NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.1

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Jasmine Grover

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.1 is provided in this article. Class 10 Maths Chapter 6 Triangles cover important concepts like the similarity of triangles, congruence of triangles and Pythagoras theorem. Chapter 6 Triangles Exercise 6.1 mainly includes questions based on the concept of similarity between two figures. 

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CBSE X Related Questions

  • 1.
    If the quadratic equation \(9x^2 + 8kx + 16 = 0\) has real and equal roots, then the value of k is

      • 3
      • –3
      • –4
      • \(\pm 3\)

    • 2.
      Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
      Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

        • 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 the Assertion (A).
        • Assertion (A) is true, but Reason (R) is false.
        • Assertion (A) is false, but Reason (R) is true.

      • 3.
        If \(\alpha, \beta\) are the zeroes of the polynomial \(p(x) = x^2 - 3x - 1\), then find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).


          • 4.
            If the pair of linear equations : \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) is consistent and dependent, then

              • \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
              • \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
              • \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
              • \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)

            • 5.
              The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :

                • $q - 9p$
                • $p - 9q$
                • $p + 9q$
                • $2p + 9q$

              • 6.
                \(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

                  • \(\frac{21}{4} \text{ cm}\)
                  • \(\frac{28}{3} \text{ cm}\)
                  • \(\frac{12}{7} \text{ cm}\)
                  • \(5.5 \text{ cm}\)

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