NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.3

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Jasmine Grover

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.3 are provided in this article. Class 10 Maths Chapter 6 Triangles covers important concepts related to properties of triangles, the similarity between triangles and important theorems. Chapter 6 Exercise 6.3 includes questions mainly based on the congruence of triangles.

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Check below the NCERT solutions pdf for Class 10 Maths Chapter 6 Exercise 6.3

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CBSE X Related Questions

  • 1.
    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

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

    • 2.
      In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

        • \(2.5\) cm
        • \(10.5\) cm
        • \(3.5\) cm
        • \(\frac{14}{3}\) cm

      • 3.
        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.


          • 4.
            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.


              • 5.
                If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).


                  • 6.
                    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 :

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

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