NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.1 Solutions

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NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.1 Solutions are based on the introduction to Circles and related terminologies.

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Check out NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.1 Solutions

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
Cyclic quadrilateral	Semi Circle	Tangent circle formula
Circles Important Question	Circles Revision Notes	Tangent to a Circle

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CBSE X Related Questions

1.
If \( \alpha \) and \( \beta \) are two zeroes of a polynomial \( f(x) = px^2 - 2x + 3p \) and \( \alpha + \beta = \alpha\beta \), then value of p is :
- \( -\frac{2}{3} \)
- \( \frac{2}{3} \)
- \( \frac{1}{3} \)
- \( -\frac{1}{3} \)
View Solution
2.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is
- \(0\)
- \(\frac{80}{81}\)
- \(1\)
- \(\frac{82}{80}\)
View Solution
3.
The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are
- \((x + 2), (x - 5)\)
- \((x + 5), (x - 2)\)
- \((x + 1), (x - 5)\)
- \((x - 1), (x + 2)\)
View Solution
4.
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
- \(4 \text{ cm}\)
- \(4\sqrt{2} \text{ cm}\)
- \(8 \text{ cm}\)
- \(2\sqrt{2} \text{ cm}\)
View Solution
5.
The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is
- 2
- \(-6\)
- \(-2\)
- 6
View Solution
6.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is
- \(\frac{7}{2}\)
- \(\frac{6}{7}\)
- \(\frac{-7}{6}\)
- \(\frac{7}{6}\)
View Solution

View All

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NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.1 Solutions

Exercise Solutions of Class 9 Maths Chapter 10 Circles

CBSE X Related Questions

1.
If \( \alpha \) and \( \beta \) are two zeroes of a polynomial \( f(x) = px^2 - 2x + 3p \) and \( \alpha + \beta = \alpha\beta \), then value of p is :

2.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

3.
The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are

4.
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

5.
The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

6.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

Similar Mathematics Concepts

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NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.1 Solutions

Exercise Solutions of Class 9 Maths Chapter 10 Circles

CBSE X Related Questions

1.If \( \alpha \) and \( \beta \) are two zeroes of a polynomial \( f(x) = px^2 - 2x + 3p \) and \( \alpha + \beta = \alpha\beta \), then value of p is :

2.For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

3.The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are

4.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

5.The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

6.If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If \( \alpha \) and \( \beta \) are two zeroes of a polynomial \( f(x) = px^2 - 2x + 3p \) and \( \alpha + \beta = \alpha\beta \), then value of p is :

2.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

3.
The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are

4.
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

5.
The value of k for which sum of the zeroes of the polynomial \(p(x) = 3x^2 - kx + 6\) is 2, is

6.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is