NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles
NCERT solutions for class 10 maths chapter 12 Areas Related to Circles - This chapter is in continuation with the chapter 'Circles' of class 9 which you have already studied. There will be a lot of questions in which class 9 concepts will be used. NCERT solutions for class 10 maths chapter 12 Areas Related to Circles cover the elaborated solutions to each and every question which is available in the practice exercises including the optional exercises. This topic is a part of the mensuration unit, and it holds 10 marks in class 10 board examinations. From the trend of the last years' exams, you can expect at least 1 question from this chapter in the board examinations. With the help of NCERT solutions for class 10 maths chapter 12 Areas Related to Circles, you won't miss any mark if a question directly comes from the practice exercises. In this chapter, you will get to know about the terms like- Sector, Chord, Segment, Radius, Diameter, etc. There is a total of 3 exercises with 35 questions in them. NCERT solutions for class 10 maths chapter 12 Areas Related to Circles are constructed in a manner so that a student can get 100% out of it. Apart from the solutions to this particular chapter, NCERT solutions are also available from class 6 to 12 for science and maths.
NCERT Solutions for Class 10 Maths Chapter 12 Areas related to circles Excercise: 12.1
Answer:
Circumference of 1st circle is given :
And the circumference of the 2nd circle is :
Thus the circumference of the new circle is
or
or
Hence the radius of the new circle is 28 cm.
Answer:
We know that the area of the circle is :
Thus area of a 1st circle :
or
And the area of a 2nd circle is :
or
According to the question area of the new circle is
or
or
Hence the area of the new circle is 10 cm.
Answer:
The radius of each scoring region can be found by adding 10.5 in respective colors.
The area of the golden region is :
Similarly area of the red region is = Area of a red score - Area of the golden region
Then the area of the blue region is :
And the area of the black region is :
Area of the white region is :
Answer:
Let the number of revolution of the wheel be n.
The circumference of the wheel is given by:-
Now, the speed of the car is given by:-
Thus distance traveled in 10 minutes is:
According to the question, we get;
or
Hence the number of revolutions made by the wheel is 4372.
(A) 2 units
(B) units
(C) 4 units
(D) 7 units
Answer:
Let the radius of the circle be r.
Then according to question, we can write:-
Perimeter = Area
Hence option (A) is correct.
NCERT Solutions for Class 10 Maths Chapter 12 Areas related to circles Excercise: 12.2
Q1 Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Answer:
We know that the area of a sector having radius r and angle is given by:-
Thus the area of the given sector is:-
Q2 Find the area of a quadrant of a circle whose circumference is 22 cm.
Answer:
We are given the circumference of the circle.
Thus,
Also, we know that the area of a sector is given by :
It is given that we need to find the area of a quadrant thus
Hence the area becomes:-
Answer:
The minute hand rotates 360 ^{ o } in one hour.
We need to find rotation in 5 min. :-
The area of the sector is given by :
Hence the area swept by minute hand in 5 minutes is _{} .
Q4 A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding : (i) minor segment
Answer:
The angle in the minor sector is 90 ^{ o } .
Thus the area of the sector is given by:-
Now the area of a triangle is:-
Thus the area of minor segment = Area of the sector - Area of a triangle
or
Q4 A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding : (ii) major sector. (Use π = 3.14)
Answer:
The area of the major sector can be found directly by using the formula :
In the case of this, the angle is 360 ^{ o } - 90 ^{ o } = 270 ^{ o } .
Thus the area is : -
Q5 In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find: (i) the length of the arc
Answer:
The length of the arc is given by:-
Hence the length of the arc is 22 cm.
Q5 In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find: (ii) area of the sector formed by the arc
Answer:
We know that the area of the sector is given by:-
Thus the area of the sector is 231 cm ^{ 2 } .
Q5 In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find: (iii) area of the segment formed by the corresponding chord
Answer:
For the area of the segment, we need to subtract the area of the triangle attached with the area of arc.
Thus consider the triangle:-
It is given that the angle of arc is 60 ^{ o } , or we can say that all angles are 60 ^{ o } (since two sides are equal). Hence it is an equilateral triangle.
Area of triangle is:-
Hence the area of segment is:-
Answer:
The area of the sector is :
Now consider the triangle, the angle of the sector is 60 ^{ 0 } .
