NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.3 Solutions

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NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3 Solutions are based on Triangles on the Same Base and Between the Same Parallel lines.

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Exercise Solutions of Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles 

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CBSE X Related Questions

  • 1.
    The dimensions of a window are $156\text{ cm} \times 216\text{ cm}$. 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.


      • 2.
        A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is $3/5$, then find the number of yellow balls.


          • 3.
            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.


              • 4.
                In the given figure, $AB \parallel DE$ and $AC \parallel DF$. Show that $\Delta ABC \sim \Delta DEF$. If $BC = 10\text{ cm}$, $EB = CF = 5\text{ cm}$ and $AB = 7\text{ cm}$, then find the length $DE$.


                  • 5.
                    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.

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

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