NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circles (Ex 12.1) Exercise 12.1

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circles (Ex 12.1) Exercise 12.1

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Access NCERT Solutions for Class 10 Mathematics Chapter 12 – Areas related to circles part-1

Access NCERT Solutions for Class 10 Mathematics Chapter 12 – Areas related to circles

Exercise (12.1)

1: The radii of two circles are 19 cm and 9 cm respectively. Find the radius of

the circle which has circumference equal to the sum of the circumferences of

the two circles.

Ans:

Given that,

Radius of 1st circle = \[{r_1}\] = 19 cm

Radius of 2nd circle = \[{r_2}\] = 9 cm

Circumference of 3rd circle = Circumference of 1st circle + Circumference of

2nd circle

Let the radius of the 3rd circle be \[r\].

Now,

Circumference of 1st circle = \[2\pi {r_1}\] 

                                             = \[2\pi (19)\] 

                                             = \[38\pi \]

Circumference of 2nd circle = \[2\pi {r_2}\] 

                                              = \[2\pi (9)\] 

                                              = \[18\pi \]

Circumference of 3rd circle = \[2\pi {r_{}}\]

Using given condition,

\[2\pi r = 38\pi  + 18\pi \]

         \[ = 56\pi \]

\[ \Rightarrow r = \dfrac{{56\pi }}{{2\pi }}\]

          \[ = 28\].

Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm.


2: The radii of 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.

Ans:

Given that,

Radius of 1st circle = \[{r_1}\] = 8 cm

Radius of 2nd  circle = \[{r_2}\] = 6 cm

Area of 3rd circle = Area of 1st circle + Area of 2nd circle

Let the radius of the 3rd circle be \[r\].

Area of 1st circle = \[\pi {r_1}^2\]

                                       \[ = \pi {(8)^2}\]

                                       \[ = 64\pi \]

Area of 2nd circle = \[\pi {r_2}\]

                                           =\[\pi {r_2}^2\]

                                           =\[36\pi \]

Using given condition,

\[\pi {r^2} = \pi {r_1}^2 + \pi {r_2}^2\]

        \[ = 64\pi  + 36\pi \]

        \[ = 100\pi \]

\[ \Rightarrow {r^2} = 100\]

     \[r =  \pm 10\] 

We know that the radius cannot be negative. Therefore, the radius of the circle having an area equal to the sum of the areas of the other two circles, is 10 cm.


3: Given figure depicts an archery target marked with its five scoring areas from the center outwards as Gold, Red, Blue, Black and White. The diameter of the region representing 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.  \[\mathbf{Use}\text{ }\pi \text{ }\!\!~\!\!\text{ }=\dfrac{22}{7}\]

(Image will be uploaded soon)

Solution 3:

(Image will be uploaded soon)

Radius of gold region (i.e., 1st circle) = \[{r_1}\]

                                                             \[ = \dfrac{{21}}{2}\]

                                                              \[ = 10.5\].

Given that each circle is 10.5 cm wider than the previous circle.

Thus, radius of 2nd circle = \[{r_2}\]

                                         = \[10.5 + 10.5\]

                                         = 21 cm

Radius of 3rd circle = \[{r_3}\]

                               = \[21{\rm{ }} + {\rm{ }}10.5\]

                               = 31.5 cm

Radius of 4th circle = \[{r_4}\]

                                            = \[31.5{\rm{ }} + {\rm{ }}10.5\]

                               = 42 cm

Radius of 5th circle = \[{r_5}\] 

                                = \[42{\rm{ }} + {\rm{ }}10.5\]

                                = 52.5 cm

According to given condition,

Area of gold region = Area of 1st circle

\[ \Rightarrow \pi {r_1}^2 = \pi {(10.5)^2}\]

\[ = 346.5c{m^2}\]

Area of red region = Area of 2nd circle − Area of 1st circle

                                           \[ = \pi {r_2}^2 - \pi {r_1}^2\]

                                           \[ = \pi {(21)^2} - \pi {(10.5)^2}\]

                                           \[ = 441\pi  - 110.25\pi \] 

                                            \[ = 330.75\pi \]

                                            \[ = 1039.5c{m^2}\]                                          

Area of blue region = Area of 3rd circle − Area of 2nd circle

                                             \[ = \pi {r_3}^2 - \pi {r_2}^2\]

                                             \[ = \pi {(31.5)^2} - \pi {(21)^2}\]

                                             \[ = 992.25\pi  - 441\pi \]

                                              \[ = 551.25\pi \]

                                               \[ = 1732.5c{m^2}\]

Area of black region = Area of 4th circle − Area of 3rd circle

                                               \[ = \pi {r_4}^2 - \pi {r_3}^2\]

