NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.2 Solutions

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NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.2 Solutions are based on two theorems, which are, (i) Equal circle chords subtend equal angles at the center. (ii) If the angles subtended by the chords at the center tend to be equal, then the chords are equal too.

Download PDF: NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.2 Solutions

Check out NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.2 Solutions

Read More: NCERT Solutions For Class 9 Maths Chapter 10 Circles

Exercise Solutions of Class 9 Maths Chapter 10 Circles

Also check other Exercise Solutions of Class 9 Maths Chapter 10 Circles

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Important Chapter Related Topics
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CBSE X Related Questions

1.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)
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2.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :
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The HCF of 960 and 432 is :
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In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.
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5.
If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).
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6.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals
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