NCERT Solutions for Class 10 Maths Chapter 12 - Exercise

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.

NCERT Solutions for Class 10 Maths

• 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

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