NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1

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NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.1 Solutions are based on Perimeter And Area Of A Circle.

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Exercise Solutions of Class 10 Maths Chapter 12 Areas Related to Circles

Also check other Exercise Solutions of Class 10 Maths Chapter 12 Areas Related to Circles

NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.2 Solutions

NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.3 Solutions

Also Check:

Chapter Related Topics
Circumference of Circle	Areas Related to Circles Formula	Areas Related to Circles Revision Notes
Area of Sector	Sector Of A Circle	Area of Segment of a Circle
Circle Formula	Area of a Circle Formula	Arc Length Formula
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Also Check:

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CBSE X Related Questions

1.
There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?
- 144
- 2
- 420
- 272
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.
In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is
- $55^\circ$
- $20^\circ$
- $35^\circ$
- $70^\circ$
View Solution
5.
If for a data, median is 5 and mode is 4, then mean is equal to :
- 7
- 11
- $\frac{11}{2}$
- $\frac{14}{3}$
View Solution
6.
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

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NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2

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You can also check

NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1

Exercise Solutions of Class 10 Maths Chapter 12 Areas Related to Circles

CBSE X Related Questions

1.
There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

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.
In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is

5.
If for a data, median is 5 and mode is 4, then mean is equal to :

6.
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.

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.1

Exercise Solutions of Class 10 Maths Chapter 12 Areas Related to Circles

CBSE X Related Questions

1.There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

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.In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is

5.If for a data, median is 5 and mode is 4, then mean is equal to :

6.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.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
There are two sections A and B of Grade X. There are 28 students in Section A and 30 students in Section B. What is the minimum number of books you will acquire for the class library so that they can be distributed equally among students of Section A or Section B ?

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.
In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is

5.
If for a data, median is 5 and mode is 4, then mean is equal to :

6.
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.