NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.3

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Jasmine Grover

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.3 are provided in this article. Class 10 Maths Chapter 6 Triangles covers important concepts related to properties of triangles, the similarity between triangles and important theorems. Chapter 6 Exercise 6.3 includes questions mainly based on the congruence of triangles.

Download PDF: NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.3


Check below the NCERT solutions pdf for Class 10 Maths Chapter 6 Exercise 6.3

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CBSE X Related Questions

  • 1.
    Three tennis balls are just packed in a cylindrical jar. If radius of each ball is \(r\), volume of air inside the jar is

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

    • 2.
      A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))

        • \(314 \sqrt{2}\) \(\text{cm}^{2}\)
        • \(314\) \(\text{cm}^{2}\)
        • \(\frac{3140}{3}\) \(\text{cm}^{2}\)
        • \(3140 \sqrt{2}\) \(\text{cm}^{2}\)

      • 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.
            Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

              • \(10^{\circ}\)
              • \(60^{\circ}\)
              • \(45^{\circ}\)
              • \(100^{\circ}\)

            • 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.
                  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

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

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