -3), (0, -2), (1, -1), (2, 0), (3, 1), (4, 2), (5, 3), (6, 4), (7,5) etc.

Linear Equations for Two Variables Class 9 Extra Questions HOTS

Question 1.
Solve for y :

Solution:

Question 2.
A and B are friends. A is elder to B by 5 years. B’s sister C is half the age of B while A’s father D is 8 years older than twice the age of B. If the present age of D is 48 years, find the present ages of A, B and C.
Solution:
Let the age of B’s sister i.e., C be x years.
∴Age of B be 2x years, age of A be (2x + 5) years.
And age of A’s father i.e., D be 2(2x) + 8 = 4x + 8 years
According to the statement of the question, we have
4x + 8 = 48
⇒ 4x = 48 – 8 = 40
⇒ x = \(\frac{40}{4}\) = 10
Age of A = (2x + 5) years i.e., (2 × 10 + 5) years = 25 years
Age of B = 2x years = 2 × 10 years = 20 years
and Age of C = \(\frac{1}{2}\) of B’s age = \(\frac{1}{2}\) × 20 years = 10 years

Linear Equations for Two Variables Class 9 Extra Questions Value Based (VBQs)

Question 1.
Mrs Sharma lost her purse containing 50 rupee and 100 rupee notes amount to 1500 in a shop. Next day shopkeeper found the purse during dusting. He immediately went to Mrs Sharma’s house and returned the purse and rupees. Mrs Sharma appreciates the shopkeeper for his act.
(i) Represent the situation as an equation and draw the graph.
(ii) What value do you learn from shopkeeper’s act ?
Solution:
(i) Let the number of 50 rupee notes be x
And the number of 100 rupee notes be y
∴ We have 50x + 100y= 1500
⇒ x + 2y = 30
(dividing each term by 50)
2y = 30 – x

We have the following table :

By plotting the points (8, 11), (10, 10) and (12, 9) on the graph and by joining them, we obtain the straight line represented by equation
(i) as shown in graph.
(ii) We should be always honest to feel good.

Question 2.
In an election, a good candidate may lose because 40% of voters do not cast their votes due to various reasons. Form an equation and draw the graph with data. From the graph, find :
(i) The total number of voters, if 720 voters cast their votes.
(ii) The number of votes cast, if the total number of voters are 1000.
(iii) What message did you get from above information ?
Solution:
(i) We have, total number of voters who do not cast their votes = 40%
⇒ Total number of voters who cast their votes = 60%
Let the total number of voters be x and number of voters cast their votes be y

Thus, we have the following table :

By plotting the points (100, 60), (200, 120), (300, 180) on the graph and by joining them, we get the graph of equation (i) as shown in figure.

From the graph, we see that:
(i) When total votes polled = 720
i.e., y = 720, the total number of voters i.e., x = 1200
Hence, total number of voters = 1200.
(ii) When total number of voters x = 1000
Number of votes cast is 600.
(iii) Every voters should cast vote to elect an honest candidate.

Here we are providing Introduction to Euclid’s Geometry Class 9 Extra Questions Maths Chapter 5 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Introduction to Euclid’s Geometry with Answers Solutions

Extra Questions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry with Solutions Answers

Introduction to Euclid’s Geometry Class 9 Extra Questions Very Short Answer Type

Class 9 Maths Chapter 5 Extra Questions With Solutions Question 1.
Give a definition of parallel lines. Are there other terms that need to be defined first ? What are they and how might you define them?
Solution:
Two coplanar lines in a plane) which are not intersecting are called parallel lines. The other term intersecting is undefined.

Introduction To Euclidean Geometry Class 9 Extra Questions Question 2.
Give a definition of perpendicular lines. Are there other terms that need to be defined first ? What are they and how might you define them?
Solution:
Two coplanar (in a plane) lines are perpendicular if the angle between them at the point
of intersection is one right angle. The other terms point of intersection and one right angle are undefined.

Euclid’s Geometry Class 9 Extra Questions Question 3.
Give a definition of line segment. Are there other terms that need to be defined first ? What are they and how might you define them ?
Solution:
A line segment PQ of a line ‘l is the continuous part of the line I with end points P and Q.

Here, continuous part of the line ‘l is undefined.

