NCERT Solutions For Class 10 Maths Chapter 12, Areas Related to Circles, available now in PDF for registered student in BYJU’S. These solutions are intended to help students to find the areas of circles which are in correlation with a circle. To make the students of 10th Class prepared for their board exams, the experts have formulated these solutions based on CBSE syllabus of NCERT book.

For students of class 10 maths facing their first real board exam, it is necessary for them to learn all the topics covered in Maths syllabus prescribed by the board. Here we have provided solutions for class 10 maths chapter 12 ‘areas related to circles’ in a broad and simplified manner.

## Introduction to NCERT Solutions Class 10 Maths Areas Related to Circles

Area related to circles, the chapter of class 10 consists of important topics such as;

- Introduction
- Perimeter and area of a circle
- Areas of sector and segment of a circle
- Areas of combinations of plane figure

We are also providing here exercises belonging to each topic to practice well in a detailed manner and understand the concepts defined in chapter 12 of class 10. Students are advised to solve previous year question papers to get an idea of paper pattern and also it could help to do time management based on the duration given to write the exams. Click here to get question paper for Class 10th Maths subject.

Many objects that we come across in our daily life are related to the circular shape in some form or the other. The chapter, ‘Areas related to circles’ covers the topics whose concepts are relevant to our real-life applications.

Students are recommended to use these material provided by BYJU’S to prepare themselves for their board exams and also for students who are appearing for competitive exams such as JEE, CAT, NEET,etc. This year, in 2019. With the help of these solutions, you can understand the concepts in depth for each question and score good marks in exams.

### NCERT Solutions Class 10 Maths Chapter 12 Exercises

**1. If perimeter of a circle is equal to that of a square, What would be the ratio of their areas.**

Here, perimeter of circle = \( 2\pi r\) and perimeter of square = 4a

∴\( 2\pi r\) = 4a \(\Rightarrow r=\frac{2a}{\pi }\)

∴ Ratio of their areas = \(\frac{\pi r^{2}}{a^{2}}\) = \(\frac{\pi }{a^{2}}\times \left( \frac{2a}{\pi } \right )^{2}\)

= \(\frac{\pi }{a^{2}}\times \left( \frac{4a^{2}}{\pi_{2} } \right )\) = \(\frac{4 }{\pi }\) = \(\frac{4 }{22 }\times 7\) = \(\frac{14}{11}\)

**2. Find the areas of circle that can be inscribed in a square of side 8 cm.**

** **Diameter of the circle = Side of the square

∴ Diameter = 8 cm

\(\Rightarrow\) Radius = \(\frac{8}{2}\) = 4 cm∴ Area of the circle = \(\pi \times4 \times4\) \(cm^{2}\)

= \(16\pi\) \(cm^{2}\)

**3. If sum of the areas of two circles with radii \(R_{1} and R_{2}\) is equal to area of a circle radius R, derive the relation among their radii.**

Since sum of areas of two circles with radii \(R_{1} and R_{2}\) = Area of circle with radii R.

\(\Rightarrow\) \(\pi R_{1}^{2}+\pi R_{2}^{2}=\pi R^{2}\) \(\Rightarrow\) \(R_{1}^{2}+R_{2}^{2}=R^{2}\) or R = \(\sqrt{R_{1}^{2}+R_{2}^{2}}\)

**4. If sum of the circumference of two circles with radii radii \(R_{1} and R_{2}\) is equal to circumference of a circle radius R, derive the relation among their radii.**

Since sum of circumference of two circles with radii \(R_{1} and R_{2}\) = Circumference of circle with radii R.

