NCERT Solutions for Class 9 Maths Chapter 5 Exercise 5.1 | 2026-27

Exercise Set 5.1 

Question 1.
Draw ∆ABC with AB = 5 cm, ∠A = 70°, and ∠B = 60°. Draw the circumcircle of ∆ABC. Is the centre inside or outside the triangle?

Solution:
Given:

AB = 5 cm
∠A = 70°
∠B = 60°

Steps of Construction:

1. Draw a line segment AB of length 5 cm.

2. At point A, construct an angle of 70° using a protractor.

3. At point B, construct an angle of 60°.

4. Let the two rays meet at point C. Join AC and BC to form ∆ABC.

5. Draw the perpendicular bisectors of any two sides of the triangle, such as AB and AC.

6. Mark their point of intersection as O. This point is the circumcentre of ∆ABC.

7. With O as the centre and OA as the radius, draw a circle passing through A, B, and C.

Since all the angles of ∆ABC are less than 90°, it is an acute-angled triangle. Therefore, its circumcentre O lies inside the triangle.

Question 2.
Draw ∆ABC with AB = 5 cm, ∠A = 100°, and AC = 4 cm. Draw the circumcircle of ∆ABC. Is the centre inside or outside the triangle?

Solution:

Given:

AB = 5 cm
∠A = 100°
AC = 4 cm

Steps of Construction:

1. Draw a line segment AB of length 5 cm.

2. At point A, construct an angle of 100°.

3. With A as the centre and a radius of 4 cm, draw an arc cutting the second arm of the angle at C.

4. Join B and C to complete ∆ABC.

5. Draw the perpendicular bisectors of any two sides, such as AB and AC.

6. Let the perpendicular bisectors meet at O. This point is the circumcentre.

7. With O as the centre and OA as the radius, draw a circle passing through A, B, and C.

Since ∠A = 100°, ∆ABC is an obtuse-angled triangle. Therefore, its circumcentre lies outside the triangle.

Question 3.
Draw ∆ABC, with AB = 6 cm, BC = 7 cm, and CA = 7 cm. Draw the circumcircle of ∆ABC. Let the circumcentre be O. Measure OA, OB, and OC.

Solution:
Given:

AB = 6 cm
BC = 7 cm
CA = 7 cm

Steps of Construction:

1. Draw a line segment AB of length 6 cm.

2. With A as the centre and radius 7 cm, draw an arc.

3. With B as the centre and radius 7 cm, draw another arc intersecting the first arc at C.

4. Join AC and BC to form ∆ABC.

5. Draw the perpendicular bisectors of any two sides of the triangle.

6. Mark their point of intersection as O. This is the circumcentre.

7. With O as the centre and OA as the radius, draw a circle passing through A, B, and C.

The circumcentre is equally distant from all three vertices. Therefore:

OA = OB = OC

On measuring, each distance is approximately 3.87 cm. These equal distances represent the radius of the circumcircle.

Question 4.
What is the least possible radius of a circle through two points A and B?

Solution:
The centre of every circle passing through A and B lies on the perpendicular bisector of AB. The least possible radius is obtained when the centre is the midpoint of AB. In this case, AB becomes the diameter of the circle.

Therefore:

Least possible radius = AB/2

Hence, the smallest circle passing through A and B has a radius equal to half the length of AB.

Think and Reflect (NCERT Textbook Page No. 93)

Question 1.
Jamuna has a circular piece of paper. She is trying to locate its centre. Amina gives her a suggestion. She follows the instructions and is thrilled to find that it works. Can you guess what Amina told her?

Solution:
Amina may have asked Jamuna to fold the circular paper exactly in half and make a crease. This crease represents a diameter of the circle. She should then unfold the paper and fold it again in another direction to form a second diameter. The point where the two creases intersect is the centre of the circular paper.

Think and Reflect (NCERT Textbook Page No. 94)

Question 1.
What are the rotational symmetries of a square? How many lines of reflection symmetry does it have? What about a regular pentagon? A regular hexagon?

Solution:

(a) Square

A square has rotational symmetry of order 4. It matches its original position after rotations of 90°, 180°, 270°, and 360°.

