NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.1

Collegedunia Team logo

Collegedunia Team

Content Curator

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.1 Solutions are based on Division of a Line Segment and Solved examples based on the concept. 

Download PDF NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.1 Solutions

Check out the solutions of NCERT Class 10 Maths Chapter 11 Constructions Exercise 11.1

Read More: NCERT Solutions For Class 10 Maths Chapter 11 Constructions

Exercise Solutions of Class 10 Maths Chapter 11 Constructions

Also check other Exercise Solutions of Class 10 Maths Chapter 11 Constructions

Also Check:

Also check:

CBSE X Related Questions

  • 1.
    There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

      • 144
      • 2
      • 420
      • 272

    • 2.
      If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

        • \(4 \text{ cm}\)
        • \(4\sqrt{2} \text{ cm}\)
        • \(8 \text{ cm}\)
        • \(2\sqrt{2} \text{ cm}\)

      • 3.
        Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)


          • 4.
            In the given figure, PA and PB are tangents to a circle centred at O. If \( \angle AOB = 130^\circ \), then \( \angle APB \) is equal to :

              • \( 130^\circ \)
              • \( 50^\circ \)
              • \( 120^\circ \)
              • \( 90^\circ \)

            • 5.
              D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.


                • 6.
                  In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

                    Comments


                    No Comments To Show