Ex 5.1 Class 9 Maths Ganita Manjari Solutions | Sid Classes

NCERT Class 9 Maths Ganita Manjari Ch 5 Exercise 5.1 solutions on circle symmetries and circumcircle constructions, explained step by step at Sid Clas

Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.1. Designed to reflect how top-scoring students write their answers in school examinations, this walkthrough makes geometric constructions, circumcircles, and circumcentre properties entirely clear.

🔑 Key Geometric Concepts Used in this Exercise:

  • Circumcircle Construction: The circle passing through all three vertices of a triangle. Its centre (circumcentre) is found by drawing perpendicular bisectors of any two sides of the triangle.
  • Position of Circumcentre: Lies inside for acute-angled triangles, outside for obtuse-angled triangles, and at the midpoint of the hypotenuse for right-angled triangles.
  • Radius Property: The distance from the circumcentre to any of the three vertices is always equal (radius of the circumcircle).

NCERT Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.1 Solutions

Question 1

Draw \(\triangle ABC\) with \(AB = 5\) cm, \(\angle A = 70^\circ\) and \(\angle B = 60^\circ\). Draw the circumcircle of \(\triangle ABC\). Is the centre inside or outside the triangle?

Solution & Construction Steps:

  1. Step 1: Draw a line segment \(AB = 5\) cm using a ruler.
  2. Step 2: At point \(A\), use a protractor to construct an angle of \(70^\circ\). Draw a ray along this angle.
  3. Step 3: At point \(B\), construct an angle of \(60^\circ\). The point where this ray intersects the first ray is vertex \(C\). (Note: \(\angle C = 180^\circ - (70^\circ + 60^\circ) = 50^\circ\)).
  4. Step 4: Draw perpendicular bisectors for side \(AB\) and side \(BC\). Let their intersection point be \(O\) (the circumcentre).
  5. Step 5: With centre \(O\) and radius equal to \(OA\) (or \(OB\), \(OC\)), draw a circle passing through vertices \(A\), \(B\), and \(C\).

Answer: Since all angles of \(\triangle ABC\) are acute (\(70^\circ, 60^\circ, 50^\circ\)), \(\triangle ABC\) is an acute-angled triangle. Therefore, the centre (circumcentre) lies inside the triangle.

A B C O Fig: Circumcircle (Centre Inside)

Question 2

Draw \(\triangle ABC\) with \(AB = 5\) cm, \(\angle A = 100^\circ\), \(AC = 4\) cm. Draw the circumcircle of \(\triangle ABC\). Is the centre inside or outside the triangle?

Solution & Construction Steps:

  1. Step 1: Draw a line segment \(AB = 5\) cm using a ruler.
  2. Step 2: At point \(A\), construct an obtuse angle of \(100^\circ\) using a protractor.
  3. Step 3: Measure a length of \(4\) cm from \(A\) along this ray and mark point \(C\). Join \(B\) to \(C\) to complete \(\triangle ABC\).
  4. Step 4: Construct perpendicular bisectors for side \(AB\) and side \(AC\). Their intersection point \(O\) is the circumcentre.
  5. Step 5: With centre \(O\) and radius \(OA\), draw the circumcircle.

Answer: Since \(\triangle ABC\) has an obtuse angle (\(\angle A = 100^\circ\)), it is an obtuse-angled triangle. Therefore, the centre (circumcentre) lies outside the triangle.

A B C O Fig: Circumcircle (Centre Outside)

Question 3

Draw \(\triangle ABC\), with \(AB = 6\) cm, \(BC = 7\) cm and \(CA = 7\) cm. Draw the circumcircle of \(\triangle ABC\). Let the circumcentre be \(O\). Measure \(OA, OB, OC\).

Solution & Construction Steps:

  1. Step 1: Draw base line segment \(BC = 7\) cm.
  2. Step 2: With \(B\) as centre and radius \(6\) cm, draw an arc. With \(C\) as centre and radius \(7\) cm, draw another arc intersecting the first arc at point \(A\). Join \(AB\) and \(AC\) to form an isosceles triangle.
  3. Step 3: Construct perpendicular bisectors of sides \(AB\) and \(BC\). Mark their intersection point as \(O\).
  4. Step 4: Draw the circumcircle with centre \(O\) passing through \(A\), \(B\), and \(C\).
  5. Step 5: Measure distances \(OA\), \(OB\), and \(OC\) using a ruler.

Observation: Upon measurement, you will find that \(\mathbf{OA = OB = OC}\). This confirms the geometric property that the distance from the circumcentre to all vertices of a triangle is equal to the radius of the circumcircle.

A B C O Fig: Isosceles Triangle Circumcircle

Question 4

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

Solution:

Let the distance between points \(A\) and \(B\) be represented by the line segment \(AB\).
Any circle passing through both points \(A\) and \(B\) must have its chord \(AB\) connecting those two points.
In any circle, the longest possible chord is the diameter. Conversely, for a given chord length \(AB\), the circle of smallest possible size occurs when \(AB\) acts as the exact diameter of the circle.
Therefore, the minimum diameter equals the length of line segment \(AB\), which means the least possible radius is half of that distance.

Answer: \(\mathbf{\frac{1}{2} \times AB}\) (Half of the distance between points \(A\) and \(B\)).

A B Centre Fig: Least Radius (AB as Diameter)

Frequently Asked Questions (FAQs)

Q1: Where does the circumcentre lie in different types of triangles?
A: The circumcentre lies inside for acute-angled triangles, outside for obtuse-angled triangles, and exactly at the midpoint of the hypotenuse for right-angled triangles.

Q2: Why are \(OA, OB,\) and \(OC\) equal in Question 3?
A: By definition, the circumcentre is equidistant from all three vertices of a triangle because each of these distances represents the radius of the circumcircle.

Q3: How do you construct the perpendicular bisector of a line segment?
A: Place the compass pointer on one endpoint, open it to more than half the length of the segment, and draw arcs above and below the line. Repeat from the other endpoint and connect the intersection points of the arcs.

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