NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.5 Solutions

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NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volume Exercise 13.5 Solutions are based on all the topics covered in the chapter, surface areas and volumes. 

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CBSE X Related Questions

  • 1.
    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.

    • 2.
      It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

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

      • 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.
            If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

              • \(-2\)
              • \(\sqrt{\frac{8}{3}}\)
              • \(1\)
              • \(2\)

            • 6.
              For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

                • \(0\)
                • \(\frac{80}{81}\)
                • \(1\)
                • \(\frac{82}{80}\)

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