Euclid’s Geometry Class 9 Extra Questions With Solutions Question 4.
Solve the equation a – 15 = 25 and state which axiom do you use here.
Solution:
a – 15 = 25
On adding 15 to both sides, we obtain
a – 15 + 15 = 25 + 15 [using Euclid’s second axiom]
a = 40

Questions On Euclid Geometry Class 9 Question 5.
Ram and Ravi have the same weight. If they each gain weight by 2 kg, how will their new weights be compared ?
Solution:
Let x kg be the weight each of Ram and Ravi.
On adding 2 kg,
Weight of Ram and Ravi will be (x + 2) kg each.
According to Euclid’s second axiom, when equals are added to equals, the wholes are equal.
So, weight of Ram and Ravi are again equal.

Chapter 5 Maths Class 9 Extra Questions Question 6.
If a point C be the mid-point of a line segment AB, then write the relation among AC, BC and AB.
Solution:
Here, C is the mid-point of AB
⇒ AC = BC
⇒ AC = BC = \(\frac{1}{2}\)AB

Euclid Geometry Class 9 Questions And Answers Question 7.
If a point P be the mid-point of MN and C is the mid-point of MP, then write the relation between MC and MN.
Solution:
Here, P is the mid-point of MN and C is the mid-point of MP.

∴ MC = \(\frac{1}{4}\)MN

Extra Questions For Class 9 Maths Chapter 5 Question 8.
How many lines does pass through two distinct points ?
Solution:
One and only one.

Euclid Geometry Class 9 Questions Question 9.
In the given figure, if AB = CD, then prove that AC = BD. Also, write the Euclid’s axiom used for proving it.

Solution:
Here, given that
AB = CD
By using Euclid’s axiom 2, if equals are added to equals, then the wholes are equal, we have
AB + BC = CD + BC
⇒ AC = BD

Introduction to Euclid’s Geometry Class 9 Extra Questions Short Answer Type 2

Euclids Geometry Class 9 Extra Questions Question 1.
Define :
(a) a square (b) perpendicular lines.
Solution:
(a) A square : A square is a rectangle having same length and breadth. Here, undefined terms are length, breadth and rectangle.
(b) Perpendicular lines : Two coplanar (in a plane) lines are perpendicular, if the angle between them at the point of intersection is one right angle. Here, the term one right angle is undefined.

Class 9 Maths Chapter 5 Important Questions Question 2.
In the given figure, name the following :
(i) Four collinear points
(ii) Five rays
(iii) Five line segments
(iv) Two-pairs of non-intersecting line segments.

Solution:
(i) Four collinear points are D, E, F, G and H, I, J, K
(ii) Five rays are DG, EG, FG, HK, IK.
(iii) Five line segments are DH, EI, FJ; DG, HK.
(iv) Two-pairs of non-intersecting line segments are (DH, EI) and (DG, HK).

Class 9 Maths Chapter 5 Extra Questions Question 3.
In the given figure, AC = DC and CB = CE. Show that AB = DE. Write the Euclid’s axiom to support this.

Solution:
We have
AC = DC
CB = CE
By using Euclid’s axiom 2, if equals are added to equals, then wholes are equal.
⇒ AC + CB = DC + CE
⇒ AB = DE.

Question 4.
In figure, it is given that AD=BC. By which Euclid’s axiom it can be proved that AC = BD?

Solution:
We can prove it by Euclid’s axiom 3. “If equals are subtracted from equals, the remainders are equal.”
We have AD = BC
⇒ AD – CD = BC – CD
⇒ AC = BD

Question 4.
If A, B and C are any three points on a line and B lies between A and C (see figure), then prove that AB + BC = AC.

Solution:
In the given figure, AC coincides with AB + BC. Also, Euclid’s axiom 4, states that things
which coincide with one another are equal to one another. So, it is evident that:
AB + BC = AC.

Question 5.
In the given figure, AB = BC, BX = BY, show that
AX = CY.

Solution:
Given that AB = BC
and BX = BY
By using Euclid’s axiom 3, equals subtracted from equals, then the remainders are equal, we have
AB – BX = BC – BY
AX = CY

Question 6.

In the above figure, if AB = PQ, PQ = XY, then AB = XY. State True or False. Justify your answer.
Solution:
True. ∵ By Euclid’s first axiom “Things which are equal to the same thing are equal to one another”.
∴ AB = PQ and XY = PQ ⇒ AB = XY

Question 7.
In the given figure, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4, write the relation between ∠1 and ∠2, using an Euclid’s axiom.

