NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.3 Solutions

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NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.3 Solutions are based on two theorems, (i) The perpendicular from the center of a circle to a chord is seen to bisect the chord. (ii) The line drawn through the center of a circle to bisect a chord is seen to be perpendicular to it.

Download PDF: NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.3 Solutions

Check out NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.3 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.
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
3.
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
4.
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
5.
It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?
- \(\frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle EDF} = \frac{AB}{ED}\)
- \(\frac{AB}{ED} = \frac{AC}{EF}\)
- \(\angle A = \angle D, \angle C = \angle F\)
- \(\frac{AB + BC}{AC} = \frac{ED + DF}{EF}\)
View Solution
6.
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

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NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.3 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.
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

3.
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 :

4.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

5.
It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

6.
The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are

Similar Mathematics Concepts

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NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.3 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.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

3.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 :

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

5.It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

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

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.
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

3.
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 :

4.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

5.
It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

6.
The graph of a quadratic polynomial f(x) passes through (5,0), (0, -1) and (-2, 0). The two factors of the polynomial are