NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.5

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.5 are provided in this article. Class 10 Maths Chapter 6 Triangles is an important chapter included under the Unit Geometry of class 10 maths syllabus. Exercise 6.5 includes questions based on Pythagoras theorem.

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Check below the NCERT solutions pdf for Class 10 Maths Chapter 6 Exercise 6.5

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Find below NCERT solutions of other exercises of class 10 maths chapter 6 Triangles:


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CBSE X Related Questions

  • 1.
    PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is

      • \(a^2 + (a + 2)^2 = (2b)^2\)
      • \(b^2 = a + 4\)
      • \(2a^2 + 1 = b^2\)
      • \(b^2 = a + 1\)

    • 2.
      Assertion (A) : \((\sqrt{3} + \sqrt{5})\) is an irrational number.
      Reason (R) : Sum of the any two irrational numbers is always irrational.

        • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
        • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
        • Assertion (A) is true, but Reason (R) is false.
        • Assertion (A) is false, but Reason (R) is true.

      • 3.
        A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))

          • \(314 \sqrt{2}\) \(\text{cm}^{2}\)
          • \(314\) \(\text{cm}^{2}\)
          • \(\frac{3140}{3}\) \(\text{cm}^{2}\)
          • \(3140 \sqrt{2}\) \(\text{cm}^{2}\)

        • 4.
          Prove that: \(\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \cdot \csc \theta (\sec \theta + \csc \theta)\)


            • 5.
              Prove that \(2 + 3\sqrt{5}\) is an irrational number given that \(\sqrt{5}\) is an irrational number.


                • 6.
                  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

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

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