Solution:
Here, ∠3 = ∠4, ∠1 = ∠3 and ∠2 = ∠4. Euclid’s first axiom says, the things which are equal to equal thing are equal to one another. So ∠1 = ∠2.,

Question 8.
In the given figure, we have ∠1 = ∠2, ∠3 = ∠4. Show that ∠ABC = ∠DBC. State the Euclid’s Axiom used.
Solution:
Here, we have 1 = ∠2 and ∠3 = ∠4. By using Euclid’s Axiom 2. If equals are added to
equals, then the wholes are equal..
∠1 + ∠3 = ∠2 + ∠4
∠ABC = ∠DBC.

Question 9.
In the figure, we have BX and \(\frac{1}{2}\) AB =\(\frac{1}{2}\) BC. Show that BX = BY.

Solution:
Here, BX = \(\frac{1}{2}\) AB and BY = \(\frac{1}{2}\) BC …(i) [given]
Also, AB = BC [given]
⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)BC …(ii)
[∵ Euclid’s seventh axiom says, things which are halves of the same thing are equal to one another]
From (i) and (ii), we have BX = BY

Question 10.
In the given figure, AC = XD, C is mid-point of AB and D is mid-point of XY. Using an Euclid’s axiom, show that AB = XY.

Solution:
∵ C is the mid-point of AB
AB = 2AC
Also, D is the mid-point of XY
XY = 2XD
By Euclid’s sixth axiom “Things which are double of same things are equal to one another.”
∴ AC = XD = 2AC = 2XD ⇒ AB = XY

Introduction to Euclid’s Geometry Class 9 Extra Questions HOTS

Question 1.
For given four distinct points in a plane, find the number of lines that can be drawn through :
(i) When all four points are collinear.
(ii) When three of the four points are collinear.
(iii) When no three of the four points are collinear.

Solution:

(i) Consider the points given are A, B, C and D.
When all the four points are collinear :
One line \(\overrightarrow{\mathrm{AD}}\).

(ii) When three of the four points are collinear :
4 lines
Here, we have four lines \(\stackrel{\leftrightarrow}{A B}, \stackrel{\leftrightarrow}{B C}, \stackrel{\leftrightarrow}{B D}, \stackrel{\leftrightarrow}{A D}\) (four).

(iii) When no three of the four points are collinear :
6 lines Here, we have
\(\stackrel{\leftrightarrow}{A B}, \stackrel{\leftrightarrow}{B C}, \stackrel{\leftrightarrow}{A C}, \stackrel{\leftrightarrow}{A D}, \stackrel{\leftrightarrow}{B D}, \stackrel{\leftrightarrow}{C D}\) (six).

Question 2.
Show that : length AH > sum of lengths of AB + BC + CD.

Solution:
We have
AH = AB + BC + CD + DE + EF + FG + GH
Clearly, AB + BC + CD is a part of AH.
⇒ AH > AB + BC + CD
Hence, length AH > sum of lengths AB + BC + CD.

Introduction to Euclid’s Geometry Class 9 Extra Questions Value Based (VBQs)

Question 1.
Rohan’s maid has two children of same age. Both of them have equal number of dresses. Rohan on his birthday plans to give both of them same number of dresses. What can you say about the number of dresses each one of them will have after Rohan’s birthday? Which Euclid’s axiom is used to answer this question? What value is Rohan depicting by doing so ? Write one more Euclid’s axiom.
Solution:
Here, Rohan’s maid has two children of same age group and both of them have equal number of dresses. Rohan on his birthday plans to give both of them same number of dresses.
∴ By using Euclid’s Axiom 2, if equals are added to equals, then the whole are equal. Thus, again both of them have equal number of dresses. Value depicted by Rohan are caring and other social values. According to Euclid’s Axiom 3, if equals are subtracted from equals, then the remainders
are equal.

Question 2.
Three lighthouse towers are located at points A, B and C on the section of a national forest to protect animals from hunters by the forest department as shown in figure. Which value is department exhibiting by locating extra towers ? How many straight lines can be drawn from A to C? State the Euclid Axiom which states the required result. Give one more. Postulate.

Solution:
One and only one line can be drawn from A to C. According to Euclid’s Postulate, “A straight line may be drawn from any point to any other point:” An other postulate : “A circle may be described with any centre and any radius.” Wildlife is a part of our environment and conservation of each of its element is important for ecological balance.