\(\Rightarrow\) 2\(\pi R_{1}+2\pi R_{2}=2\pi R\) \(\Rightarrow\) \(R_{1}+ R_{2}= R\)

**5. If the circumference of a circle and the perimeter of a square are equal, then write the relation between their radii.**

Since sum of circumference of a circle = Perimeter of a square

\(\Rightarrow\) \(2\pi r=4\times side\) \(\Rightarrow\) \(r=\frac{2}{\pi }\times side\)∴ Area of circle = \(\pi \times \frac{4}{\pi ^{2}}\times (side)^{2}\)

= \(\frac{4}{\pi}(side)^{2}\)

= \(\frac{28}{22}(side)^{2}\)

Area of square = \((side)^{2}\)

Since \(\frac{28}{22}(side)^{2}\) > \((side)^{2}\)

**6. To build a single circular park equal in area to the sum of areas of two circular parks of diameter 8 m and 6 m in a locality. What would be the radius of new park?**

Here, area of single park = Area of the park with (d= 8m) + Area of the park with (d= 6m)

\(\Rightarrow\) \(\pi R^{2}= \pi (4)^{2}+\pi (3)^{2}\) \(\Rightarrow\) \(\pi R^{2}= \pi (16+9)\) \(\Rightarrow\) \(R^{2}= 25\) \(\Rightarrow\) R = 5 m.

**7. Find the area of a square that can be inscribed in a circle of radius 6 cm.**

Since square is inscribed in a circle

\(\Rightarrow\) Diagonal of square = Diametre of circle \(\Rightarrow\) \(\sqrt{2}\times side\) = 12Side = \(\frac{12}{\sqrt{2}}\) cm

Now, area of the square = \(\frac{12}{\sqrt{2}} \times \frac{12}{\sqrt{2}}\) = 72 \(cm^{2}\)

**8. Find the radius of a circle whose circumference is equal to the sum of the circumference of two circles of diameter 24 cm and 16 cm.**

From the above, we obtain

2\(\pi R = 2\pi R_{1}+2\pi R_{2}\) \(\Rightarrow\) \(R =R_{1}+R_{2}\) \(\Rightarrow\) R = 12 + 8 = 20 cm

**9. Find the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 20 cm and 15 cm.**

From the above, we obtain

\(\pi R^{2}= \pi R_{1}^{2}+\pi R_{2}^{2}\) \(\Rightarrow\) \(R^{2}=R_{1}^{2}+R_{2}^{2}\) \(\Rightarrow\) \(R^{2}=(20)^{2}+(15)^{2}\) = 400 + 225 = 625 \(\Rightarrow\) R = 25 cm

**10. Find the area of the largest triangle that can be inscribed in a semicircle of radius r.**

Here, among all altitudes, radius is the largest altitude

∴ \( Area of \bigtriangleup PQR = \frac{1}{2}\times 2r \times r\)

= \(r^{2} sq.units\)

Hence, area of the largest triangle inscribed in a semicircle r is \(r^{2} sq.units\).

**11. In the given figure , a square of diagonal 14 cm is inscribed in a circle. Find the area of the shaded region.**

Here, diagonal of the square = 14 cm

\(\Rightarrow\) \(\sqrt{2}\times side\) = 14** **\(\Rightarrow\) side = \(\frac{14}{\sqrt{2}}\) cm

∴ Area of the square = \(\frac{8}{\sqrt{2}}\times \frac{8}{\sqrt{2}}\) = 98 sq.cm

Diameter of the circle = 14 cm

∴ Radius of the circle = 7 cm.

Area of the circle = \(\pi (4)^{2}\) = \(\frac{22}{7}\times 49\) = 154 sq.cm

Thus, area of the shaded region = 154 – 98

= 56 sq.cm

**12. The wheel of a motor cycle is of radius 28cm. How many revolutions per minitue must the wheel make so as to keep a speed of 60 Km/h?**

Here, speed of the motor cycle = 60 Km/h

= \(\frac{60000}{60}\) m/min

= 1000 m/min

Radius of the wheel (r) = 28 cm = \(\frac{28}{100}\) m.

