Infinitely many circles can pass through two given points. Among them, the smallest circle is formed when the two points become the endpoints of the diameter. In this case, the centre lies at the midpoint of the line segment joining the points. Therefore, the least possible radius is equal to half tRead more
Infinitely many circles can pass through two given points. Among them, the smallest circle is formed when the two points become the endpoints of the diameter. In this case, the centre lies at the midpoint of the line segment joining the points. Therefore, the least possible radius is equal to half the distance between the two given points.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
No point can be equally distant from three collinear points. The perpendicular bisectors of AB and BC remain parallel, so they never meet to form a circumcentre. Without a common circumcentre, a circle cannot pass through all three collinear points. Also, a straight line can intersect a circle at aRead more
No point can be equally distant from three collinear points. The perpendicular bisectors of AB and BC remain parallel, so they never meet to form a circumcentre. Without a common circumcentre, a circle cannot pass through all three collinear points. Also, a straight line can intersect a circle at a maximum of two distinct points, never at three distinct points.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
Yes, it is possible to construct other triangles congruent to the given triangle on the same circumcircle. These triangles have identical side lengths and angles but occupy different positions on the circle. Although their orientation changes, they remain congruent and share the same circumcircle.Read more
Yes, it is possible to construct other triangles congruent to the given triangle on the same circumcircle. These triangles have identical side lengths and angles but occupy different positions on the circle. Although their orientation changes, they remain congruent and share the same circumcircle.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
We factor this variable fraction expression by recognizing the underlying square identity. The first term is the square of p/4, and the final term 16/p square is the square of 4/p. The middle term is negative, and when we calculate two times p/4 times 4/p, the variables cancel out perfectly to leaveRead more
We factor this variable fraction expression by recognizing the underlying square identity. The first term is the square of p/4, and the final term 16/p square is the square of 4/p. The middle term is negative, and when we calculate two times p/4 times 4/p, the variables cancel out perfectly to leave just the number two. This gives us the final factored expression (p/4 – 4/p) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
We observe that this long expression has six separate terms, which indicates it comes from a three-term square identity. The three perfect square terms are m square/9, k square/4 and 9n square, which are the squares of m/3, k/2 and 3n respectively. The remaining three terms perfectly match the doublRead more
We observe that this long expression has six separate terms, which indicates it comes from a three-term square identity. The three perfect square terms are m square/9, k square/4 and 9n square, which are the squares of m/3, k/2 and 3n respectively. The remaining three terms perfectly match the double cross-products of these bases. Therefore, the complete factored form is (m/3 + k/2 + 3n) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
What is the least possible radius of a circle through two points A and B?
Infinitely many circles can pass through two given points. Among them, the smallest circle is formed when the two points become the endpoints of the diameter. In this case, the centre lies at the midpoint of the line segment joining the points. Therefore, the least possible radius is equal to half tRead more
Infinitely many circles can pass through two given points. Among them, the smallest circle is formed when the two points become the endpoints of the diameter. In this case, the centre lies at the midpoint of the line segment joining the points. Therefore, the least possible radius is equal to half the distance between the two given points.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See lessA, B and C are three collinear points. Can you find a point P such that PA = PB = PC? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?
No point can be equally distant from three collinear points. The perpendicular bisectors of AB and BC remain parallel, so they never meet to form a circumcentre. Without a common circumcentre, a circle cannot pass through all three collinear points. Also, a straight line can intersect a circle at aRead more
No point can be equally distant from three collinear points. The perpendicular bisectors of AB and BC remain parallel, so they never meet to form a circumcentre. Without a common circumcentre, a circle cannot pass through all three collinear points. Also, a straight line can intersect a circle at a maximum of two distinct points, never at three distinct points.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See lessThe circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?
Yes, it is possible to construct other triangles congruent to the given triangle on the same circumcircle. These triangles have identical side lengths and angles but occupy different positions on the circle. Although their orientation changes, they remain congruent and share the same circumcircle.Read more
Yes, it is possible to construct other triangles congruent to the given triangle on the same circumcircle. These triangles have identical side lengths and angles but occupy different positions on the circle. Although their orientation changes, they remain congruent and share the same circumcircle.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See lessFactor using suitable identities: p square/16 – 2 + 16/p square
We factor this variable fraction expression by recognizing the underlying square identity. The first term is the square of p/4, and the final term 16/p square is the square of 4/p. The middle term is negative, and when we calculate two times p/4 times 4/p, the variables cancel out perfectly to leaveRead more
We factor this variable fraction expression by recognizing the underlying square identity. The first term is the square of p/4, and the final term 16/p square is the square of 4/p. The middle term is negative, and when we calculate two times p/4 times 4/p, the variables cancel out perfectly to leave just the number two. This gives us the final factored expression (p/4 – 4/p) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessFactor using suitable identities: m square/9 + mk/3 + k square/4 + 3nk + 2mn + 9n square
We observe that this long expression has six separate terms, which indicates it comes from a three-term square identity. The three perfect square terms are m square/9, k square/4 and 9n square, which are the squares of m/3, k/2 and 3n respectively. The remaining three terms perfectly match the doublRead more
We observe that this long expression has six separate terms, which indicates it comes from a three-term square identity. The three perfect square terms are m square/9, k square/4 and 9n square, which are the squares of m/3, k/2 and 3n respectively. The remaining three terms perfectly match the double cross-products of these bases. Therefore, the complete factored form is (m/3 + k/2 + 3n) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See less