Here we are providing Oh, I Wish I’d Looked After Me Teeth Extra Questions and Answers Class 9 English Literature Reader, Extra Questions for Class 9 English was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-english/

Oh, I Wish I’d Looked After Me Teeth Extra Questions and Answers Class 9 English Literature

Oh, I Wish I’d Looked After Me Teeth Extra Questions and Answers Short Answer Type

Answer the following questions briefly.

Oh, I Wish I Looked After Me Teeth Question Answers Cbse Question 1.
What were the “perils” that the narrator spotted in her teeth? How had they been caused?
Answer:
The “perils” refer to the cavities and tooth decay that she is suffering from. They have been caused by her eating too many sweets as a child.

Oh I Wish I Looked After Me Teeth Question Answers Cbse Question 2.
When did the narrator have “more teeth than fillin’”? What does this tell us about her present condition?
Answer:
As a child, she had more teeth but now she had lost most of her teeth and had to have fillings on her remaining teeth.

Oh I Wish I’d Looked After Me Teeth Question 3.
What does the narrator mean when she says, “My conscience gets horribly pricked”? Why does she feel like this?
Answer:
It means she is feeling very guilty because she realises that she, herself, is responsible for her tooth decay. She also feels guilty at the thought of the number of sweets she has had in the past.

Oh I Wish I Looked After My Teeth Question 4.
How do we know that the narrator had been careless about taking care of her teeth?
Answer:
We know the narrator had been careless about taking care of her teeth as she has mentioned that she had bashed her teeth lightly because she had thought brushing teeth was a waste of time.

Question 5.
Why has the narrator described a filling as a “murder”?
Answer:
The narrator has done this to express the pain that she feels every time that she has to undergo a filling.

Question 6.
Why does the narrator have to look up the dentist’s nose?
Answer:
The narrator does this because she has to lie in the dentist’s chair while he works on her teeth.

Question 7.
What are molars? What is the word that the narrator uses in the poem to describe her teeth? What does the dentist do to them?
Answer:
Molars are teeth that are used for grinding food. She has also called them “choppers”. The dentist drills holes in the teeth and reconstructs them.

Question 8.
What is the mood of the poem?
Answer:
The poem is written in a humorous style, making light of a painful experience, that is, a visit to the dentist.

Question 9.
Why does the narrator say “it was a time of reckonin’” for her now? Why is it ironical?
Answer:
The narrator feels that it was time for her to face the consequences of teasing her mother in the past. It is ironical because in the past she had made fun of her mother’s false teeth and now, she would soon have to get them for herself, too.

Question 10.
What is the rhyme scheme of the poem? Is the title appropriate? Justify your answer.
Answer:
The rhyme scheme is aabba. Yes, the title is appropriate because it expresses the feeling of remorse that the narrator experiences when she has to visit a dentist to treat her decaying teeth. It also highlights her guilt at not having taken proper care of them.

Oh, I Wish I’d Looked After Me Teeth Long Answer Type

Question 1.
Read the following statement where “I” refers to “you” “I can’t afford to, after what Jack’s done to his teeth.” What is it, you think you can’t afford and why? Write a diary entry of not less than 125 words.
Answer:
24 August 20xx
Jack, my friend had not come to school today, so I just dropped in to see him after school. He was in bed with the left side of his face all swollen and in pain. He had a toothache. I never thought tooth aches could be so painful!
He had gone to the dentist who had extracted one of his teeth on the lower jaw because it had cavities. He told Jack it was because he ate so many toffees and did not brush his teeth properly. He also told him that he would have to take out two more teeth after the swelling came down!

I am frightened! I love eating sweets as much as Jack does. But I suppose I can’t afford to like them so much anymore, not after seeing the pain that Jack is suffering. I will have to resist the temptation and cut down on the number of chocolates and toffees I eat. Also, I will have to brush my teeth with greater care if I don’t want cavities in my teeth!

Question 2.
In line 35, the poet has misspelled the word “amalgum”. Why do you think she has done that? Discuss.
Answer:
The word has been misspelt deliberately to create a pun with the word “gum”. On one hand, the word “amalgam” refers to the mixture of mercury and filling used by the dentist to make fillings while the word “gum” refers to the tissues in the jaw area in which the teeth lie embedded.

Oh, I Wish I’d Looked After Me Teeth Reference to Context

Read the extracts given below and answer the following questions.

Question 1.
“Oh, I wish I’d looked after me teeth,
And spotted the perils beneath.
All the toffees I chewed,
And the sweet sticky food,
Oh, I wish I’d looked after me teeth.”

