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

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.
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.
View Solution
2.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is
- 34
- 43
- 38.5
- 41.5
View Solution
3.
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
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 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
6.
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
- \(10^{\circ}\)
- \(60^{\circ}\)
- \(45^{\circ}\)
- \(100^{\circ}\)
View Solution

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

Exercise Solutions of Class 9 Maths Chapter 10 Circles

CBSE X Related Questions

1.
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.

2.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

3.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

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

6.
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

Similar Mathematics Concepts

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

Exercise Solutions of Class 9 Maths Chapter 10 Circles

CBSE X Related Questions

1.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.

2.Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

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

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

6.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

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
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.

2.
Mean and Median of a frequency distribution are 43 and 40 respectively. The value of mode is

3.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

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

6.
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