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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CBSE X Related Questions

  • 1.
    It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

      • \(\frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle EDF} = \frac{AB}{ED}\)
      • \(\frac{AB}{ED} = \frac{AC}{EF}\)
      • \(\angle A = \angle D, \angle C = \angle F\)
      • \(\frac{AB + BC}{AC} = \frac{ED + DF}{EF}\)

    • 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.
        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

        • 4.
          If \( \alpha \) and \( \beta \) are two zeroes of a polynomial \( f(x) = px^2 - 2x + 3p \) and \( \alpha + \beta = \alpha\beta \), then value of p is :

            • \( -\frac{2}{3} \)
            • \( \frac{2}{3} \)
            • \( \frac{1}{3} \)
            • \( -\frac{1}{3} \)

          • 5.
            Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

              • 34
              • 43
              • 38.5
              • 41.5

            • 6.
              In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

                • $6$
                • $4$
                • $\pm 4$
                • $3$

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