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


        • 3.
          A kite is flying at a height of \(60 \text{ m}\) above the ground level. Ravi, standing at the roof of the house is holding the string straight and observes the angle of elevation of kite as \(30^{\circ}\). From the bottom of the same building, the angle of elevation of kite is \(45^{\circ}\). Find the length of the string and height of roof from the ground. (Use \(\sqrt{3} = 1.73\))


            • 4.
              In the given figure, point D divides the side BC of $\Delta ABC$ in the ratio $1 : 2$. Find length AD.


                • 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.
                      \(ABCD\) 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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