Circumference of the wheel = \(2\times \frac{22}{7}\times \frac{28}{100}\) m

= 1.76 m

Number of revolutions per minute = \(\frac{1000}{1.76}\)

= 568.18

**13. A cow is tied with a rope of length 21 m at the corner of a rectangular field of dimensions 27 m \(\times\) 23 m. Find the area of the field in which the cow can graze.**

Length of the rope = 21 m

∴ Area of the field in which the cow can graze

= \(\frac{90°}{360°}\times \frac{22}{7}\times 21\times 21\)

= 346.5 \(m^{2}\)

**14. In the given fig, arcs are drawn by taking vertices P,Q and R of an equilateral triangle of side 12 cm to intersect the sides QR, RR and PQ at their respective mid-points X, Y and Z.** **Find the area of the shaded region.**

** **** **Side of an equilateral \(m^{2\triangle }\) PQR = 12 cm

Since, arcs are drawn by taking vertices P,Q and R to intersect the sides QR, RR and PQ at their mid-points X, Y and Z.

∴ Radii of each sector (r) = 6 cm

Also, \(m^{2\triangle }\) PQR is an equilateral, therefore

\(\angle P= \angle Q=\angle R= 60°\)Required area of the shaded region

= \(3\times \frac{\theta }{360°}\times\pi r^{2}\)

= \(3\times \frac{60° }{360°}\times3.14\times 6^{2}\)

= \(3\times \frac{1}{6}\times3.14\times36\)

= 56.52 \(cm^{2}\)

**15. In the given fig, arcs have been drawn with radii 7 cm each and with centers A, B and C.**

Find the area of the shaded region.

** **

Here, radii of each sector at A, B and C is 7 cm

Let \(\angle A=\theta _{1}°\angle B=\theta _{2}°, \angle C=\theta _{3}°\)

Required area of shaded region

= \(\frac{\theta_{1}°}{360°}\pi r^{2}+\frac{\theta_{2}°}{360°}\pi r^{2}+\frac{\theta_{3}°}{360°}\pi r^{2}\)

= \(\frac{\pi r^{2}}{360°}(\theta _{1}°+\theta _{2}°+\theta _{3}°)\)

= \(\frac{22\times 7\times 7}{7\times 360°}\times 180°\) \([Since\, \theta _{1}°+\theta _{2}°+\theta _{3}°=180°]\)

= 77 \(cm^{2}\)

** **

** **

**16. A circular park is surrounded by a road 20 m wide. If the radius of the park is 100 m, find the area of the road.**

** **

Here, radius of the circular park (r) = 100 m

Width of the road = 20 m

∴ Radius of the outer circle (R) = 100 + 20 = 120 m

Required area of road

= \(\pi R^{2}+\pi r^{2}\)

= \(\pi\) (R + r) (R – r)

= \(\frac{22}{7}\) (120 + 100) (120 – 100)

= \(\frac{22}{7}\times 220\times 20\) = 13816 \(m^{2}\)

**17. A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of \(90°\) at its center. Find the radius of the circle.**

Here, length of wire (arc of required circle) = 22 cm

And central angle(\(\theta\)) = \(90°\)

Let r be the radius of the required circle.

∴\( \frac{\theta }{360°}\times 2\pi r\) = length of the arc

∴\( \frac{60°}{360°}\times 2\times \frac{22}{7} r\) = 20

\(\Rightarrow r=\frac{22\times 7\times 360°}{60°\times 2\times 22}\) = 21 cm** **

**18. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand during the time period 6:05 A.M and 6:40 A.M.**

Angle swept by the minute hand in one minute = \(6°\)

∴Angle swept by the minute hand in 35 minute = \((6\times 35)°= 210°\)

Length of the minute hand = 7 cm

Area swept by the minute hand = \(\frac{210°}{360°}\times \frac{22}{7}\times 7\times 7\)

= 89.83 \(cm^{2}\)

**19. Find the area of the sector of a circle of radius 7 cm, if the corresponding arc length is 4 cm.**

Radius of the circle (r) = 7 cm

Length of the arc = 3.5 cm

∴ \(\frac{\theta }{360°}\times 2\pi r\) = 4 cm

\(\Rightarrow \frac{\theta }{360°}\times 2\times \frac{22}{7}\times 7\) = 4 \(\Rightarrow \frac{\theta }{360°}\) = \(\frac{4\times 7}{2\times 22\times 7}\) = \(\frac{1}{11}\)Now, area of the corresponding arc

= \(\frac{1}{11}\times \frac{22}{7}\times 7\times 7\)

