NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4

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Jasmine Grover

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4 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. This chapter covers important concepts related to triangles. Exercise 6.4 mainly includes questions based on areas of similar triangles.

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Check below the NCERT solutions pdf for Class 10 Maths Chapter 6 Exercise 6.4

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

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

    • 2.
      In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

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

      • 3.
        Which of the following graphs represents a polynomial with both zeroes being positive?


        • 4.
          The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

            • 2
            • \(-6\)
            • \(-2\)
            • 6

          • 5.
            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

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

            • 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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