Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.1 Solutions – I’m Up and Down, and Round and Round

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.

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.1 Solutions - I’m Up and Down, and Round and Round

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.1 NCERT Solutions

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.

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°.

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.

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.

NCERT Solutions for Class 9 Maths Chapter 5 - Related Links