It has 4 lines of reflection symmetry:

2 lines through the midpoints of opposite sides

2 diagonal lines through opposite vertices

(b) Regular Pentagon

A regular pentagon has rotational symmetry of order 5. It matches its original position after rotations of 72°, 144°, 216°, 288°, and 360°.

It has 5 lines of reflection symmetry. Each line passes through a vertex and the midpoint of the opposite side.

(c) Regular Hexagon

A regular hexagon has rotational symmetry of order 6. It matches its original position after rotations of 60°, 120°, 180°, 240°, 300°, and 360°.

It has 6 lines of reflection symmetry:

3 lines through pairs of opposite vertices

3 lines through the midpoints of opposite sides

Question 2.
What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?

Solution:
The longest chord of a circle is its diameter.

Diameter = 2 × Radius

= 2 × 5

= 10 units

Therefore, the longest chord is 10 units long.

There is no smallest chord because two points on the circle can be chosen as close to each other as desired. The chord length can approach zero but cannot become zero for two distinct points.

Question 3.
The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points?

(Hint: We know that any point that is equidistant from two given points A and B lies on the perpendicular bisector of AB. Does this make the perpendicular bisector the locus? For this, we have to show that all the points on the perpendicular bisector are equidistant from A and B.)

Solution:
Let A and B be two fixed points. Any point P that is equally distant from A and B satisfies:

PA = PB

According to the perpendicular bisector theorem, every point lying on the perpendicular bisector of AB is equidistant from A and B. Conversely, every point equidistant from A and B lies on the perpendicular bisector of AB.

Therefore, the locus of all points equidistant from two given points A and B is the perpendicular bisector of the line segment AB.

Think and Reflect (NCERT Textbook Page No. 95)

Question 1.
How many circles pass through two points on a plane?

Solution:
Infinitely many circles can pass through two given points on a plane. The centre of each circle can be chosen at any suitable point on the perpendicular bisector of the line segment joining the two points.

Question 2.
Are there circles of all possible radii passing through A and B? What is the radius of the smallest circle passing through A and B? What is the radius of the largest circle passing through A and B?

Solution:
No, circles of every possible radius cannot pass through A and B. The radius cannot be less than half the distance AB.

The smallest circle is formed when its centre is the midpoint of AB. Its radius is:

AB/2

There is no largest circle. As the centre moves farther away along the perpendicular bisector of AB, the radius continues to increase without any upper limit.

Question 3.
As you move away from segment AB along its perpendicular bisector, do the radii of the circles containing A and B increase or decrease?

Solution:
The centre of every circle passing through A and B lies on the perpendicular bisector of AB. As the centre moves farther away from the midpoint of AB, its distance from A and B increases. Therefore, the radii of the circles increase.

Question 4.
As you go along the perpendicular bisector, will the circle drawn from that point through A and B appear more curved or less curved?

Solution:
As the centre moves farther away along the perpendicular bisector, the radius of the circle becomes larger. A circle with a larger radius appears flatter near A and B. Therefore, the circle will appear less curved.

Question 5.
You are given two points, A and B, on a plane. How many squares can you draw on the same plane with A and B on the boundary? How many squares can you draw on the plane with A and B as the corners of the square?

Solution:

(a) When A and B lie on the boundary:

Infinitely many squares can be drawn because their size, position, and orientation can be varied while keeping both A and B somewhere on the boundary.

(b) When A and B are corners:

Three squares can be drawn.

If A and B are adjacent vertices, two squares can be constructed on opposite sides of AB.

If A and B are opposite vertices, one square can be constructed by taking AB as its diagonal.

Quick Guide for NCERT for Chapter 5 Maths Class 9

Exercise 5.1 Solutions

Exercise 5.1 gives you a chance to learn circle geometry by actually working with figures rather than only reading definitions. You’ll construct triangles and circumcircles, identify the position of a circumcentre, explore chords and symmetry, and use the idea of a locus to understand geometric relationships.

For Class 9 Maths Chapter 5 Exercise 5.1, pay attention to the reason behind each construction. Practising the figures yourself and then checking the Class 9 Maths Chapter 5 Exercise 5.1 solutions can help you spot small errors in measurements, construction steps, or reasoning before exams.

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