NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.4

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Jasmine Grover

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NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.4 is given in this article. Class 10 Maths Chapter 2 Exercise 2.3 has 5 exercise questions that cover various important concepts of polynomials such as degree of a polynomial, zeroes of a polynomial and roots of polynomials

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Class 10 Chapter 2 Polynomials Topics:

CBSE Class 10 Maths Study Guides:

CBSE X Related Questions

  • 1.
    The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.


      • 2.
        Assertion (A) : The system of linear equations $3x - 5y + 7 = 0$ and $-6x + 10y + 14 = 0$ is inconsistent.
        Reason (R) : When two linear equations don't have unique solution, they always represent parallel lines.

          • 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.
          PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.


            • 4.
              In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.


                • 5.
                  A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.


                    • 6.
                      Prove that: $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$

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