NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.6 (Optional)

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Jasmine Grover

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.6 are provided in this article. Class 10 Maths Chapter 6 Triangles is included under the Unit Geometry of class 10 maths syllabus. This chapter contains a total of 6 exercises. Exercise 6.6 is an optional exercise which includes questions based on different concepts covered in the chapter.

Download PDF: NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.6 (Optional)


Check below the NCERT solutions pdf for Class 10 Maths Chapter 6 Exercise 6.6 (Optional)

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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.
    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.
      Two water taps together can fill a tank in $8\frac{8}{9}$ hours. The tap of larger diameter takes 4 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.


        • 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.
              \(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

                • \(\frac{21}{4} \text{ cm}\)
                • \(\frac{28}{3} \text{ cm}\)
                • \(\frac{12}{7} \text{ cm}\)
                • \(5.5 \text{ cm}\)

              • 5.
                The HCF of 960 and 432 is :

                  • 48
                  • 54
                  • 72
                  • 36

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

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