Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.1 Solutions

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.1 Solutions explain how to construct the circumcircle of a triangle using perpendicular bisectors.
These solutions also explain the position of the circumcentre and the least possible radius of a circle through two points.

Chapter 5, I’m Up and Down, and Round and Round, begins with circles seen in nature, such as raindrops on water, plant stems, sunflowers, the moon and the sun during an eclipse. The chapter then moves from visual circular shapes to exact geometry terms like circle, centre, radius, chord, diameter, circumcircle and circumcentre. Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.1 Solutions are based on constructing triangles and drawing their circumcircles. These Class 9 Maths Chapter 5 Exercise 5.1 Solutions help students practise construction steps, angle checking and the logic behind a circle through three non-collinear points. The textbook states that a circle is the set of all points equidistant from a given point, and that point is called the centre.

Key Takeaways

• Circumcircle: A circle passing through all three vertices of a triangle is called its circumcircle.
• Circumcentre: The circumcentre is the point where the perpendicular bisectors of two sides meet.
• Acute Triangle: The circumcentre lies inside an acute-angled triangle.
• Obtuse Triangle: The circumcentre lies outside an obtuse-angled triangle.

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.1 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.1 Solutions for Circumcircle Construction

Exercise Set 5.1 has four questions on drawing triangles, constructing circumcircles, measuring equal radii and finding the least radius through two fixed points. The textbook asks students to draw circumcircles for given triangles and identify whether the centre lies inside or outside the triangle.

Q1. Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?

The centre lies inside the triangle because all three angles of ΔABC are acute.

Steps:

1. Draw AB = 5 cm.
2. At A, draw a ray making ∠A = 70°.
3. At B, draw a ray making ∠B = 60°.
4. Let both rays meet at C.
5. ΔABC is now formed.
6. Find ∠C using the angle sum property.

Equation:

∠A + ∠B + ∠C = 180°

70° + 60° + ∠C = 180°

130° + ∠C = 180°

∠C = 180° - 130°

∠C = 50°

1. Draw the perpendicular bisector of AB.
2. Draw the perpendicular bisector of AC.
3. Let both perpendicular bisectors meet at O.
4. With O as centre and OA as radius, draw a circle.
5. The circle passes through A, B and C.

Answer:

The circumcentre O lies inside ΔABC because 70°, 60° and 50° are all acute angles.

Ganita Manjari Class 9 Chapter 5 Exercise 5.1: Circumcentre Method

A circle through three non-collinear points is unique, so every triangle has one fixed circumcircle. The textbook explains this using perpendicular bisectors of sides and equal distances from the centre.

Q2. Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?

The centre lies outside the triangle because ∠A = 100° makes ΔABC an obtuse-angled triangle.

Steps:

1. Draw AB = 5 cm.
2. At A, draw a ray making ∠A = 100° with AB.
3. On this ray, mark point C such that AC = 4 cm.
4. Join B to C.
5. ΔABC is now formed.
6. Draw the perpendicular bisector of AB.
7. Draw the perpendicular bisector of AC.
8. Let both perpendicular bisectors meet at O.
9. With O as centre and OA as radius, draw the circumcircle.
10. The circle passes through A, B and C.

Answer:

The circumcentre O lies outside ΔABC because one angle is greater than 90°.

Class 9 Maths Ganita Manjari Chapter 5 Solutions: Measuring Equal Radii

The circumcentre is equidistant from all three vertices of a triangle. In Exercise 5.1, this property is checked by measuring OA, OB and OC after drawing the circumcircle.

Q3. Draw ΔABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.

OA, OB and OC are equal because they are radii of the same circumcircle.

Steps:

1. Draw AB = 6 cm.
2. With A as centre and radius 7 cm, draw an arc.
3. With B as centre and radius 7 cm, draw another arc.
4. Let both arcs meet at C.
5. Join AC and BC.
6. ΔABC is now formed.
7. Draw the perpendicular bisector of AB.
8. Draw the perpendicular bisector of AC.
9. Let both perpendicular bisectors meet at O.
10. With O as centre and OA as radius, draw the circumcircle.
11. Measure OA, OB and OC with a scale.

Equation:

OA = OB = OC

Approximate measurement from construction:

OA = OB = OC ≈ 4 cm

Answer:

OA, OB and OC are equal because O is the centre of the circumcircle passing through A, B and C.

Class 9 Maths Chapter 5 Exercise 5.1 Solutions: Radius of Circle Through Two Points

The least radius of circle through two points is obtained when the two points are endpoints of a diameter. The centre of this smallest circle is the midpoint of the line segment joining the two points.

Q4. What is the least possible radius of a circle through two points A and B?

The least possible radius is half of AB.

Steps:

1. A circle through A and B must have its centre equidistant from A and B.
2. The centres of all such circles lie on the perpendicular bisector of AB.
3. The smallest circle is formed when AB is the diameter.
4. The centre is the midpoint of AB.

Equation:

Least possible radius = AB ÷ 2

or

Least possible radius = AB/2

Answer:

The least possible radius of a circle through two points A and B is AB/2.

I’m Up and Down and Round and Round Class 9: Concepts Used in Exercise 5.1

The chapter title I’m Up and Down and Round and Round Class 9 introduces circles through real circular forms before moving to mathematical definitions. The textbook defines radius as the distance from the centre to any point on the circle and chord as a line segment joining two points on a circle.

Circumcircle Class 9

A circumcircle Class 9 question usually asks students to draw a circle passing through all vertices of a triangle.

Key fact:

Every triangle has one circumcircle.

Reason:

The three vertices of a triangle are non-collinear points.

Circumcentre Class 9

A circumcentre Class 9 question usually asks students to locate the centre of the circumcircle.

Key fact:

The circumcentre is the intersection point of perpendicular bisectors of the sides of a triangle.

Equation:

OA = OB = OC

Meaning:

O is equally distant from A, B and C.

Circle Through Three Non-Collinear Points

A circle through three non-collinear points is unique.

Reason:

The perpendicular bisectors of two sides meet at one point.

That point becomes the centre of the circle.

Perpendicular Bisector of Chord

The perpendicular bisector of chord idea is useful because every point on it is equidistant from the chord’s endpoints.

In Exercise 5.1, this idea helps locate the centre of a circle passing through two or three points.

Class 9 Maths Circles Solutions

Class 9 Maths circles solutions in this exercise focus on construction accuracy, equal radii and circumcentre placement.

The main checking step is:

OA = OB = OC

Quick Concept Table for Exercise 5.1

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5