** **= 14 \(cm^{2}\)

** **

**20. ****A circular pond is of diameter 20 m. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs.30 per \(m^{2}\).**

** **

Radius of the circular pond r = \(\frac{20}{2}\) = 10 m

Width of the path = 2 m

∴ Radius of the outer circle (R) = 10 + 2 = 12 m

Area of the path = \(\pi R^{2}-\pi r^{2}\)

= \(\pi\) (R + r) (R – r)

= \(\frac{22}{7}\) (12 + 10) (12 – 10)

= \(\frac{22}{7} \times 22\times 2\) = 138.16 \(m^{2}\)

Cost of constructing the path @ Rs.30 per \(m^{2}\)

** **= Rs.4144.8

** **

**21. Find the number of revolutions made by a circular tier of area 1.32 \(m^{2}\) in rolling a distance of 165 m.**

** **

** **Area of a circular wheel = 1.32 \(m^{2}\)
\(\pi r^{2}\) = 1.32

r = \(\sqrt{0.42}\) = 0.648 m

Distance traveled in one rotation = \(2\pi r\)

= \(2\times \frac{22}{7}\times 0.648\)

= 4.07 m

Total distance to be travelled = 165

∴Number of rotations = \(\frac{165}{4.07}\) = 40.54

** **

**22. Find the difference of the areas of sectors of angle \(90°\) and its corresponding major sector of a circle of radius 14 cm.**

** **

Radius of the circle (r) = 14 cm

Angle of the sector \(\theta = 90°\)

∴ Area of the sector = \(\frac{\theta }{360°} \times \pi r^{2}\)

= \(\frac{90°}{360°} \times \frac{22}{7}\times 14\times 14\)

= 154 \(cm^{2}\)

Area of major sector

= \(\frac{360° – 90°}{360°}\times \frac{22}{7}\times 14\times 14\)

= 462 \(cm^{2}\)

** ∴** The required distance = 308 cm²

**23. Find the area of the shaded region in the give figure.**

** **

Here, unshaded region is combination of two semicircles and one rectangle.

∴ Area of unshaded region = Area of rectangle with dimensions (26-3-3-) cm by (12-4-4) cm + 2 area of semicircle with radii \(\left ( \frac{12-4-4}{2} \right )\)

= \(16\times 4+ 2\times \frac{1}{2}\times \pi \times 2\times 2\)

** **** **= (64+4\(\pi\)) \(cm^{2}\)

** **

Required area of shaded region

= \(26\times 12-(64+4\pi )\)

= 312 – 64 – 4\(\pi\)

= (248 – 4\(\pi\)) \(cm^{2}\)

**24. Find the area of the minor segment of a circle of radius 28 cm, when the angle of the corresponding sector is \(90°\).**

Radius of the circle (r) = 28 cm

Central angle (\(\theta\)) = \(90°\)

Area of minor segment

** **** = Area of sector PRQ – Area of \(△\)PRQ**

= \(\frac{90°}{360°}\times \frac{22}{7}\times 28\times 28 -\frac{1}{2}\times 28\times 28\times sin 90°\) \(90°\)

= 616 – 392

= 224 \(cm^{2}\)

**25. The diameter of front and rear wheels of a tractor 70 cm and 2 m respectively. Find the number of rotations that rear wheel will make in covering a distance in which the wheel makes 1200 rotations.**

** **

Radius of the front wheel (r) = 35 cm

Number of rotations = 1200

∴ Total distance covered by front wheel = 1200 2\(\pi\)r

= \(1200\times 2\times \frac{22}{7}\times 35\)

= 264000 cm

Radius of the rear wheel (R) = \(\frac{200}{2}\) = 100 cm

** Distance covered by the rear wheel in one rotation = \(2\times \frac{22}{7}\times 100\)**

** **

**26. Number of rotations = \(\frac{264000 \times 7}{2\times 22\times 100}\) = 420****Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.**

** **

** **Length of the rope = 7 m

Sides of the triangular field are AB = 16 m, BC = 17 m and AC = 15 m

Let cow, buffalo and horse be tied at vertices A, B and C respectively.

