Searching for the NCERT Class 9 Maths Ganita Manjari Chapter 1 End of Chapter Exercise Solutions? In this post, we provide comprehensive, step-by-step answers for the final review exercise of Chapter 1. These solutions follow the new 2026-27 NCERT Ganita Manjari textbook and are perfect for students looking to revise the entire chapter before their exams.
Ganita Manjari Class 9 Chapter 1 End of Chapter Exercise Solutions
Question 1 Solution
\What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
Answer:
The x-axis and y-axis intersect at the origin.
Therefore:
x-coordinate = 0
y-coordinate = 0
So, the point of intersection is (0, 0).
Question 2 Solution
Point W has x-coordinate equal to –5. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?
Answer:
If point W has x-coordinate –5, then any point on the line through W parallel to the y-axis will also have x-coordinate –5.
Therefore, the coordinates of H will be of the form:
H = (–5, y), where y can be any real number.
- If y > 0, then H lies in Quadrant II.
- If y < 0, then H lies in Quadrant III.
- If y = 0, then H lies on the x-axis.
So, H can lie in Quadrant II, Quadrant III, or on the x-axis.
Question 3 Solution
Consider the points R (3, 0), A (0, –2), M (–5, –2) and P (–5, 2). If they are joined in the same order, predict:
(i) Two sides of RAMP that are perpendicular to each other.
Answer:
AM joins A(0, –2) to M(–5, –2), so it is horizontal.
MP joins M(–5, –2) to P(–5, 2), so it is vertical.
A horizontal line and a vertical line are perpendicular.
Therefore:
AM ⟂ MP
The two perpendicular sides are AM and MP.
(ii) One side of RAMP that is parallel to one of the axes.
Answer:
AM is parallel to the x-axis because both points A and M have the same y-coordinate (–2).
MP is parallel to the y-axis because both points M and P have the same x-coordinate (–5).
Therefore:
- AM is parallel to the x-axis
- MP is parallel to the y-axis
(iii) Two points that are mirror images of each other in one axis. Which axis will this be?
Answer:
Comparing M(–5, –2) and P(–5, 2):
- They have the same x-coordinate.
- Their y-coordinates are equal in magnitude but opposite in sign.
Therefore, they are mirror images of each other in the x-axis.
The points M and P are mirror images of each other in the x-axis.
Question 4 Solution
Plot point Z (5, –6) on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides.
Answer:
Take:
- I = (5, 0)
- Z = (5, –6)
- N = (0, –6)
Triangle IZN is right-angled because:
- IZ is vertical
- ZN is horizontal
Now finding lengths:
IZ:
= 6 units
ZN:
= 5 units
IN:
= √[(5 – 0)² + (0 – (–6))²]
= √(25 + 36)
= √61 units
Therefore:
- IZ = 6 units
- ZN = 5 units
- IN = √61 units
Question 5 Solution
What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?
Answer:
If we did not have negative numbers, coordinates could only be zero or positive.
Therefore:
- Only Quadrant I could be used properly.
- Negative parts of axes would not exist.
- Quadrants II, III and IV could not be represented.
Hence, such a system would NOT allow us to locate all points on a 2-D plane.
Question 6 Solution
Are the points M (–3, –4), A (0, 0) and G (6, 8) on the same straight line? Suggest a method to check this without plotting and joining the points.
Answer:
Using the distance formula:
MA = √[(0 + 3)² + (0 + 4)²]
= √(3² + 4²)
= √(9 + 16)
= √25 = 5
AG = √[(6 − 0)² + (8 − 0)²]
= √(36 + 64)
= √100 = 10
MG = √[(6 + 3)² + (8 + 4)²]
= √(9² + 12²)
= √(81 + 144)
= √225 = 15
Now:
MA + AG = 5 + 10 = 15 = MG
Therefore, the points M, A and G lie on the same straight line.
Question 7 Solution
Use your method from Problem 6 to check if the points R (–5, –1), B (–2, –5) and C (4, –12) are on the same straight line.
Answer:
Using the distance formula:
RB = √[(–2 + 5)² + (–5 + 1)²]
= √(3² + (–4)²)
= √(9 + 16)
= √25 = 5
BC = √[(4 + 2)² + (–12 + 5)²]
= √(6² + (–7)²)
= √(36 + 49)
= √85
RC = √[(4 + 5)² + (–12 + 1)²]
= √(9² + (–11)²)
= √(81 + 121)
= √202
Now:
RB + BC ≠ RC
Therefore, the points R, B and C do not lie on the same straight line.
Question 8 Solution
(i) A right-angled isosceles triangle
Answer:
One possible set of vertices is:
- O = (0, 0)
- A = (4, 0)
- B = (0, 4)
Since:
- OA = 4 units
- OB = 4 units
- OA ⟂ OB
Therefore, triangle OAB is a right-angled isosceles triangle.
(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV
Answer:
One possible set of vertices is:
- O = (0, 0)
- P = (–3, –4)
- Q = (3, –4)
Since:
- P lies in Quadrant III
- Q lies in Quadrant IV
- OP = OQ = 5 units
Therefore, triangle OPQ is an isosceles triangle.
