Class 10 Maths Case Study Questions Chapter 12 Areas Related to Circles
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Case study Questions in the Class 10 Mathematics Chapter 12 are very important to solve for your exam. Class 10 Maths Chapter 12 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Class 10 Maths Case Study Questions Chapter 12 Areas Related to Circles
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In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passagebased as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.
Areas Related to Circles Case Study Questions With Answers
Here, we have provided casebased/passagebased questions for Class 10 Maths Chapter 12 Areas Related to Circles
Case Study/PassageBased Questions
Question 1:
(a) 700 cm  (b) 729 cm  (c) 732 cm  (d) 735 cm 
Answer: (b) 729 cm2
(ii) Area of rectangle left for car parking is
(a) 64 cm  (b) 76 cm  (c) 81 cm  (d) 100 cm 
Answer: (c) 81 cm2
(iii) Radius of semicircle is
(a) 6.75 cm  (b) 7 cm  (c) 7.75 cm  (d) 8.75 cm 
Answer: (a) 6.75 cm
(iv) Area of a semicircle is
(a) 61.59 cm  (b) 66.29 cm  (c) 70.36 cm  (d) 71.59 cm 
Answer: (d) 71.59 cm2
(v) Find the area of the shaded region
(a) 660.82 cm  (b) 666.82 cm  (c) 669.89 cm  (d) 700 cm 
Answer: (b) 666.82 cm2
Question 2:
A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some brooch are shown below. Observe them carefully.
Design A: Brooch A is made with silver wire in the form of a circle with a diameter of 28mm. The wire is used for making 4 diameters which divides the circle into 8 equal parts.
Design B: Brooch b is made of two colors – Gold and silver. The outer part is made of Gold. The circumference of the silver part is 44mm and the gold part is 3mm wide everywhere.
Refer to Design A
1. The total length of silver wire required is
Answer: b) 200 mm
2. The area of each sector of the brooch is
Answer: c) 77 mm2
Refer to Design B
3. The circumference of outer part (golden) is
a) 48.49 mm
c) 72.50 mm
d) 62.86 mm
Answer: d) 62.86 mm
4. The difference of areas of golden and silver parts is
a) 18 π
b) 44 π
c) 51 π
d) 64 π
Answer: c) 51 π
5. A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80 mm?
Answer: c) 4
Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 12 Areas Related to Circles with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Areas Related to Circles Case Study and PassageBased Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate
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 CBSE Class 10 Study Material
CBSE Class 10 Maths Case Study Questions for Chapter 10  Circles (Published By CBSE)
Check cbse class 10 maths case study questions for chapter 10  circles. these questions have been published by the board for class 10 mathematics..
CBSE: Case study questions for CBSE Class 10 Maths Chapter 10  Circles are provided here which students can practice to get familiarised with the new format of questions. These questions have been published by CBSE itself. Answers to all the questions have been provided for the convenience of students. Case study questions are helpful for the preparation of the Class 10 Maths Exam 20212022.
Case Study Questions for Class 10 Maths Chapter 10  Circles
CASE STUDY 1:
A Ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a rotating upright wheel with multiple passengercarrying components (commonly referred to as passenger cars, cabins, tubs, capsules, gondolas, or pods) attached to the rim in such a way that as the wheel turns, they are kept upright, usually by gravity.
After taking a ride in Ferris wheel, Aarti came out from the crowd and was observing her friends who were enjoying the ride . She was curious about the different angles and measures that the wheel will form. She forms the figure as given below.
1. In the given figure find ∠ROQ
Answer: c) 150
2. Find ∠RQP
Answer: a) 75
3. Find ∠RSQ
Answer: b) 75
4. Find ∠ORP
Answer: a) 90
CASE STUDY 2:
Varun has been selected by his School to design logo for Sports Day Tshirts for students and staff . The logo design is as given in the figure and he is working on the fonts and different colours according to the theme. In given figure, a circle with centre O is inscribed in a ΔABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. The lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively.
1. Find the length of AD
Answer: a) 7
2. Find the Length of BE
Answer: b) 5
3. Find the length of CF
Answer: d) 3
4. If radius of the circle is 4cm, Find the area of ∆OAB
Answer: c) 24
5. Find area of ∆ABC
Answer: b) 60
Also Check:
Tips to Solve Case Study Based Questions Accurately
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Case Study Questions for Class 10 Maths Chapter 12 Areas Related to Circles
 Last modified on: 1 year ago
 Reading Time: 4 Minutes
Case Study Questions
Question 1:
A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some broochs are shown below. Observe them carefully
Design A: Brooch A is made with silver wire in the form of a circle with diameter 28 mm. A wire used for making 4 diameters which divide the circle into 8 equal parts.
Design B: Brooch B is made of two colours gold and silver. Outer part is made with gold. The circumference of silver part is 44 mm and the gold part is 3 mm wide everywhere.
Based on the above information, answer the following questions:
Refer to design A (i) The total length of silver wire required is (a) 180 mm (b) 200 mm (c) 250 mm (d) 280 mm
(ii) The area of each sector of the brooch is (a) 44 mm 2 (b) 52 mm 2 (c) 77 mm 2 (d) 68 mm 2
Refer to design B (iii) The circumference of outer part (golden) is (a) 48.49 mm (b) 82.2 mm (c) 72.50 mm (d) 62.86 mm
(iv) The difference of areas of golden and silver parts is (a) 18π (b) 44π (c) 51π (d) 64π
(v) A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80π mm ? (a) 2 (b) 3 (c) 4 (d) 5
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CBSE 10th Standard Maths Subject Areas Related to Circles Case Study Questions With Solution 2021
CBSE 10th Standard Maths Subject Areas Related to Circles Case Study Questions With Solution 2021 QB365  Question Bank Software May22 , 2021
QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get more marks in Exams
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Case Study Questions
(ii) Find the area of the red region.
(iii) Find the radius of the circle formed by combining the red and silver region.
(iv) Find the area of the silver region.
(v) Area of the circular path formed by two concentric circles of radii r 1 and r 2 (r 1 > r 2 ) =
(\({r}_{1}^{2}\) + \({r}_{2}^{2}\)) sq. units  (\({r}_{1}^{2}\)  \({r}_{2}^{2}\)) sq. units 
(\({r}_{1}^{2}\) + \({r}_{2}^{2}\)) sq. units  (\({r}_{1}^{2}\)  \({r}_{2}^{2}\)) sq. units 
(ii) Area of rectangle left for car parking is
(iii) Radius of semicircle is
(iv) Area of a semicircle is
(v) Find the area of the shaded region
(ii) Area of quadrant BCD is
(iii) Find the area of \(\Delta\) CEF.
(iv) Area of the shaded portion is
(v) Area of the unshaded portion is
(ii) Area of the region grazed by the cow is
\((a) \frac{\angle A}{360^{\circ}} \times \pi \times(3.5)^{2}\)  \((b) \frac{\angle B}{360^{\circ}} \times \pi \times(24)^{2}\)  \((c) \frac{\angle C}{360^{\circ}} \times \pi \times(3.5)^{2}\) 
(iii) Area of region grazed by the buffalo and the horse is
\((a) \frac{(\angle A+\angle C)}{360^{\circ}} \times \pi \times(5.5)^{2}\)  \((b) \frac{(\angle B+\angle C)}{360^{\circ}} \times \pi \times(5.6)^{2}\) 
\((c) \frac{(\angle A+\angle C)}{360^{\circ}} \times \pi \times(3.5)^{2}\)  \((d) \frac{(\angle B+\angle C)}{360^{\circ}} \times \pi \times(3.5)^{2}\) 
(iv) Total area grazed by the cow, the buffalo and the horse is
(v) Find the area of the field that cannot be grazed.
