In a triangle, any vertex can be found by adding the two adjacent midpoints and subtracting the opposite midpoint. To find vertex A (opposite D), we calculate A = E + F – D, resulting in (1, 7). To find vertex B (opposite E), we calculate B = D + F – E, resulting in (–1, –1). To find vertex C (opposRead more
In a triangle, any vertex can be found by adding the two adjacent midpoints and subtracting the opposite midpoint. To find vertex A (opposite D), we calculate A = E + F – D, resulting in (1, 7). To find vertex B (opposite E), we calculate B = D + F – E, resulting in (–1, –1). To find vertex C (opposite F), we calculate C = D + E – F, resulting in (11, 3). These are the three vertices.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
(i) The city is represented as a coordinate grid where North-South streets are vertical lines and East-West streets are horizontal lines. (ii) (a) Based on the convention that an ordered pair represents a specific crossing, only one unique intersection exists for the point (4, 3), where the 4th NortRead more
(i) The city is represented as a coordinate grid where North-South streets are vertical lines and East-West streets are horizontal lines. (ii) (a) Based on the convention that an ordered pair represents a specific crossing, only one unique intersection exists for the point (4, 3), where the 4th North-South street meets the 3rd East-West street. (b) Likewise, there is only one unique intersection referred to as (3, 4). These represent two different physical locations in the city.
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Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
(i) Both circles are within the screen because their centers plus or minus their radii stay between 0 and 800 for width and 0 and 600 for height. (ii) The distance between center A (100, 150) and center B (250, 230) is calculated as 170 pixels. The sum of their radii is 80 plus 100, which equals 180Read more
(i) Both circles are within the screen because their centers plus or minus their radii stay between 0 and 800 for width and 0 and 600 for height. (ii) The distance between center A (100, 150) and center B (250, 230) is calculated as 170 pixels. The sum of their radii is 80 plus 100, which equals 180 pixels. Since the distance between the centers is less than the combined radii, the two circular icons must overlap or intersect.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
By calculating the distances between the points, we find that all four sides are equal to the square root of 10 units. To confirm it is a square, we check the diagonals: both AC and BD are equal to the square root of 20 units. Since all sides are equal and both diagonals are equal, ABCD is a perfectRead more
By calculating the distances between the points, we find that all four sides are equal to the square root of 10 units. To confirm it is a square, we check the diagonals: both AC and BD are equal to the square root of 20 units. Since all sides are equal and both diagonals are equal, ABCD is a perfect square. The area is calculated by squaring the side length, which gives a total of 10 square units.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
CBSE Class 9 Maths Ganita Manjari (2026–27) is a newly introduced textbook under NEP 2020. It focuses on activity-based and conceptual learning. The book covers coordinates, linear polynomials, number systems, algebraic identities, integers, mensuration, probability and sequences. It helps studentsRead more
CBSE Class 9 Maths Ganita Manjari (2026–27) is a newly introduced textbook under NEP 2020. It focuses on activity-based and conceptual learning. The book covers coordinates, linear polynomials, number systems, algebraic identities, integers, mensuration, probability and sequences. It helps students understand mathematical concepts through real-life situations, logical reasoning and problem-solving approaches, making learning more engaging and meaningful for better understanding.
Chapter Names:
Chapter 1 – Orienting Yourself: The Use of Coordinates
Chapter 2 – Introduction to Linear Polynomials
Chapter 3 – The World of Numbers
Chapter 4 – Exploring Algebraic Identities
Chapter 5 – I’m Up and Down and Round and Round
Chapter 6 – Measuring Space: Perimeter and Area
Chapter 7 – The Mathematics of Maybe: Introduction to Probability
Chapter 8 – Predicting What Comes Next: Exploring Sequences and Progressions
For more NCERT Solutions of Class 9 Maths Ganita Manjari All Chapter Solutions (2026-27):
The midpoints of the sides of triangle ABC are the points D, E and F. Given that the coordinates of D, E and F are (5, 1), (6, 5) and (0, 3), respectively, find the coordinates of A, B and C.