This implies it is an equilateral triangle. (As two sides are equal so will have the same angle. This possible only when all angles are equal i.e., 60 ^{ o } .)
Thus, the area of the triangle is:-
or
Hence area of the minor segment :
And the area of the major segment is :
or
or
or
Answer:
For the area of the segment, we need the area of sector and area of the associated triangle.
So, the area of the sector is :
or
Now, consider the triangle:-
Draw a perpendicular from the center of the circle on the base of the triangle (let it be h).
Using geometry we can write,
or
Similarly,
or
Thus the area of the triangle is :
or
Hence the area of segment is: .
Q8 A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig.). Find (i) the area of that part of the field in which the horse can graze.
Answer:
The part grazed by the horse is given by = Area of sector
Q8 A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig.). Find (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use )
Answer:
When the length of the rope is 10 m, the area grazed will be:-
Hence the change in the grazing area is given by :
Q9 A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. Find: (i) the total length of the silver wire required.
Answer:
The total wire required will be for 5 diameters and the circumference of the brooch.
The circumference of the brooch:-
Hence the total wire required will be:- _{ . }
Q9 A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 12.12. Find : (ii) the area of each sector of the brooch.
Answer:
The total number of lines present in the brooch is 10 (line starting from the centre).
Thus the angle of each sector is 36 ^{ o } .
The area of the sector is given by:-
Answer:
It is given that the umbrella has 8 ribs so the angle of each sector is 45 ^{ o } .
Thus the area of the sector is given by:-
Hence the area between two consecutive ribs is _{} .
Answer:
The area cleaned by one wiper is:-
or
or
Hence the required area (area cleaned by both blades) is given by:-
Answer:
The area of the sector is given by:-
In this case, the angle is 80 ^{ o } .
Thus the area is:-
or
Answer:
The angle of each of the six sectors is 60 ^{ o } at the center.
Area of the sector is given by:-
or
or
And the area of the equilateral triangle associated with segment:-
or
Hence the area of segment is :
Thus the total area of design is :
So, the total cost for the design is:-
Q14 Tick the correct answer in the following :
Area of a sector of angle p (in degrees) of a circle with radius R is
(A)
(B)
(C)
(D)
Answer:
We know that the area of the sector is given by:-
Hence option (d) is correct.
NCERT Solutions for Class 10 Maths Chapter 12 Areas related to circles Excercise: 12.3
Answer:
We know that RPQ is 90 ^{ o } as ROQ is the diameter.
RQ can be found using the Pythagoras theorem.
or
or
Now, the area of the shaded region is given by = Area of semicircle - Area of PQR
Area of a semicircle is:-
or
or
And, the area of triangle PQR is :
Hence the area of the shaded region is : _{}
Answer:
The area of a shaded region can be easily found by using the formula of the area of the sector.
Area of the shaded region is given by : Area of sector OAFC - Area of sector OBED
Answer:
Area of the shaded region is given by = Area of the square - Area of two semicircles.
Area of square is :
And the area of the semicircle is:-
Hence the area of the shaded region is given by : _{}
Answer:
Area of the shaded region is given by = Area of triangle + Area of the circle - Area of the sector
Area of the sector is : -
or
And, the area of the triangle is :
And, the area of the circle is :
or
or
Hence the area of the shaded region is:-
or
Answer:
Consider the quadrant in the given figure:- We have an angle of the sector as 90 ^{ o } and radius 1 cm.
Thus the area of the quadrant is:-
or
And the area of the square is :
And, the area of the circle is:-
Hence the area of the shaded region is: = Area of the square - Area of the circle - 4 (Area of quadrant)
or
or
Answer:
Assume the center of the circle to be point C and AD as the median of the equilateral triangle.
Then we can write:-
or
Thus
Consider ABD,
or
or
Thus the area of an equilateral triangle is:-
or
And the area of the circle is :
or
Hence the area of the design is:-
Answer:
It is clear from the figure that the area of all sectors is equal (due to symmetry).
Also, the angle of the sector is 90 ^{ o } and the radius is 7 cm.
Thus the area of the sector is:-
or
And, the area of the square is :
Hence the area of the shaded region is :
Q8 Fig. depicts a racing track whose left and right ends are semicircular.
(i) the distance around the track along its inner edge
Answer:
The distance around the track is Length of two straight lines + Length of two arcs.