(ii) Find the area of the circle.
(Hi) If D lies at the middle of arc BC, then area of region COD is
(iv) Area of the \(\Delta\) BAC is
(v) Find .the area of the polluted region.
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Areas Related to Circles Class 10 Notes CBSE Maths Chapter 12 (Free PDF Download)
Exam  Focused Revision Notes for CBSE Class 10 Maths Chapter 12  Areas Related to Circles
We will analyze the Introduction of Area Related to circles in this chapter and also find the equation of any circle whose centre and radius are given. We’ll find the circumference, area of a circle and circular paths. Also, we’ll study derivations and understand the formulae for perimeter and area of a sector of a circle. We’ll use the above formula to find the perimeter and the area of a sector. Little deeper into the chapter, we’ll find the areas of some combined figures involving circles, sectors, triangles, squares, rectangles and also solve daily life problems on the basis of perimeters and areas of different plane figures.
The revision notes for Chapter 12 Areas Related to Circles are developed according to the NCERT curriculum by the experts in Vedantu who have vast knowledge on the subject. The solutions are developed in a step by step manner to highlight the important formulas and shortcuts. These Areas Related to Circles Class 10 Notes are carefully designed to provide the students with a great learning experience and to make them understand the concepts much faster. The solutions to the important questions of CBSE Class 10 Maths Notes Chapter 12 Areas Related to Circles are available in free PDF versions, students can use these PDFs at Vedantu. Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Register Online for Class 10 Science tuition on Vedantu.com to score more marks in CBSE board examination. Vedantu.com is No.1 Online Tutoring Company in India Provides you Free PDF download of NCERT Solutions for Class 10 Maths solved by Expert Teachers as per NCERT (CBSE) Book guidelines.
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 Chapter 12  Areas Related to Circles Notes 



Access Class 10 Mathematics Chapter 12 – Areas Related to Circles Notes in 30 Minutes
The path of a point moving in such a way that its distance from a fixed point is always the same is called a circle. That fixed point is called the centre of that circle and that path is called the locus of that point. The fixed distance between the centre and path is called the radius of that circle. We can see many examples of circles around us like bangles, round chapatis, dial watch, sun, etc.
Perimeter of a circle – Perimeter is the circumferential length of a closed shape or a polygon. In case of circle, if we travel once around a circle, then the length covered gives us the perimeter of circumference. Circumference of a circle always bears a constant ratio with its diameter, which is denoted by a Greek letter $\pi $. Mathematically,
$\pi =\frac{Circumference}{Diameter}$
$\Rightarrow Circumference=\pi \times diameter$
$\Rightarrow Circumference=\pi \times d$
$\Rightarrow Circumference=\pi \times 2r$ (Where $r$ is the radius of circle and $d=2r$).
Area of a circle – The space covered or occupied by a polygon in a twodimensional plane is called the area. In case of a circle, it is the space occupied withing its boundary or the perimeter. If, $r$ is the radius of a given circle, then the formula for finding the area is given as;
$Area=\pi \times {{r}^{2}}$
Pi ($\pi $):
The value of $\pi $ was given by the great Indian mathematician Aryabhatta. He gave an approximate value of $\pi $ as $\pi =\frac{62832}{20000}$ which is almost equal to $3.1416$. It should be noted that $\pi $ is an irrational number as its value is nonterminating and nonrecurring. For calculation purposes, we often take the value of $\pi $ as $\frac{22}{7}$ which in turn is a rational number.
Semicircle:
When a circle is cut into half along a diameter, semicircle is formed as shown below. Its perimeter consists of length of half a circle and the length of a diameter. If the semicircle is open, then diameter length is not added. If the length of diameter is given by $d$ and radius is given by $r$ then perimeter is given by,
$Perimeter=\pi r+d$ (For closed semicircle)
$Perimeter=\pi r$ (For open semicircle)
And the area of a semicircle is just half the area of a circle and is mathematically given as $\frac{\pi {{r}^{2}}}{2}$.
Similarly, area of a quadrant of a circle is given by $\frac{\pi {{r}^{2}}}{4}$.
Sector of a Circle:
The portion of a circle enclosed within an arc and two radii of that circle is called as sector.
Let us take the central angle between the radii is $\theta $ which is ${{360}^{\circ }}$ for a complete circle. Now let the length of that arc be $l$. Then the length $l$ can be found out using the following relation,
$l=\tfrac{\theta }{{{360}^{\circ }}}\times 2\pi r$.
Now, perimeter of sector is given as $2r+l$.
Similarly, area of sector is given by $\frac{\theta }{{{360}^{\circ }}}\times \pi {{r}^{2}}$.
Segment of a Circle:
The part of the circular region enclosed between a chord and the corresponding arc of that circle is called the segment of a circle. The chord having centre of the circle as a point on it is the diameter and also the longest chord of the circle and divides the circle into two equal halves. When the chord is not the diameter, then the portion consisting the centre of circle is called the major segment and the other region is called the minor segment.
In the diagram above the chord, $BC$ divides the circle in two segments. Such as;
Area of minor segment$=$Area of sector $ABDC$$$Area of $\Delta ABC$.
And area of major segment$=$Area of circle$$Area of minor segment.
Here, area of $\Delta ABC$ can be found out using the formula $\frac{1}{2}{{r}^{2}}\sin \theta $.
And the area od sector $ABDC$ is given by $\frac{\theta }{{{360}^{\circ }}}\times \pi {{r}^{2}}$.
Hence, the area of segment $ACB=\left( \frac{\theta }{{{360}^{\circ }}}\times \pi {{r}^{2}} \right)\left( \frac{1}{2}{{r}^{2}}\sin \theta \right)$
$={{r}^{2}}\left[ \frac{\pi \theta }{{{360}^{\circ }}}\frac{\sin \theta }{2} \right]$.
Area of a Ring:
Ring is the region between two concentric circles having different radii. Let the radius of larger circle be $R$ and radius of smaller circle be $r$.
Hence the area of the ring is given by;
$\pi {{R}^{2}}\pi {{r}^{2}}$
$=\pi \left( {{R}^{2}}{{r}^{2}} \right)$.
Comprehensive Revision Notes for CBSE Class 10 Maths Chapter 12: Areas Related to Circles
Areas Related to Circles Class 10 Notes are prepared by Vedantu to help you revise your questions in this chapter. The following chapter presents several new concepts relating to a circle, for example, lines that cross the circle at different points forming components such as tangents, chords and diameters. This chapter helps you create a solid geometry basis for higher education and to achieve good results in the examinations. In real life also circles and their different properties, such as radius, diameter, circumference and area have applications.
So, the Introduction of Area Related to circles chapter is one of the important topics for Class 10 students from in higher studies point of view. Depending upon the properties & applications of the circles, few topics are designed in higher classes. So, let’s look into the important concepts of the circles which are discussed in this chapter:
Introduction
Area of a circle, circumference of a circle.
