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

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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)

Read Also: NCERT Solutions for Class 10 Maths Chapter 6 Triangles


Find below NCERT solutions of other exercises of class 10 maths chapter 6 Triangles:

Class 10 Chapter 6 Triangles Related Links:

Class 10 Maths Study Guides:

CBSE X Related Questions

  • 1.
    PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is

      • \(a^2 + (a + 2)^2 = (2b)^2\)
      • \(b^2 = a + 4\)
      • \(2a^2 + 1 = b^2\)
      • \(b^2 = a + 1\)

    • 2.
      Three tennis balls are just packed in a cylindrical jar. If radius of each ball is \(r\), volume of air inside the jar is

        • \(2\pi r^3\)
        • \(3\pi r^3\)
        • \(5\pi r^3\)
        • \(4\pi r^3\)

      • 3.
        A trader has three different types of oils of volume \(870 \text{ l}\), \(812 \text{ l}\) and \(638 \text{ l}\). Find the least number of containers of equal size required to store all the oil without getting mixed.


          • 4.
            For any natural number n, \( 5^n \) ends with the digit :

              • 0
              • 5
              • 3
              • 2

            • 5.
              The dimensions of a window are 156 cm \(\times\) 216 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.


                • 6.
                  An ice-cream cone of radius \(r\) and height \(h\) is completely filled by two spherical scoops of ice-cream. If radius of each spherical scoop is \(\frac{r}{2}\), then \(h : 2r\) equals

                    • \(1 : 8\)
                    • \(1 : 2\)
                    • \(1 : 1\)
                    • \(2 : 1\)

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