                                                \[ = \pi {(42)^2} - \pi {(31.5)^2}\]

                                                \[ = 1764\pi  - 992.25\pi \]

                                                 \[ = 771.75\pi \]

                                                 \[ = 2425.5c{m^2}\]

Area of white region = Area of 5th circle − Area of 4th circle

                                                \[ = \pi {r_5}^2 - \pi {r_4}^2\]

                                                \[ = \pi {(52.5)^2} - \pi {(42)^2}\]

                                                \[ = 2756.25\pi  - 1764\pi \]

                                                \[ = 992.25\pi \]

                                                \[ = 3118.5c{m^2}\]

Therefore, areas of gold, red, blue, black, and white regions are \[346.5c{m^2}\], \[1039.5c{m^2}\], \[1732.5c{m^2}\], \[2425.5c{m^2}\] and \[3118.5c{m^2}\] respectively.


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 traveling at a speed of 66 km per hour?

Ans:

Given that,

Diameter of the wheel of the car = 80 cm

Radius of the wheel of the car = \[r\] =40 cm

Speed of car = 66 km/hour

We know that,

Circumference of wheel = \[2\pi r\]

                                        = \[2\pi (40)\] 

                                        = \[80\pi cm\]

Speed of car \[ = \dfrac{{66 \times 100000}}{{60}}cm/\min \]

                              \[ = 1,10,000cm/\min \]

Now, distance travelled by the car in 10 minutes \[ = 110000 \times 10\]

\[ = 11,00,000cm\]

Let the number of revolutions of the wheel of the car be n.

We know that,

Distance travelled in 10 minutes = n × Distance travelled in 1 revolution (i.e., circumference)

\[ \Rightarrow 1100000 = n \times 80\pi \]

                        \[ = \dfrac{{35000}}{8}\]

                        \[ = 4375\]

Therefore, each wheel of the car will make 4375 revolutions.


5: Tick the correct answer 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

Ans:

Given that, 

the circumference and the area of the circle are equal.

Let the radius (to be find) of the circle be \[r\]

Thus, 

Circumference of circle \[ = 2\pi r\] and 

Area of circle \[ = \pi {r^2}\]

According to given condition,

\[2\pi r = \pi {r^2}\]

\[ \Rightarrow 2 = r\]

Therefore, the radius of the circle is 2 units.

Hence, the correct answer is A.                                


NCERT Solutions for Class 10 Maths Chapter 12 Areas Related To Circles Exercise 12.1

Opting for the NCERT solutions for Ex 12.1 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 12.1 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 12 Exercise 12.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 10 Maths Chapter 12 Exercise 12.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 10 Maths Chapter 12 Exercise 12.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

FAQs (Frequently Asked Questions)

1. Which are the best sums in Exercise 12.1 of 10th Maths for Exams?

There is no such thing as best sums. It is because the easier ones help you recall your concepts and the more advanced ones test your problem-solving skills and how you apply your concepts.The Chapter 12 Areas Related To Circles is a chapter that requires your utmost attention or else you might not be able to grasp the concepts clearly. Kindly practice all the sums irrespective of the difficulty level.

2. Which questions are tough questions in Exercise 12.1 Class 10 Maths?

The questions of NCERT Class 10 Maths Exercise 12.1 are all basic questions which the students can easily solve if their concepts are clear. There are not many complex questions in that exercise. Practice all of them to revisit your concepts and clarify doubts if any existed. You can also visit the official website of Vedantu or the Vedantu app if more guidance or aid is required to understand the chapter. The PDFs of  NCERT Solutions are available here at free of cost for students to access.

3. Which questions are counted as the easiest ones from Exercise 12.1 of 10th Maths?

If students' concepts are clear they can simply solve the problems in NCERT Class 10 Maths Exercise 12.1. In that exercise, there aren't a lot of challenging questions. All of them should be practised to refresh your memory and clear any doubts you may have had. If you need additional help understanding the chapter, you can go visit the Vedantu website or download the Vedantu app.

4. What is a Circle?

Two-dimensional figures are produced by all points in the same plane that are equidistant from the fixed point. The fixed point is called the circle's centre, and the fixed distance from the circle's centre is called the circle's radius. This is the basic definition of the shape called a circle. The geometry related to it along with its applications and formulas are of utmost importance.

5. What are the Exercises in Class 10 Chapter 12?

In the previous chapters, we covered the basic concept of circles as well as other essentials. This is where we'll continue our examination of circles in more detail. Two theorems and two exercises are there in the chapter. All the exercises need to be given equal importance as all of them are structured efficiently to test your concepts as well as to make you familiar with the exam format.

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