Segment of a Circle
Minor arc and Major Arc
Sector of a Circle
Angle of a Sector
Length of an arc of a sector
Area of a Sector of a Circle
Area of a Triangle
Area of a Segment of a Circle
Visualizations
Areas of different plane figures
Areas of Combination of Plane figures
A circle is defined as a collection of points separated by a fixed distance, known as the radius, from a fixed point, known as the centre.
When a line and a circle are both in the same plane, the line and circle will not intersect. At a certain point, the line can come close to touching the circle. That kind of line is known as the tangent to the circle. The line is the secant for the circle as it intersects the circle at two points.
Tangent to a Circle
Tangent to a circle is the line that touches the circle at a single point. The point of tangency is the intersection of a tangent and a circle. The tangent is perpendicular to the circle's radius, in which it intersects. Any curved shape can have tangents. Since tangent is a line, it has its own equation.
The tangent will touch the circle only at one point.
We can name the line that contains the radius through a point of contact as ‘normal’ to circle at the point.
Condition of Tangency
The tangent is called only if it touches a curve at a single point. If not it is said to be simply called a line. So depending on the point of tangency, and also where it falls with respect to a circle, we can specify the criteria for tangent as follows:
When the point lies inside of the circle.
When the point lies on the circle.
When the point lies outside of the circle.
The circumference of a circle, also known as its perimeter, is the measurement of the circle's boundary. The area of a circle, on the other hand, determines the region it occupies. The circumference of a circle is its length when we open it and draw a straight line through it. It's normally expressed in units like centimetres or metres.
⇒ Circumference (or) perimeter of a circle = 2πR
Area of a circle is nothing but the region occupied by the circle in a 2D plane. It can be determined by using the formula, A = πr 2 . (Here, r is the radius of a circle) This formula is useful while measuring the area occupied by a circular field or a plot.
Perimeter of Semicircle
The perimeter of a semicircle is nothing but the sum of half of the circumference of a circle and the diameter. We know that the perimeter of a circle is 2πr or πd. So, the perimeter of a semicircle will be ½ (πd) + d or πr + 2r, in which r is the radius.
A sector is a section of a circle between its two radii and the adjacent arc. A semicircle that represents half a circle is the most common sector of the circle. A circle that has a sector can be further divided into 2 regions called a Major Sector and a Minor Sector. You can find all the important topics explained in CBSE Class 10 Maths Notes Chapter 12 Areas Related to Circles PDF.
Benefits of Studying Vedantu’s Revision Notes:
Mathematics can be a difficult subject for Class 10 students to achieve good grades in, but if they prepare methodically by having revision notes, they can easily achieve more marks in their Maths exam.
Areas Related to Circles Class 10 Notes will assist you in predicting the types of questions that could be asked during the examination.
Solutions are split into various sections of the exam for a better understanding of the subjects.
You can get a better understanding of the topics in simple language with our Revision notes.
Solutions from Vedantu are errorfree and wellorganized.
The questions are categorised such as short questions, long answer type questions, all the sections of the question paper in your school exams are thoroughly covered. If you solve these exercises extensively using Vedantu platforms as a reference source, you get full conceptual clarity in question and answer format. It is advisable to practise these questions because the activities in this chapter cover the course indepth and are equally appropriate for quick review right before your exams.
Tips on How to Prepare for Exams Using Chapter 12 Areas Related to Circles
The tips given below will help students to prepare for their exams by using the free PDF of Areas Related to Circles Class 10 Notes available on Vedantu.
Every question should be carefully read before attempting. Since there are some tough questions, there is a risk that we would give the incorrect answer if the questions are not completely understood.
The basic formulas for finding circumference and area should be memorised in the circle chapter since they are fundamental formulas for solving any problems.
Vedantu's Notes PDF includes several exercises and practise problems. To get good grades on your examinations, students can solve and practise these exercises several times.
These solutions and concepts have been developed by Subject Experts to address your questions & doubts at the same time. This strategy will also allow you to increase your studying effectiveness in your selfstudy hours. For all your queries relating to 'Area Related to Circles,' Vedantu wants to provide you with a onestop solution. These solutions are truly informative and provide you with realistic tips and tricks for correctly solving problems.
Vedantu's Areas Related to Circles Class 10 Notes for CBSE Maths Chapter 12 offer a comprehensive and valuable resource for students studying this topic. The free PDF download provided by Vedantu is a fantastic opportunity for learners to access highquality study material without any financial burden. The notes cover essential concepts, formulas, and solved examples, enhancing students' understanding and problemsolving skills. With Vedantu's userfriendly approach, learners can grasp intricate concepts easily, fostering a deeper appreciation for the subject. Whether preparing for exams or seeking clarity on challenging topics, these notes serve as a reliable and effective aid, empowering students to excel in their academic journey.
FAQs on Areas Related to Circles Class 10 Notes CBSE Maths Chapter 12 (Free PDF Download)
1. Why Should I Refer to Vedantu for CBSE Class 10 Maths Notes Chapter 12 Areas Related to Circles?
Ans: Vedantu has a strong faculty with vast experience in teaching Math subjects. Until offering solutions to major issues, these experts conduct extensive research. Subject matter experts developed these solutions and principles to simultaneously answer your concerns and doubts. This technique would also help you research more efficiently during your selfstudy hours. As a result, when students use the free PDF, they can learn and enjoy the subject.
2. What are the Topics that are Covered in the Introduction of the Area Related to Circles?
Ans: Circles chapter is one of the important chapters which has more weightage in CBSE exams. Around 5% of weightage is given to this chapter in the board exam and also 23% weightage is given JEE exams. The chapter covers Introduction, Area of a Circle, Circumference of a Circle, Segment of a Circle, Minor arc and Major Arc, Sector of a Circle, Angle of a Sector, Length of an arc of a sector, Area of a Sector of a Circle, Area of a Triangle, Area of a Segment of a Circle, Visualizations, Areas of different plane figures, Areas of Combination of Plane figures.
3. How can I Access a Free PDF of Areas Related to Circles Class 10 Notes?
Ans: Vedantu provides free PDF solutions to Chapter 12 Areas Related to Circles. If you solve these exercises extensively using Vedantu platforms as a reference source, you get full conceptual clarity in question and answer format. This free PDF also includes additional problems for students to practise alongside the exercise problems.
4. What are some of the important points to remember for Areas Related to Circles Class 10 Notes CBSE Maths Chapter 12?
Here are some of the important points to remember for Areas Related to Circles Class 10 Notes CBSE Maths Chapter 12:
The radius of a circle is the distance from the center of the circle to any point on its circumference.
The diameter of a circle is twice the radius.
The circumference of a circle is the distance around the circle.
The area of a circle is the amount of space enclosed by the circle.
The formula for the circumference of a circle is 2πr.
The formula for the area of a circle is πr².
5. What are some of the properties and applications of circles?
Properties:
All radii of a circle are equal in length.
All chords of a circle that pass through its center are equal in length.
The angle in a semicircle is a right angle.
The sum of the angles in a triangle inscribed in a circle is 180°.
Applications:
Circles have many applications in the real world, such as:
Sports equipment
Architecture
Engineering
CBSE Case Study Questions for Class 10 Maths Area Related to Circles Free PDF
Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Area Related to Circles in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!
I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.
MCQ Set 1 
Mcq set 2 , checkout our case study questions for other chapters.
 Chapter 10: Circles Case Study Questions
 Chapter 11: Construction Case Study Questions
 Chapter 13: Surface Area and Volumes Case Study Questions
 Chapter 14: Statistics Case Study Questions
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Solve every question of NCERT by hand, without looking at the solution.