Length of the arc is -
Thus the length of the inner track is :
Q8 Fig. depicts a racing track whose left and right ends are semicircular.
Answer:
The area of track = Area of outer structure - Area of inner structure.
Area of outer structure is: = Area of square + Area of 2 semicircles
And, area of inner structure: = Area of inner square + Area of 2 inner semicircles
Thus the area of the track is :
Hence the area of the track is 4320 m ^{ 2 } .
Answer:
Firstly, the area of the smaller circle is :
Now, the area of :-
or
And, the area of the bigger semicircle is :
Hence the area of the shaded region is:-
Therefore the area of the shaded region is 66.5 cm ^{ 2 } .
Answer:
Area of an equilateral triangle is:-
Now, consider the sector:- Angle of the sector is 60 ^{ o } and the radius is 100 cm.
Thus the area of the sector:-
Thus the area of the shaded region is :
Answer:
Since one side of the square has 3 circles, thus the side of the square is 42 cm.
Area of the square :
And, area of a circle :
Hence the area of the remaining portion is :
Answer:
The quadrant OACB is a sector with angle 90 ^{ o } and radius 3.5 cm.
Thus the area of the quadrant is:-
Hence the area of the quadrant is _{} .
Q12 In Fig, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (ii) shaded region.
Answer:
For area of shaded region we need to find area of the triangle.
Area of triangle is:-
Hence the area of the shaded region is = Area of the quadrant - Area of triangle
Answer:
In the given figure we need to find the radius of the circle:-
Consider OAB,
Thus area of quadrant:-
Also, the area of the square is :
Area of the shaded region is :
Answer:
Area of the shaded region is = Area of larger sector - Area of smaller sector
Hence the area of the shaded region is
Answer:
Consider ABC,
Area of triangle is :
Now, area of sector is :
And area of semicircle is : -
Hence the area of the shaded region is : _{}
Answer:
It is clear from the figure that the required area (designed area) is the area of the intersection of two sectors.
Area of the sector is:-
And, area of the triangle:-
Hence the area of the designed region is :
NCERT solutions for class 10 maths chapter wise
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables |
Chapter 4 | NCERT solutions for class 10 maths chapter 4 Quadratic Equations |
Chapter 5 | NCERT solutions for class 10 chapter 5 Arithmetic Progressions |
Chapter 6 | |
Chapter 7 | NCERT solutions for class 10 maths chapter 7 Coordinate Geometry |
Chapter 8 | NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry |
Chapter 9 | NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | NCERT solutions for class 10 chapter maths chapter 12 Areas Related to Circles |
Chapter 13 | NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes |
Chapter 14 | |
Chapter 15 |
NCERT solutions of class 10 subject wise
How to use NCERT solutions for class 10 maths chapter 12 Areas Related To Circles ?
Before coming to this chapter please make sure that you are comfortable with the basic concepts of circles, squares, rectangles, triangles, etc.
Learn some formulae related to sectors and segments from the NCERT textbook.
Go through some examples given in the textbook to understand the approach of solving the problems.
Once you have done the above-said points then its time to put your efforts into practice by practicing the practice exercises.
During the practice, you can take the help of NCERT solutions for class 10 maths chapter 12 Areas Related to Circles.
Keep working hard & happy learning!
Frequently Asked Question (FAQs) - NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles
Question: How does the NCERT solutions are helpful in CBSE board exam?
Answer:
As most of the questions in CBSE board exam are directly asked from NCERT textbook, so you must know the NCERT textbook very well. NCERT solutions are provided in a very simple language as well as according with CBSE guidelines.
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Answer:
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Answer:
Here you will get the detailed NCERT solutions for class 10 maths by clicking on the link.
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Answer:
NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.
Question: Which is the best book for CBSE class 10 maths ?
Answer:
NCERT textbook is the best book for CBSE class 10 maths. Most of the questions in CBSE class 10 board exam are directly asked from NCERT textbook. All you need to do is rigorous practice of the questions given in the NCER textbook.
Question: Which are the most difficult chapter in the NCERT class 10 maths ?
Answer:
Students consider trigonometry that is introduction to trigonometry and application of trigonometry is most difficult chapter in CBSE class 10 maths. With the regular practice you will get conceptual clarity and will be able to have a strong grip on Trigonometry.