Question 9 Solution
State whether M is the midpoint of segment ST in each case.
(i) S(−3, 0), M(0, 0), T(3, 0)
SM = 3
MT = 3
Since SM = MT, M is the midpoint.
(ii) S(2, 3), M(3, 4), T(4, 5)
SM = √2
MT = √2
Since SM = MT, M is the midpoint.
(iii) S(0, 0), M(0, 5), T(0, −10)
SM = 5
MT = 15
Since SM ≠ MT, M is NOT the midpoint.
(iv) S(−8, 7), M(0, −2), T(6, −3)
SM = √145
MT = √37
Since SM ≠ MT, M is NOT the midpoint.
Question 10 Solution
Find the coordinates of B given that M (–7, 1) is the midpoint of A (3, –4) and B (x, y).
Answer:
Using midpoint formula:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Given:
- A = (3, –4)
- M = (–7, 1)
- B = (x, y)
For x-coordinate:
(3 + x)/2 = –7
3 + x = –14
x = –17
For y-coordinate:
(–4 + y)/2 = 1
–4 + y = 2
y = 6
Therefore, the coordinates of B are:
B = (–17, 6)
Question 11 Solution
Let P and Q be points of trisection of AB, with P closer to A and Q closer to B. Find the coordinates of P and Q for A (4, 7) and B (16, –2).
Answer:
Given:
- A = (4, 7)
- B = (16, –2)
Since P and Q trisect the line segment AB:
- P divides AB in the ratio 1:2
- Q divides AB in the ratio 2:1
Coordinates of P:
P = (8, 4)
Coordinates of Q:
Q = (12, 1)
Therefore:
- P = (8, 4)
- Q = (12, 1)
Question 12 Solution
(i) Show that A (1, –8), B (–4, 7) and C (–7, –4) lie on a circle centered at the origin.
Answer:
Finding distance from origin:
OA = √(1² + (–8)²)
= √(1 + 64)
= √65
OB = √((–4)² + 7²)
= √(16 + 49)
= √65
OC = √((–7)² + (–4)²)
= √(49 + 16)
= √65
Since all three distances are equal, the points lie on the same circle centered at the origin.
Radius of the circle = √65
(ii) Check whether D (–5, 6) and E (0, 9) lie inside, on, or outside the circle.
Answer:
OD = √((–5)² + 6²)
= √(25 + 36)
= √61
OE = √(0² + 9²)
= 9
Radius = √65 ≈ 8.06
- √61 < √65, so D lies inside the circle.
- 9 > √65, so E lies outside the circle.
Question 13 Solution
The midpoints of the sides of triangle ABC are D(5,1), E(6,5), and F(0,3). Find the coordinates of A, B and C.
Answer:
Let:
- A = (x₁, y₁)
- B = (x₂, y₂)
- C = (x₃, y₃)
Using midpoint formula:
D is midpoint of BC:
x₂ + x₃ = 10
y₂ + y₃ = 2
E is midpoint of CA:
x₃ + x₁ = 12
y₃ + y₁ = 10
F is midpoint of AB:
x₁ + x₂ = 0
y₁ + y₂ = 6
Solving these equations:
A = (1, 7)
B = (–1, –1)
C = (11, 3)
Question 14 Solution
(i) Draw a model of the city.
Answer:
Given:
- Scale: 1 cm = 200 m
- 10 streets in North–South direction
- 10 streets in East–West direction
Draw:
- 10 vertical parallel lines
- 10 horizontal parallel lines
- Each line 1 cm apart
This forms a square grid representing the city.
(ii)(a) How many street intersections can be referred to as (4, 3)?
Answer:
Only one intersection can be called (4, 3).
(ii)(b) How many street intersections can be referred to as (3, 4)?
Answer:
Only one intersection can be called (3, 4).
Question 15 Solution
A computer graphics screen is 800 pixels wide and 600 pixels high. Two circular icons are drawn. Determine whether any part lies outside the screen and whether the circles intersect.
Answer:
- Circle A: Center = (100,150), Radius = 80
- Circle B: Center = (250,230), Radius = 100
(i) Both circles lie completely inside the screen.
(ii) Yes, the two circles intersect each other.
Question 16 Solution
Plot the points A (2,1), B (–1,2), C (–2,–1), and D (1,–2). Is ABCD a square? Find its area.
Answer:
Using distance formula:
AB = √[(–1 − 2)² + (2 − 1)²]
= √(9 + 1)
= √10
BC = √10
CD = √10
DA = √10
All four sides are equal.
Now diagonals:
AC = √20
BD = √20
Diagonals are equal.
Therefore, ABCD is a square.
Area of square:
= side²
= (√10)²
= 10 square units
Conclusion
We hope these Ganita Manjari Class 9 Chapter 1 End-of-Chapter Exercise solutions helped you understand the chapter better. Keep visiting Sid Classes for more Class 9 Maths solutions.