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Sit through chapter wise FULLY INVIGILATED TESTS
Practice MCQ Questions (Very Important)
Practice Assertion Reason & Case Study Based Questions
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CBSE Class 10 Maths: Case Study Questions of Chapter 12 Areas Related to Circles PDF Download
Case study Questions in the Class 10 Mathematics Chapter 12 are very important to solve for your exam. Class 10 Maths Chapter 12 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case studybased questions for Class 10 Maths Chapter 12 Areas Related to Circles
In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason . There will be a few questions based on case studies and passagebased as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.
Areas Related to Circles Case Study Questions With answers
Here, we have provided casebased/passagebased questions for Class 10 Maths Chapter 12 Areas Related to Circles
Case Study/PassageBased Questions
Question 1:
(a) 700 cm  (b) 729 cm  (c) 732 cm  (d) 735 cm 
Answer: (b) 729 cm2
(ii) Area of rectangle left for car parking is
(a) 64 cm  (b) 76 cm  (c) 81 cm  (d) 100 cm 
Answer: (c) 81 cm2
(iii) Radius of semicircle is
(a) 6.75 cm  (b) 7 cm  (c) 7.75 cm  (d) 8.75 cm 
Answer: (a) 6.75 cm
(iv) Area of a semicircle is
(a) 61.59 cm  (b) 66.29 cm  (c) 70.36 cm  (d) 71.59 cm 
Answer: (d) 71.59 cm2
(v) Find the area of the shaded region
(a) 660.82 cm  (b) 666.82 cm  (c) 669.89 cm  (d) 700 cm 
Answer: (b) 666.82 cm2
Question 2:
A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some brooch are shown below. Observe them carefully.
Design A: Brooch A is made with silver wire in the form of a circle with a diameter of 28mm. The wire is used for making 4 diameters which divides the circle into 8 equal parts.
Design B: Brooch b is made of two colors – Gold and silver. The outer part is made of Gold. The circumference of the silver part is 44mm and the gold part is 3mm wide everywhere.
Refer to Design A
1. The total length of silver wire required is
Answer: b) 200 mm
2. The area of each sector of the brooch is
Answer: c) 77 mm2
Refer to Design B
3. The circumference of outer part (golden) is
a) 48.49 mm
c) 72.50 mm
d) 62.86 mm
Answer: d) 62.86 mm
4. The difference of areas of golden and silver parts is
a) 18 π
b) 44 π
c) 51 π
d) 64 π
Answer: c) 51 π
5. A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80 mm?
Answer: c) 4
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 Chapter 12: Areas Related To Circles
NCERT Solutions For Class 10 Maths Chapter 12 Areas Related to Circles
Ncert solutions for class 10 maths chapter 12 – cbse free pdf download.
* According to the CBSE Syllabus 202324, this chapter has been renumbered as Chapter 11.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles are important study resources needed for the students in Class 10. These NCERT Solutions for Class 10 Maths help the students understand the types of questions that will be asked in the CBSE Class 10 Maths board exams. Moreover, providing solutions to all areas related to circles help students in preparing for the CBSE exams in an effective way.
Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 12 Areas Related to Circles
Download most important questions for class 10 maths chapter – 12 areas related to circles.
The NCERT Solutions for Class 10 Maths , available at BYJUS, will help students ace their board exams by preparing for them well in advance. Get free Maths NCERT Solutions for Class 10 of this chapter and clear all conceptual doubts. To assist the students of Class 10 in being better prepared for their board exams, the experts have formulated these NCERT Solutions based on the latest update on the CBSE syllabus for 202324.
 Chapter 1 Real Numbers
 Chapter 2 Polynomials
 Chapter 3 Pair of Linear Equations in Two Variables
 Chapter 4 Quadratic Equations
 Chapter 5 Arithmetic Progressions
 Chapter 6 Triangles
 Chapter 7 Coordinate Geometry
 Chapter 8 Introduction to Trigonometry
 Chapter 9 Some Applications of Trigonometry
 Chapter 10 Circles
 Chapter 11 Constructions
 Chapter 12 Areas Related to Circles
 Chapter 13 Surface Areas and Volumes
 Chapter 14 Statistics
 Chapter 15 Probability
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles
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Access answers of Maths NCERT Class 10 Chapter 12 – Areas Related to Circles
Class 10 maths chapter 12 exercise: 12.1 (page no: 230).
Exercise: 12.1 (Page No: 230)
1. The radii of the two circles are 19 cm and 9 cm, respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.
The radius of the 1 st circle = 19 cm (given)
∴ circumference of the 1 st circle = 2π×19 = 38π cm
The radius of the 2 nd circle = 9 cm (given)
∴ circumference of the 2 nd circle = 2π×9 = 18π cm
The sum of the circumference of two circles = 38π+18π = 56π cm
Now, let the radius of the 3 rd circle = R
∴ the circumference of the 3 rd circle = 2πR
It is given that sum of the circumference of two circles = circumference of the 3 rd circle
Hence, 56π = 2πR
Or, R = 28 cm.
2. The radii of the two circles are 8 cm and 6 cm, respectively. Find the radius of the circle having an area equal to the sum of the areas of the two circles.
The radius of 1 st circle = 8 cm (given)
∴ area of 1 st circle = π(8) 2 = 64π
The radius of 2 nd circle = 6 cm (given)
∴ area of 2 nd circle = π(6) 2 = 36π
The sum of 1 st and 2 nd circle will be = 64π+36π = 100π
Now, assume that the radius of 3 rd circle = R
∴ area of the circle 3 rd circle = πR 2
It is given that the area of the circle 3 rd circle = Area of 1 st circle + Area of 2 nd circle
Or, πR 2 = 100πcm 2
R 2 = 100cm 2
So, R = 10cm
3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing the Gold score is 21 cm, and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
The radius of 1 st circle, r 1 = 21/2 cm (as diameter D is given as 21 cm)
So, area of gold region = π r 1 2 = π(10.5) 2 = 346.5 cm 2
Now, it is given that each of the other bands is 10.5 cm wide,
So, the radius of 2 nd circle, r 2 = 10.5cm+10.5cm = 21 cm
∴ area of red region = Area of 2 nd circle − Area of gold region = (πr 2 2 −346.5) cm 2
= (π(21) 2 − 346.5) cm 2
= 1386 − 346.5
= 1039.5 cm 2
The radius of 3 rd circle, r 3 = 21 cm+10.5 cm = 31.5 cm
The radius of 4 th circle, r 4 = 31.5 cm+10.5 cm = 42 cm
The Radius of 5 th circle, r 5 = 42 cm+10.5 cm = 52.5 cm
For the area of n th region,
A = Area of circle n – Area of the circle (n1)
∴ area of the blue region (n=3) = Area of the third circle – Area of the second circle
= π(31.5) 2 – 1386 cm 2
= 3118.5 – 1386 cm 2
= 1732.5 cm 2
∴ area of the black region (n=4) = Area of the fourth circle – Area of the third circle
= π(42) 2 – 1386 cm 2
= 5544 – 3118.5 cm 2
= 2425.5 cm 2
∴ area of the white region (n=5) = Area of the fifth circle – Area of the fourth circle
= π(52.5) 2 – 5544 cm 2
= 8662.5 – 5544 cm 2
= 3118.5 cm 2
4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
The radius of car’s wheel = 80/2 = 40 cm (as D = 80 cm)
So, the circumference of wheels = 2πr = 80 π cm
Now, in one revolution, the distance covered = circumference of the wheel = 80 π cm
It is given that the distance covered by the car in 1 hr = 66km
Converting km into cm, we get,
Distance covered by the car in 1hr = (66×10 5 ) cm
In 10 minutes, the distance covered will be = (66×10 5 ×10)/60 = 1100000 cm/s
∴ distance covered by car = 11×10 5 cm
Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels)
=( 11×10 5 )/80 π = 4375.
5. Tick the correct solution in the following and justify your choice. If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units
(B) π units
(C) 4 units
(D) 7 units
Since the perimeter of the circle = area of the circle,
So, option (A) is correct, i.e., the radius of the circle is 2 units.
Exercise: 12.2 (Page No: 230)
1. Find the area of a sector of a circle with a radius 6 cm if the angle of the sector is 60°.
It is given that the angle of the sector is 60°
We know that the area of sector = (θ/360°)×πr 2
∴ area of the sector with angle 60° = (60°/360°)×πr 2 cm 2
= (36/6)π cm 2
= 6×22/7 cm 2 = 132/7 cm 2
2. Find the area of a quadrant of a circle whose circumference is 22 cm.
Circumference of the circle, C = 22 cm (given)
It should be noted that a quadrant of a circle is a sector which is making an angle of 90°.
Let the radius of the circle = r
As C = 2πr = 22,
R = 22/2π cm = 7/2 cm
∴ area of the quadrant = (θ/360°) × πr 2
Here, θ = 90°
So, A = (90°/360°) × π r 2 cm 2
= (49/16) π cm 2
= 77/8 cm 2 = 9.6 cm 2
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Length of minute hand = radius of the clock (circle)
∴ Radius (r) of the circle = 14 cm (given)
Angle swept by minute hand in 60 minutes = 360°
So, the angle swept by the minute hand in 5 minutes = 360° × 5/60 = 30°
Area of a sector = (θ/360°) × πr 2
Now, the area of the sector making an angle of 30° = (30°/360°) × πr 2 cm 2
= (1/12) × π14 2
= (49/3)×(22/7) cm 2
= 154/3 cm 2
4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
(i) minor segment
(ii) major sector. (Use π = 3.14)
Here, AB is the chord which is subtending an angle 90° at the centre O.
It is given that the radius (r) of the circle = 10 cm
(i) Area of minor sector = (90/360°)×πr 2
= (¼)×(22/7)×10 2
Or, the Area of the minor sector = 78.5 cm 2
Also, the area of ΔAOB = ½×OB×OA
Here, OB and OA are the radii of the circle, i.e., = 10 cm
So, the area of ΔAOB = ½×10×10
Now, area of minor segment = area of the minor sector – the area of ΔAOB
= 78.5 – 50
= 28.5 cm 2
(ii) Area of major sector = Area of the circle – Area of he minor sector
= (3.14×10 2 )78.5
= 235.5 cm 2
5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord
Radius = 21 cm
(i) Length of an arc = θ/360°×Circumference(2πr)
∴ Length of an arc AB = (60°/360°)×2×(22/7)×21
= (1/6)×2×(22/7)×21
Or Arc AB Length = 22cm
(ii) It is given that the angle subtended by the arc = 60°
So, the area of the sector making an angle of 60° = (60°/360°)×π r 2 cm 2
= 441/6×22/7 cm 2
Or, the area of the sector formed by the arc APB is 231 cm 2
(iii) Area of segment APB = Area of sector OAPB – Area of ΔOAB
Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔOAB is an equilateral triangle. So, its area will be √3/4×a 2 sq. Units.
The area of segment APB = 231(√3/4)×(OA) 2
= 231(√3/4)×21 2
Or, the area of segment APB = [231(441×√3)/4] cm 2
6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)
Radius = 15 cm
Area of sector OAPB = (60°/360°)×πr 2 cm 2
= 225/6 πcm 2
Now, ΔAOB is equilateral as two sides are the radii of the circle and hence equal and one angle is 60°
So, Area of ΔAOB = (√3/4) ×a 2
Or, (√3/4) ×15 2
∴ Area of ΔAOB = 97.31 cm 2
Now, the area of minor segment APB = Area of OAPB – Area of ΔAOB
Or, the area of minor segment APB = ((225/6)π – 97.31) cm 2 = 20.43 cm 2
Area of major segment = Area of the circle – Area of the segment APB
Or, area of major segment = (π×15 2 ) – 20.4 = 686.06 cm 2
7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)
Radius, r = 12 cm
Now, draw a perpendicular OD on chord AB, and it will bisect chord AB.
So, AD = DB
Now, the area of the minor sector = (θ/360°)×πr 2
= (120/360)×(22/7)×12 2
= 150.72 cm 2
Consider the ΔAOB,
∠ OAB = 180°(90°+60°) = 30°
Now, cos 30° = AD/OA
√3/2 = AD/12
Or, AD = 6√3 cm
We know OD bisects AB. So,
AB = 2×AD = 12√3 cm
Now, sin 30° = OD/OA
Or, ½ = OD/12
∴ OD = 6 cm
So, the area of ΔAOB = ½ × base × height
Here, base = AB = 12√3 and
Height = OD = 6
So, area of ΔAOB = ½×12√3×6 = 36√3 cm = 62.28 cm 2
∴ area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB
= 150.72 cm 2 – 62.28 cm 2 = 88.44 cm 2
8. A horse is tied to a peg at one corner of a squareshaped grass field of side 15 m by means of a 5 m long rope (see Fig. 12.11). Find
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)
As the horse is tied at one end of a square field, it will graze only a quarter (i.e. sector with θ = 90°) of the field with a radius 5 m.
Here, the length of the rope will be the radius of the circle, i.e. r = 5 m
It is also known that the side of the square field = 15 m
(i) Area of circle = πr 2 = 22/7 × 5 2 = 78.5 m 2
Now, the area of the part of the field where the horse can graze = ¼ (the area of the circle) = 78.5/4 = 19.625 m 2
(ii) If the rope is increased to 10 m,
Area of circle will be = πr 2 =22/7×10 2 = 314 m 2
Now, the area of the part of the field where the horse can graze = ¼ (the area of the circle)
= 314/4 = 78.5 m 2
∴ increase in the grazing area = 78.5 m 2 – 19.625 m 2 = 58.875 m 2
9. A brooch is made with silver wire in the form of a circle with a diameter 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:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Diameter (D) = 35 mm
Total number of diameters to be considered= 5
Now, the total length of 5 diameters that would be required = 35×5 = 175
Circumference of the circle = 2πr
Or, C = πD = 22/7×35 = 110
Area of the circle = πr 2
Or, A = (22/7)×(35/2) 2 = 1925/2 mm 2
(i) Total length of silver wire required = Circumference of the circle + Length of 5 diameter
= 110+175 = 285 mm
(ii) Total Number of sectors in the brooch = 10
So, the area of each sector = total area of the circle/number of sectors
∴ Area of each sector = (1925/2)×1/10 = 385/4 mm 2
10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
The radius (r) of the umbrella when flat = 45 cm
So, the area of the circle (A) = πr 2 = (22/7)×(45) 2 =6364.29 cm 2
Total number of ribs (n) = 8
∴ The area between the two consecutive ribs of the umbrella = A/n
6364.29/8 cm 2
Or, The area between the two consecutive ribs of the umbrella = 795.5 cm 2
11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Radius (r) = 25 cm
Sector angle (θ) = 115°
Since there are 2 blades,
The total area of the sector made by wiper = 2×(θ/360°)×π r 2
= 2×(115/360)×(22/7)×25 2
= 2×158125/252 cm 2
= 158125/126 = 1254.96 cm 2
12. To warn ships of underwater rocks, a lighthouse spreads a redcoloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.
(Use π = 3.14)
Let O bet the position of the lighthouse.
Here, the radius will be the distance over which light spreads.
Given radius (r) = 16.5 km
Sector angle (θ) = 80°
Now, the total area of the sea over which the ships are warned = Area made by the sector
Or, Area of sector = (θ/360°)×πr 2
= (80°/360°)×πr 2 km 2
= 189.97 km 2
13. A round table cover has six equal designs, as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm 2 . (Use √3 = 1.7)
Total number of equal designs = 6
AOB= 360°/6 = 60°
The radius of the cover = 28 cm
Cost of making design = ₹ 0.35 per cm 2
Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔAOB is an equilateral triangle. So, its area will be (√3/4)×a 2 sq. units
Here, a = OA
∴ Area of equilateral ΔAOB = (√3/4)×28 2 = 333.2 cm 2
Area of sector ACB = (60°/360°)×πr 2 cm 2
= 410.66 cm 2
So, the area of a single design = the area of sector ACB – the area of ΔAOB
= 410.66 cm 2 – 333.2 cm 2 = 77.46 cm 2
∴ area of 6 designs = 6×77.46 cm 2 = 464.76 cm 2
So, total cost of making design = 464.76 cm 2 ×Rs.0.35 per cm 2
= Rs. 162.66
14. Tick the correct solution in the following:
The area of a sector of angle p (in degrees) of a circle with radius R is
(A) p/180 × 2πR
(B) p/180 × π R 2
(C) p/360 × 2πR
(D) p/720 × 2πR 2
The area of a sector = (θ/360°)×πr 2
Given, θ = p
So, the area of sector = p/360×πR 2
Multiplying and dividing by 2 simultaneously,
= (p/360)×2/2×πR 2
= (2p/720)×2πR 2
So, option (D) is correct.
Exercise: 12.3 (Page No: 234)
1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
Here, P is in the semicircle, and so,
So, it can be concluded that QR is the hypotenuse of the circle and is equal to the diameter of the circle.
Using the Pythagorean theorem,
QR 2 = PR 2 +PQ 2
Or, QR 2 = 7 2 +24 2
QR= 25 cm = Diameter
Hence, the radius of the circle = 25/2 cm
Now, the area of the semicircle = (πR 2 )/2
= (22/7)×(25/2)×(25/2)/2 cm 2
= 13750/56 cm 2 = 245.54 cm 2
Also, the area of the ΔPQR = ½×PR×PQ
=(½)×7×24 cm 2
Hence, the area of the shaded region = 245.54 cm 2 84 cm 2
= 161.54 cm 2
2. Find the area of the shaded region in Fig. 12.20, if the radii of the two concentric circles with centre O are 7 cm and 14 cm, respectively and AOC = 40°.
Angle made by sector = 40°,
Radius the inner circle = r = 7 cm, and
Radius of the outer circle = R = 14 cm
Area of the sector = (θ/360°)×πr 2
So, Area of OAC = (40°/360°)×πr 2 cm 2
= 68.44 cm 2
Area of the sector OBD = (40°/360°)×πr 2 cm 2
= (1/9)×(22/7)×7 2 = 17.11 cm 2
Now, the area of the shaded region ABDC = Area of OAC – Area of the OBD
= 68.44 cm 2 – 17.11 cm 2 = 51.33 cm 2
3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Side of the square ABCD (as given) = 14 cm
So, the Area of ABCD = a 2
= 14×14 cm 2 = 196 cm 2
We know that the side of the square = diameter of the circle = 14 cm
So, the side of the square = diameter of the semicircle = 14 cm
∴ the radius of the semicircle = 7 cm
= (22/7×7×7)/2 cm 2
∴ he area of two semicircles = 2×77 cm 2 = 154 cm 2
Hence, the area of the shaded region = Area of the Square – Area of two semicircles
= 196 cm 2 154 cm 2
4. Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as the centre.
It is given that OAB is an equilateral triangle having each angle as 60°
The area of the sector is common in both.
The radius of the circle = 6 cm
Side of the triangle = 12 cm
Area of the equilateral triangle = (√3/4) (OA) 2 = (√3/4)×12 2 = 36√3 cm 2
Area of the circle = πR 2 = (22/7)×6 2 = 792/7 cm 2
Area of the sector making angle 60° = (60°/360°) ×πr 2 cm 2
= (1/6)×(22/7)× 6 2 cm 2 = 132/7 cm 2
Area of the shaded region = Area of the equilateral triangle + Area of the circle – Area of the sector
= 36√3 cm 2 +792/7 cm 2 132/7 cm 2
= (36√3+660/7) cm 2
5. From each corner of a square of side 4 cm, a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 12.23. Find the area of the remaining portion of the square.
Side of the square = 4 cm
The radius of the circle = 1 cm
Four quadrants of a circle are cut from the corner, and one circle of radius are cut from the middle.
Area of the square = (side) 2 = 4 2 = 16 cm 2
Area of the quadrant = (πR 2 )/4 cm 2 = (22/7)×(1 2 )/4 = 11/14 cm 2
∴ Total area of the 4 quadrants = 4 ×(11/14) cm 2 = 22/7 cm 2
Area of the circle = πR 2 cm 2 = (22/7×1 2 ) = 22/7 cm 2
Area of the shaded region = Area of the square – (Area of the 4 quadrants + Area of the circle)
= 16 cm 2 (22/7) cm 2 – (22/7) cm 2
= 68/7 cm 2
6. In a circular table cover of radius 32 cm, a design is formed, leaving an equilateral triangle ABC in the middle, as shown in Fig. 12.24. Find the area of the design.
The radius of the circle = 32 cm
Draw a median AD of the triangle passing through the centre of the circle.
⇒ BD = AB/2
Since, AD is the median of the triangle
∴ AO = Radius of the circle = (2/3) AD
⇒ (2/3)AD = 32 cm
⇒ AD = 48 cm
By Pythagoras’ theorem,
AB 2 = AD 2 +BD 2
⇒ AB 2 = 48 2 +(AB/2) 2
⇒ AB 2 = 2304+AB 2 /4
⇒ 3/4 (AB 2 )= 2304
⇒ AB 2 = 3072
⇒ AB= 32√3 cm
Area of ΔADB = √3/4 ×(32√3) 2 cm 2 = 768√3 cm 2
Area of the circle = πR 2 = (22/7)×32×32 = 22528/7 cm 2
Area of the design = Area of the circle – Area of ΔADB
= (22528/7 – 768√3) cm 2
7. In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
Side of square = 14 cm
Four quadrants are included in the four sides of the square.
∴ radius of the circles = 14/2 cm = 7 cm
Area of the square ABCD = 14 2 = 196 cm 2
Area of the quadrant = (πR 2 )/4 cm 2 = (22/7) ×7 2 /4 cm 2
= 77/2 cm 2
Total area of the quadrant = 4×77/2 cm 2 = 154cm 2
Area of the shaded region = Area of the square ABCD – Area of the quadrant
= 196 cm 2 – 154 cm 2
8. Fig. 12.26 depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find
(i) the distance around the track along its inner edge
(ii) the area of the track.
Width of the track = 10 m
Distance between two parallel lines = 60 m
Length of parallel tracks = 106 m
DE = CF = 60 m
The radius of the inner semicircle, r = OD = O’C
= 60/2 m = 30 m
The radius of the outer semicircle, R = OA = O’B
= 30+10 m = 40 m
Also, AB = CD = EF = GH = 106 m
Distance around the track along its inner edge = CD+EF+2×(Circumference of inner semicircle)
= 106+106+(2×πr) m = 212+(2×22/7×30) m
= 212+1320/7 m = 2804/7 m
Area of the track = Area of ABCD + Area EFGH + 2 × (area of outer semicircle) – 2 × (area of inner semicircle)
= (AB×CD)+(EF×GH)+2×(πr 2 /2) 2×(πR 2 /2) m 2
= (106×10)+(106×10)+2×π/2(r 2 R 2 ) m 2
= 2120+22/7×70×10 m 2
9. In Fig. 12.27, AB and CD are two diameters of a circle (with centre O) perpendicular to each other, and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
The radius of larger circle, R = 7 cm
The radius of smaller circle, r = 7/2 cm
Height of ΔBCA = OC = 7 cm
Base of ΔBCA = AB = 14 cm
Area of ΔBCA = 1/2 × AB × OC = (½)×7×14 = 49 cm 2
Area of larger circle = πR 2 = (22/7)×7 2 = 154 cm 2
Area of larger semicircle = 154/2 cm 2 = 77 cm 2
Area of smaller circle = πr 2 = (22/7)×(7/2)×(7/2) = 77/2 cm 2
Area of the shaded region = Area of the larger circle – Area of the triangle – Area of the larger semicircle + Area of the smaller circle
Area of the shaded region = (1544977+77/2) cm 2
= 133/2 cm 2 = 66.5 cm 2
10. The area of an equilateral triangle ABC is 17320.5 cm 2 . With each vertex of the triangle as the centre, a circle is drawn with a radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the area of the shaded region (Use π = 3.14 and √3 = 1.73205).
ABC is an equilateral triangle.
∴ ∠ A = ∠ B = ∠ C = 60°
There are three sectors, each making 60°.
Area of ΔABC = 17320.5 cm 2
⇒ √3/4 ×(side) 2 = 17320.5
⇒ (side) 2 =17320.5×4/1.73205
⇒ (side) 2 = 4×10 4
⇒ side = 200 cm
Radius of the circles = 200/2 cm = 100 cm
Area of the sector = (60°/360°)×π r 2 cm 2
= 1/6×3.14×(100) 2 cm 2
= 15700/3cm 2
Area of 3 sectors = 3×15700/3 = 15700 cm 2
Thus, the area of the shaded region = Area of an equilateral triangle ABC – Area of 3 sectors
= 17320.515700 cm 2 = 1620.5 cm 2
11. On a square handkerchief, nine circular designs, each of a radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.
Number of circular designs = 9
The radius of the circular design = 7 cm
There are three circles on one side of the square handkerchief.
∴ side of the square = 3×diameter of circle = 3×14 = 42 cm
Area of the square = 42×42 cm 2 = 1764 cm 2
Area of the circle = π r 2 = (22/7)×7×7 = 154 cm 2
Total area of the design = 9×154 = 1386 cm 2
Area of the remaining portion of the handkerchief = Area of the square – Total area of the design = 1764 – 1386 = 378 cm 2
12. In Fig. 12.30, OACB is a quadrant of a circle with centre O and a radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB
(ii) shaded region
Radius of the quadrant = 3.5 cm = 7/2 cm
(i) Area of the quadrant OACB = (πR 2 )/4 cm 2
= (22/7)×(7/2)×(7/2)/4 cm 2
= 77/8 cm 2
(ii) Area of the triangle BOD = (½)×(7/2)×2 cm 2
Area of the shaded region = Area of the quadrant – Area of the triangle BOD
= (77/8)(7/2) cm 2 = 49/8 cm 2
= 6.125 cm 2
13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)
Side of square = OA = AB = 20 cm
The radius of the quadrant = OB
OAB is the rightangled triangle
By Pythagoras’ theorem in ΔOAB,
OB 2 = AB 2 +OA 2
⇒ OB 2 = 20 2 +20 2
⇒ OB 2 = 400+400
⇒ OB 2 = 800
⇒ OB= 20√2 cm
Area of the quadrant = (πR 2 )/4 cm 2 = (3.14/4)×(20√2) 2 cm 2 = 628cm 2
Area of the square = 20×20 = 400 cm 2
Area of the shaded region = Area of the quadrant – Area of the square
= 628400 cm 2 = 228cm 2
14. AB and CD are, respectively, arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If ∠AOB = 30°, find the area of the shaded region.
The radius of the larger circle, R = 21 cm
The radius of the smaller circle, r = 7 cm
Angle made by sectors of both concentric circles = 30°
Area of the larger sector = (30°/360°)×πR 2 cm 2
= (1/12)×(22/7)×21 2 cm 2
= 231/2cm 2
Area of the smaller circle = (30°/360°)×πr 2 cm 2
= 1/12×22/7×7 2 cm 2
Area of the shaded region = (231/2) – (77/6) cm 2
= 616/6 cm 2 = 308/3cm 2
15. In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm, and a semicircle is drawn with BC as a diameter. Find the area of the shaded region .
The radius of the quadrant ABC of the circle = 14 cm
AB = AC = 14 cm
BC is the diameter of the semicircle.
ABC is the rightangled triangle.
By Pythagoras’ theorem in ΔABC,
BC 2 = AB 2 +AC 2
⇒ BC 2 = 14 2 +14 2
⇒ BC = 14√2 cm
Radius of the semicircle = 14√2/2 cm = 7√2 cm
Area of the ΔABC =( ½)×14×14 = 98 cm 2
Area of the quadrant = (¼)×(22/7)×(14×14) = 154 cm 2
Area of the semicircle = (½)×(22/7)×7√2×7√2 = 154 cm 2
Area of the shaded region =Area of the semicircle + Area of the ΔABC – Area of the quadrant
= 154 +98154 cm 2 = 98cm 2
16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.
AB = BC = CD = AD = 8 cm
Area of ΔABC = Area of ΔADC = (½)×8×8 = 32 cm 2
Area of quadrant AECB = Area of quadrant AFCD = (¼)×22/7×8 2
= 352/7 cm 2
Area of shaded region = (Area of quadrant AECB – Area of ΔABC) = (Area of quadrant AFCD – Area of ΔADC)
= (352/7 32)+(352/7 32) cm 2
= 2×(352/732) cm 2
= 256/7 cm 2
The 12th Chapter of NCERT Solutions for Class 10 Maths covers the concepts of the perimeter (circumference) and area of a circle and applies this knowledge in finding the areas of two special ‘parts’ of a circular region known as sector and segment.
Areas Related to Circles is a part of Mensuration, and the unit holds a total weightage of 10 marks in the CBSE exams. In the board examination, one question is sometimes asked from this chapter.