(a) What does the narrator wish for in the first line?
Answer:
The narrator wishes she had taken more care of her teeth.

(b) What “perils” did the narrator face?
Answer:
The “perils” that the narrator faced was the threat of tooth decay and cavities.

(c) What had given rise to these perils?
Answer:
Eating too many sweets had given rise to these perils.

Question 2.
“I wish I’d been that much more willin’
When I had more tooth there than fillin’
To pass up gobstoppers.
From respect to me choppers,
And to buy something else with me shillin”

(a) Explain: “When I had more tooth there than fillin”
Answer:
The narrator talks about a time when the narrator did not have so many cavities in her teeth and did not require so much filling.

(b) What are gobstoppers?
Answer:
Gobstoppers are a type of hard sweet or toffee which is usually round.

(c) Why should she have given up gobstoppers?
Answer:
She should have given up gobstoppers to protect her teeth.

Question 3.
“I wish I’d been that much more willin’
When I had more tooth there than fillin’
To pass up gobstoppers.
From respect to me choppers,
And to buy something else with me shillin”

(a) What is a shillin’?
Answer:
A shilling is a coin that was used in United Kingdom earlier.

(b) Explain “To pass up gobstoppers”?
Answer:
To stop buying or eating gobstoppers.

(c) What is the feeling expressed by the narrator in these lines?
Answer:
The feeling expressed by the narrator is the given lines, is that of regret, guilt and remorse.

Question 4.
“When I think of the lollies I licked, 
And the liquorice all- sorts I picked,
Sherbet dabs, big and little,
All that hard peanut brittle,
My conscience gets horribly pricked.”

(a) What does the line—“My conscience gets horribly pricked” signify?
Answer:
The given line signifies that the narrator is feeling guilty and remorseful.

(b) Why has the narrator listed the sweets she ate?
Answer:
The narrator listed all the sweets she ate to show that she had eaten all kinds of sweets.

(c) What has been the result of the narrator’s fondness for sweets?
Answer:
The result of the narrator’s fondness for sweets is that now she is suffering from tooth decay and cavities and her mouth is full of fillings.

Question 5.
“Oh I showed them the toothpaste all right,
I flashed it about late at night,
But up-and-down brushin’ And pokin’ and fussin’
Didn’t seem worth the time—I could bite!”

(a) Explain “Showed them the toothpaste”
Answer:
She bloodshed her teeth.

(b) Explain “pokin’ and fussin”
Answer:
The terms indicate brushing carefully.

Question 6.
“If I’d known, I was paving the way
To cavities, caps and decay,
The murder of fillin’s Injections and drillin’s,
I’d have thrown all me sherbet away”

(a) What are the narrator’s feelings regarding her visits to the dentist?
Answer:
The narrator finds the visits uncomfortable.

(b) What was it that the narrator did not realise when she ate those sweets?
Answer:
The narrator says she did not know about the damage the sweets would cause her teeth.

(c) Explain ‘paving the way’?
Answer:
Paving the way means making way or in this context it means that all the sweets the narrator was eating was making way for cavities and decay to set in.

Question 7.
“So I lay in the old dentist’s chair, ‘
And I gaze up his nose in despair,
And his drill it do whine,
In these molars of mine.
‘Two amalgum,’ he’ll say, ‘for in there’.”

(a) Why i.s the narrator lying in the “old dentist’s chair”?
Answer:
The narrator is lying in the “old dentist’s chair” as she is getting treatment for her tooth decay.

(b) Explain “drill it do whine”.
Answer:
The dentist’s drill makes a loud whining sound.

(c) What is the dentist doing to her molars?
Answer:
The dentist is drilling her tooth and filling it with tooth filling, he is extracting a rotten tooth and planting a tooth in its place.

Question 8.
“How I laughed at my mother’s false teeth,
As they foamed in the waters beneath.
But now comes the reckonin’
It’s me they are beckonin’
Oh, I wish I’d looked after me teeth.”

(a) How had the narrator behaved when her mother had lost all her teeth?
Answer:
The narrator laughed and made fun of her false teeth.

(b) Explain “As they foamed in the waters beneath”.
Answer:
The narrator’s mother had her false teeth preserved in the water.

(c) “But now comes the reckonin’/It’s me they are beckonin’”. Explain.
Answer:
The narrator feels she will also have to wear false teeth like her mother.