In a triangle, any vertex can be found by adding the two adjacent midpoints and subtracting the opposite midpoint. To find vertex A (opposite D), we calculate A = E + F – D, resulting in (1, 7). To find vertex B (opposite E), we calculate B = D + F – E, resulting in (–1, –1). To find vertex C (opposRead more
In a triangle, any vertex can be found by adding the two adjacent midpoints and subtracting the opposite midpoint. To find vertex A (opposite D), we calculate A = E + F – D, resulting in (1, 7). To find vertex B (opposite E), we calculate B = D + F – E, resulting in (–1, –1). To find vertex C (opposite F), we calculate C = D + E – F, resulting in (11, 3). These are the three vertices.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
See lessA city has two main roads which cross each other at the centre of the city. These two roads are along the North–South (N–S) direction and East–West (E–W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction. (i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines. (ii) There are street intersections in the model. Each street intersection is formed by two streets — one running in the N–S direction and another in the E–W direction. Each street intersection is referred to in the following manner: If the second street running in the N–S direction and 5th street in the E–W direction meet at some crossing, then we call this street intersection (2, 5). Using this convention, find: (a) how many street intersections can be referred to as (4, 3). (b) how many street intersections can be referred to as (3, 4).
(i) The city is represented as a coordinate grid where North-South streets are vertical lines and East-West streets are horizontal lines. (ii) (a) Based on the convention that an ordered pair represents a specific crossing, only one unique intersection exists for the point (4, 3), where the 4th NortRead more
(i) The city is represented as a coordinate grid where North-South streets are vertical lines and East-West streets are horizontal lines. (ii) (a) Based on the convention that an ordered pair represents a specific crossing, only one unique intersection exists for the point (4, 3), where the 4th North-South street meets the 3rd East-West street. (b) Likewise, there is only one unique intersection referred to as (3, 4). These represent two different physical locations in the city.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
See lessA computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine: (i) whether any part of either circle lies outside the screen. (ii) whether the two circles intersect each other.
(i) Both circles are within the screen because their centers plus or minus their radii stay between 0 and 800 for width and 0 and 600 for height. (ii) The distance between center A (100, 150) and center B (250, 230) is calculated as 170 pixels. The sum of their radii is 80 plus 100, which equals 180Read more
(i) Both circles are within the screen because their centers plus or minus their radii stay between 0 and 800 for width and 0 and 600 for height. (ii) The distance between center A (100, 150) and center B (250, 230) is calculated as 170 pixels. The sum of their radii is 80 plus 100, which equals 180 pixels. Since the distance between the centers is less than the combined radii, the two circular icons must overlap or intersect.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
See lessPlot the points A (2, 1), B (–1, 2), C (–2, –1) and D (1, –2) in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?
By calculating the distances between the points, we find that all four sides are equal to the square root of 10 units. To confirm it is a square, we check the diagonals: both AC and BD are equal to the square root of 20 units. Since all sides are equal and both diagonals are equal, ABCD is a perfectRead more
By calculating the distances between the points, we find that all four sides are equal to the square root of 10 units. To confirm it is a square, we check the diagonals: both AC and BD are equal to the square root of 20 units. Since all sides are equal and both diagonals are equal, ABCD is a perfect square. The area is calculated by squaring the side length, which gives a total of 10 square units.
For Detailed Solutions:
Visit NCERT Solutions for Class 9 Ganita Manjari Chapter 1 Orienting Yourself: The Use of Coordinates Question Answer:
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-1/
See lessNCERT Solutions for Class 9 Maths Ganita Manjari for 2026-27.
CBSE Class 9 Maths Ganita Manjari (2026–27) is a newly introduced textbook under NEP 2020. It focuses on activity-based and conceptual learning. The book covers coordinates, linear polynomials, number systems, algebraic identities, integers, mensuration, probability and sequences. It helps studentsRead more
CBSE Class 9 Maths Ganita Manjari (2026–27) is a newly introduced textbook under NEP 2020. It focuses on activity-based and conceptual learning. The book covers coordinates, linear polynomials, number systems, algebraic identities, integers, mensuration, probability and sequences. It helps students understand mathematical concepts through real-life situations, logical reasoning and problem-solving approaches, making learning more engaging and meaningful for better understanding.
Chapter Names:
Chapter 1 – Orienting Yourself: The Use of Coordinates
Chapter 2 – Introduction to Linear Polynomials
Chapter 3 – The World of Numbers
Chapter 4 – Exploring Algebraic Identities
Chapter 5 – I’m Up and Down and Round and Round
Chapter 6 – Measuring Space: Perimeter and Area
Chapter 7 – The Mathematics of Maybe: Introduction to Probability
Chapter 8 – Predicting What Comes Next: Exploring Sequences and Progressions
For more NCERT Solutions of Class 9 Maths Ganita Manjari All Chapter Solutions (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/
See less