List of Exercises in Class 10 Maths Chapter 12
Exercise 12.1 Solutions (5 Solved Questions)
Exercise 12.2 Solutions (14 Solved Questions)
Exercise 12.3 Solutions (16 Solved Questions)
Chapter 12 of Maths NCERT Solutions for Class 10 is about parts of circles, their measurements and areas of plane figures. BYJU’S subject experts have prepared solutions for each question adhering to the CBSE syllabus (202324).
Class 10 Maths NCERT Chapter 12 Area related to circles consists of important topics such as
12.1  Introduction 
12.2  Perimeter and Area of a Circle 
12.3  Areas of Sector and Segment of a Circle 
12.4  Areas of Combinations of Plane Figures 
12.5  Summary 
Key Features of NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles
 NCERT Solutions help students strengthen their concepts in circlerelated areas.
 The solutions are explained using diagrams which make learning more interactive and comprehensive.
 The language used in NCERT Solutions is easy and understandable.
 The stepbystep solving approach helps students to clear their basics.
 Help students solve complex problems at their own pace.
Students can also refer to the NCERT Solutions of other classes and subjects. These solutions are prepared by wellexperienced teachers at BYJU’S focusing on providing clarity on key concepts and problemsolving skills.
Students can also get a good grip on the important concepts by referring to other study materials which are provided at BYJU’S.
 RD Sharma Solutions for Class 10 Maths Chapter 15 Areas Related to Circles
Disclaimer –
Dropped Topics –
12.1 Introduction 12.2 Perimeter and area of a circle — A review 12.4 Areas of combinations of plane figures
Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 12
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NCERT Solutions of Class 10 Maths Ch 12 Areas Related to Circles
 Exercise 12.1
 Exercise 12.2
 Exercise 12.3
How many exercises in Chapter 12 Areas related to Circles
What is area of a circle, what is perimeter of semicircle, what is circumference of a circle or perimeter of a circle, contact form.
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CBSE 10th Standard Maths Subject Areas Related to Circles Case Study Questions With Solution 2021 By QB365 on 22 May, 2021 QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions .
The case study on Areas Related to Circles Class 10 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Areas Related to Circles case study questions are very easy to grasp from the PDF  download links are given on this page.
Here, we have provided casebased/passagebased questions for Class 10 Maths Chapter 12 Areas Related to Circles. Case Study/PassageBased Questions. Question 1: Mr Ramanand purchased a plot QRUT to build his house. He leave space of two congruent semicircles for gardening and a rectangular area of breadth 3 em for car parking.
Case study questions are helpful for the preparation of the Class 10 Maths Exam 20212022. Case Study Questions for Class 10 Maths Chapter 10  Circles. CASE STUDY 1: A Ferris wheel (or a big ...
CBSE 10th Standard Maths Subject Circles Case Study Questions With Solution 2021 Answer Keys. Case Study Questions. (i) (b): Here, OS the is radius of circle. Since radius at the point of contact is perpendicularto tangent. So, \ (\angle\) OSA = 90°. (ii) (d): Since, length of tangents drawn from an external point to a circle are equal.
Case Study Questions for Class 10 Maths Chapter 10 Circles. Question 1: A student draws two circles that touch each other externally at point K with centres A and B and radii 6 cm and 4 cm, respectively as shown in the figure. Based on the above information, answer the following questions. Download the APP Now!
Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam of 202223. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards. ... Chapter12 Areas related to Circles. This chapter problem ...
Updated for new NCERT  20232024 Boards. NCERT Solutions of all exercise questions and examples of Chapter 11 Class 10 Areas related to Circle. Answers to all questions are available with video free at teachoo. In this chapter, we will. Click on an exercise link below to learn from the NCERT Book way.
You can check your knowledge by solving case studybased questions for Class 10 Maths Chapter 10 Circles. In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passagebased as well. In that, a paragraph will be given, and then the MCQ ...
These tests are unlimited in nature…take as many as you like. You will be able to view the solutions only after you end the test. TopperLearning provides a complete collection of case studies for CBSE Class 10 Maths Areas Related to Circles chapter. Improve your understanding of biological concepts and develop problemsolving skills with ...
Case Study Questions for Class 10 Maths Chapter 12 Areas Related to Circles Case Study Questions Question 1: A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some broochs are shown below. Observe them carefully Design … Continue reading Case Study Questions for Class 10 Maths Chapter 12 Areas Related to Circles
QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get more marks in Exams  Complete list of 10th Standard CBSE question papers, syllabus, exam tips, study material, previous year exam question papers, centum ...
In this video i solved the case study based questi... Hello my dear students This video is from the cbse material which is provided by the cbse on its website . CBSE Exam, class 10
Here are some of the important points to remember for Areas Related to Circles Class 10 Notes CBSE Maths Chapter 12: The radius of a circle is the distance from the center of the circle to any point on its circumference. The diameter of a circle is twice the radius. The circumference of a circle is the distance around the circle.
Practice Questions For Class 10 Maths Chapter 12 Areas Related to Circles. A calf is tied with a rope of length 6 m at the corner of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5m, find the increase in the area of the grassy lawn in which the calf can graze. Find the radius of a circle whose circumference is ...
Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Area Related to Circles in order to fully complete your preparation.They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!. I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams.
Area of a Sector of a Circle. The area of a sector is given by. ( θ/360°) ×πr2. where ∠θ is the angle of this sector (minor sector in the following case) and r is its radius. Area of a sector. Example: Suppose the sector of a circle is 45° and radius is 4 cm, then the area of the sector will be: Area = (θ/360°) × πr 2.
Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 12 Areas Related to Circles with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Areas Related to Circles Case Study and Passage Based Questions with Answers, feel free to ...
Access answers of Maths NCERT Class 10 Chapter 12  Areas Related to Circles Class 10 Maths Chapter 12 Exercise: 12.1 (Page No: 230) Exercise: 12.1 (Page No: 230) 1. The radii of the two circles are 19 cm and 9 cm, respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles ...
Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now. Case Study on Areas Related to Circles Class 10 Maths: Here, you will get Case Study Questions on Class 10 Areas Related to Circles PDF at free of cost. Along with you can also download Areas Related to Circles case study ...
File Name : C:\Class  X (Maths)/Final/Chap12 28.011.06 12.2 Perimeter and Area of a Circle — A Review Recall that the distance covered by travelling once around a circle is its perimeter, usually called its circumference. You also know from your earlier classes, that circumference of a circle bears a constant ratio with its diameter.
Question 4  Case Based Questions (MCQ)  Chapter 11 Class 10 Areas related to Circles. Last updated at April 16, 2024 by Teachoo. In a workshop, brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the given figure.
Answer. Area of the sector making angle θ = (θ / 360°)×π r 2. Area of the sector making angle 60° = (60° / 360°)×π r 2 cm 2. = (1 / 6)×6 2 π = 36/6 π cm 2 = 6 × 22/7 cm 2 = 132/7 cm 2. 2. Find the area of a quadrant of a circle whose circumference is 22 cm. Answer. Quadrant of a circle means sector is making angle 90°.
Case Study Class 10 Maths Areas Related to Circle  Free download as PDF File (.pdf), Text File (.txt) or read online for free. A farmer has a rectangular field of length 60 m and breadth 30 m. A pit of diameter 14 m is dug in the center, 12 m deep. The pit soil is spread evenly in the field. Questions ask about the pit volume, field area before and after pit, and level rise in the field.