** **

**∴ Area of** \(\bigtriangleup\) ABC = \(\sqrt{24(24-15)(24-16)(24-17)}\)

= \(\sqrt{24\times 9\times 8\times 7}\) = \(24\sqrt{21}\) \(m^{2}\)

Hence, area of the field which cannot be grazed by three animals = \((24\sqrt{21}-77)\) \(m^{2}\)

**27. Find the area of the flower bed ( with semi- circular ends) as shown in the figure.**

** **** **

Here, the flower bed is a combination of two semicircular ends and rectangle.

Radius of semicircular ends = 7 cm

Length and breadth of rectangular field are 35 cm and 8 cm

∴ Area of the shaded region

= \(2\times \frac{1}{2}\pi r^{2}+l\times b\)

= \(\frac{22}{7}\times 7\times 7 + 35\times 8\)

= 154 + 280

= 434 \(cm^{2}\)

Hence, area of the flower bed is 434 \(cm^{2}\).

** **** **

**28. On a square cardboard sheet of area 196 \(cm^{2}\), four congruent circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to two circular plates. Find the area of the square sheet not covered by circular plates.**

** **

** Area of the square PQRS = 196 \(cm^{2}\)**

** Side of the square = \(\sqrt{196}\) = 14 cm**

** **

** ∴ Diameter of each circular plate = 7 cm**

** And radius of each circular plate = 3.5 cm**

** Area of four circular plates = \(4\times \frac{22}{7}\times 3.5\times 3.5\)**

** = 154 \(cm^{2}\)**

** ****Area of the square sheet not covered by the circular plate = 196 – 154 = 42 \(cm^{2}\).**

** **

** 29. ****The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120° and 40°. Find the areas of two sectors as well as the lengths of those corresponding arcs.**

Here, radii of two circles are \(r_1 = 7cm\) and \(r_2 = 21cm\)

Central angles are \(\Theta _1= 120°\) and \(\Theta _2= 40°\)

∴ Areas of two sectors are

\(\frac{120°}{360°}\times\frac{22}{7}\times 7\times 7\) and \(\frac{40°}{360°}\times\frac{22}{7}\times 21\times 21\) \(\Rightarrow 51.33 \,cm^{2}\, and\, 154\,cm^{2}\)Lengths of corresponding arcs are

\(\frac{120°}{360°}\times\frac{22}{7}\times 7\times 2\) and \(\frac{40°}{360°}\times\frac{22}{7}\times 21\times 2\) \(\Rightarrow 14.67 cm and 14.67 cm

Arc lengths of two sectors of two different circles may be equal (14.67 cm), but their areas need not be equal.

**30. Floor of a room is of dimension 8 m × 5 m and it is covered with circular tiles of diameters 40 cm each as shown in figure. Find the area of floor that remains uncovered with tiles.**

** **

** **

** Radii of each circular tile = 20 cm**

** ∴ Number of circular tiles along its length ****= 20**

** Number of circular tiles along its breadth ****= 12.5**

** Total number of circular tiles =20× 12.5 = 250**

** Area of each circular tile =3.14×20×20**

** = 1256 cm²**

** Area of 250 circular tile = 250 × 1256**

** = 314000 cm²**

** Area of the rectangle floor = 800 × 500**

** = 400000 cm²**

** **

** Thus, area of the floor that remains uncovered with tiles = 400000 – 314000**

** = 86000 cm²**

** = 8.6 cm²**

CBSE (Central Board of Secondary Education) in India follow the NCERT curriculum to conducts its examinations for class 10 and class 12 respectively. Class 10 is one of the crucial stages in students academic life. The NCERT Solutions Class 10 Maths Areas Related to Circles is given, so that students can understand the concepts of this chapter in depth.

This Class 10 Chapter 12 covers the concepts of perimeter (circumference) and area of a circle and apply this knowledge in finding the areas of two special ‘parts’ of a circular region (or briefly of a circle) known as sector and segment is provided here.

Some key points about the chapter is given here:

- The circumference of a circle bears a constant ratio with its diameter. This constant ratio is denoted by the Greek letter π (read as ‘pi’).
- The portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle and the portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle.