# NCERT Solutions for Class 10 Maths Chapter 12 - Areas Related to Circles

**Important Points**

**1. Circle :** It is a collection of all points in a plane which are at a constant distance from a fixed point.

(i) Area of cicle = πr^{2}

$$\text{(ii)\space}\text{Area of semi-circle =}\frac{\pi r^{2}}{2}\\\text{(iii) Area of quadrant of circle =}\frac{\pi r^{2}}{4}$$

(iv) Circumference of circle = 2πr = πd

where d= 2r = diameter of circle

(v) Circumference of semi-circle = πr

(vi) Perimeter of semi-circle = πr + 2r

**2. Area of Sector of Circle :** The circular region enclosed by two radii and corresponding arc is called a sector of the circle. The arc AB is called the length of the arc. The region OAB is called the minor sector and the region OACDBO is called the major sector.

$$\text{(i) Length of an arc of sector AB\space =}\space\frac{\theta}{360\degree}×2\pi r\\\text{(ii) Area of minor sector OAB =}\frac{\theta}{360\degree}×\pi r^{2}\\\text{(iii) Area of major sector OAB =}\frac{360\degree-\theta}{360\degree}×\pi r^{2}$$

**3. Area of Segment of Circle :** Let AB be a chord of the circle having centre O. The regrion covered by the line and the curve i.e., region ABC is a segment of the circle.

The line AB covered a smaller region ABC is called the minor segment of a circle and the major region ABEDA covered by a line is called the major segment of a circle.

(i) Area of minor segment

= Area of sector OACBO – Area of ΔOAB

$$=\frac{\theta}{360\degree}×\pi r^{2}-\text{Area of OAB}$$

(ii) Area of major segment

= Area of circle – Area of minor segment

$$=\pi r^{2}-\bigg(\frac{\theta}{360\degree}×\pi r^{2}-\text{Area of OAB}\bigg)$$

**Note :** The area of the sector and segment means, we have to determine the area of minor sector and minor segment.

**4. Important Useful Formulae**

(i) Area of square = (Side)^{2}

$$\text{(ii) Area of triangle =}\space\frac{1}{2}×\text{Base × Height}\\\text{(iii) Area of an equilateral triangle =}\frac{\sqrt{3}}{4}\text{(Side)}^{2}\\\text{(iv) Area of trapezium =}\frac{1}{2}\text{(Sum of parallel side)}\\ × \text{Distance between two parallel sides}$$

(v) Area of ring = π(R^{2} – r^{2})

where, R and r be radius of outer ring and inner ring, respectively.

**Exercise 12.1**

$$\text{• Unless stated otherwise, use}\space\pi=\frac{22}{7}.$$

**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.**

**Sol.** Let r_{1} = 19 cm and r_{2} = 9 cm and r be the radius of new circle.

∴ Circumference of Ist circle,

C_{1} = 2πr_{1} = 2π × 19 = 38 π cm

and circumference of IInd circle,

C_{2} = 2πr_{2} = 2 × π × 9 = 18 π cm

According to the given condition,

Circumference of new circle = Circumference of Ist circle + Circumference of IInd circle

∴ 2πr = 38π + 18π

⇒ 2πr = 56π

$$\Rarr\space r=\frac{56\pi}{2\pi}=28\text{cm}$$

Hence, radius of new circle is 28 cm.

**2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.**

**Sol.** Let r_{1} = 8 cm and r_{2} = 6 cm and r be the radius of new circle.

Area of 1st circle,

A_{1} = πr_{1}^{2} = π(8)^{2}

= 64π cm

Area of IInd circle,

A_{2} = πr_{2}^{2} = π(6)^{2}

= 36π cm

According to the given condition,

Area of new circle = A_{1} + A_{2}

∴ πr^{2} = 64 π + 36 π

⇒ πr^{2} = 100 π

⇒ r^{2} = 100

⇒ r = 10 cm

(r cannot be negative)

Hence, radius of required new circle is 10 cm.

**3. Figure 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 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.**

**Sol.** Given, diameter of gold = 21 cm

$$\therefore\space\text{Radius of gold circle =}\frac{21}{2}=10.5\space\text{cm}\\\therefore\space\text{Area of gold region = π(r)}^2 =\frac{22}{7}(10.5)^{2}\\=\frac{22}{7}×110.25$$

= 346.5 cm^{2}

∴ Radius for red region = 10.5 + 10.5

= 21 cm

(By condition)

∵ Area of ring = π (R^{2} – r^{2})

where, R = radius of outer ring

r = radius of inner ring

$$\text{Area for red region =}\\\frac{22}{7}[(21)^{2}-(10.5)^{2}]\\=\frac{22}{7}(21)^{2}-\frac{22}{7}(10.5)^{2}$$

= 1386 – 346.5

= 1039.5 cm ^{2}

Radius for blue region = 21 + 10.5 = 31.5

Area of ring = π(R^{2} – r^{2})

where, R = radius of outer ring

r = radius of inner ring

$$\therefore\space\text{Area for blue region =}\\=\frac{22}{7}[(31.5)^{2}-(21)^{2}]\\=\frac{22}{7}(31.5)^{2}-\frac{22}{7}(21)^{2}$$

= 3118.5 – 1386

= 1732.5 cm^{2}

Now,

Radius for black region = 31.5 + 10.5 = 42

(By condition)

∵ Area of ring = π(R^{2} – r^{2})

where, R = radius of outer ring

r = radius of inner ring

$$\therefore\space\text{Area for black region =}\\\frac{22}{7}[(42)^{2}-(31.5)^{2}]\\=\frac{22}{7}(42)^{2}-\frac{22}{7}(31.5)^{2}$$

= 5544 – 3118.5

= 2425.5 cm^{2}

Now,

radius for white region = 42 + 10.5 = 52.5

(By condition)

∵ Area of ring = π(R^{2} – r^{2})

where, R = radius of outer ring

r = radius of inner ring

$$\therefore\space\text{Area for black region =}\\\frac{22}{7}[(52.5)^{2}-(42)^{2}]\\=\frac{22}{7}(52.5)^{2}-\frac{22}{7}(42)^{2}$$

= 8662.5 – 5544

= 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 ?**

**Sol.** Given, diameter of the wheels of a car d_{1} = 80 cm

$$\therefore\space\text{Radius, r}=\frac{d_1}{2}=\frac{80}{2}=40\text{cm}$$

∴ Circumference of wheel of a car

$$=2\pi r=2×\frac{22}{7}×40\\=\frac{1760}{7}\text{cm}$$

Since speed of car = 66 kmh^{–1}

$$=\frac{66×1000}{60}\text{m min}^{\space-1}$$

= 1100 m min^{–1}

= 110000 cm min^{–1}

∴ Wheel of car moves in 1 min = 110000 cm

∴ Wheel of car moves in 10 min = 1100000 cm

∴ Number of complete revolution

$$=\frac{1100000}{1760/7}=\frac{7700000}{1760}=4375$$

Hence, the wheel makes complete revolutions in 10 min is 4375.

**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**

**Sol.** (a) 2 units

Area of circle = πr^{2}

and area of perimeter = 2πr

According to the given condition,

Area of circle = Area of perimeter

∴ πr^{2} = 2πr

⇒ r = 2 units

**Exercise 12.2**

$$\text{• Unless stated otherwise, use }\pi=\frac{22}{7}.$$

**1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.**

$$\textbf{Sol.}\space\because\text{Area of sector =}\frac{\theta\pi r^{2}}{360\degree}\\\text{Given, r = 6 cm, θ = 60°}\\\therefore\space\text{Area of sector = 60}\degree×\frac{22}{7}×\frac{(6)^{2}}{360\degree}=\frac{132}{7}\text{cm}^{2}$$

**2. Find the area of a quadrant of a circle whose circumference is 22 cm.**

**Sol.** Since, circumference of circle = 22

⇒ 2πr = 22

$$\Rarr\space 2×\frac{22}{7}r=22\\\therefore\space\text{Radius of circle, r =}\frac{7}{2}$$

∴ Area of a quadrant of a circle

$$=\frac{\pi r^{2}}{4}=\frac{22}{7}×\frac{\bigg(\frac{7}{2}\bigg)^{2}}{4}\\=\frac{11}{14}×\frac{49}{4}=\frac{77}{8}\text{cm}^{2}$$

**3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.**

**Sol.** Given, length of the minute hand of clock = 14 cm

It means radius of circle is r = 14 cm

∵ The angle moves by the minute hand in 30 min = 180°

∴ The angle moves by the minute hand in

$$\text{1 min}=\frac{180\degree}{30\degree}=6\degree$$

∴ The angle moves by the minute hand in 5 min = 30°

∴ Using the formula,

$$\text{Area of sector of circle =}\frac{\theta\pi r^{2}}{360\degree}=30\degree×\frac{22×(14)^{2}}{7×360\degree}\\=\frac{22×14×2}{12}=\frac{616}{12}\\=\frac{154}{3}\text{cm}^{2}$$

**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)**

**Sol.** Given, radius of circle AO = 10 cm

A perpendicular is drawn from centre of circle to the chord of the circle, which bisects the chord.

∴ AD = DC

Also, ∠AOD = ∠COD = 45°

∴ ∠AOC = ∠AOD + ∠COD

= 45° + 45° = 90°

In right ΔAOD,

$$\text{sin 45\degree}=\frac{\text{AD}}{\text{AO}}\\\Rarr\space\frac{1}{\sqrt{2}}=\frac{\text{AD}}{10}\\\Rarr\space\text{ AD =5}\sqrt{2}\text{cm}\\\text{and\space}\text{cos 45}\degree=\frac{\text{OD}}{10}\\\Rarr\space\frac{1}{\sqrt{2}}=\frac{\text{OD}}{10}\\\text{⇒ = OD =5}\sqrt{2}$$

Now, AC = 2AD

$$=2×5\sqrt{2}=10\sqrt{2}\space\text{cm}\\\text{Now, area of ΔAOC =}\frac{1}{2}\text{AC×OD}\\=\frac{1}{2}×10\sqrt{2}×5\sqrt{2}$$

= 50 cm^{2}

$$\text{Now, area of sector =}\frac{\theta\pi r^{2}}{360\degree}\\=\frac{90\degree}{360\degree}×3.14×(10)^{2}\\=\frac{314}{4}=78.5\space\text{cm}^{2}$$

**(i)** ∴ Area of minor segment AEC = Area of sector OAEC – Area of ΔAOC

= 78.5 – 50 = 28.5 cm^{2}

**(ii)** ∴ Area of major sector OAFGCO = Area of circle – Area of sector OAEC

= πr^{2} – 78.5

= 3.14 × (10)^{2} – 78.5

= 314 – 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.**

**Sol.** Since, arc ABC subtends an angle 60° to the centre.

∴ θ = 60°

and radius, r = 21 cm

$$\text{(i)\space Length of arc ABC =}\frac{\theta}{360\degree}×2\pi r\\=\frac{60\degree}{360\degree}×2×\frac{22}{7}×21\\=\frac{44}{6}×3=22\space\text{cm}$$

(ii) Area of the sector formed by the arc

$$=\frac{\theta}{360\degree}×\pi r^{2}\\=\frac{60\degree}{360\degree}×\frac{22}{7}×(21)^{2}\\=\frac{22}{6×7}×21×21\\=\frac{9702}{42}=231\space\text{cm}^{2}$$

(iii) Now, draw a perpendicular drawn from centre of the circle to the chord.

∴ ΔOAD = ΔODC

∴ ∠AOD = ∠DOC = 30°

and AD = DC

In right ΔAOD,

$$\text{sin 30\degree}=\frac{\text{AD}}{\text{AO}}\\\Rarr\space\frac{1}{2}=\frac{\text{AD}}{21}$$

⇒ AD = 10.5 cm

$$\text{and \space cos 30\degree}=\frac{\text{OD}}{\text{AO}}\\\Rarr\space\frac{\sqrt{3}}{2}=\frac{\text{OD}}{21}\\\Rarr\space\text{OD = 10.5}\sqrt{3}\space\text{cm}$$

∴ AC = 2 AD

= 2 × 10.5 = 21 cm

$$\therefore\space\text{Area of ΔOAC =}\frac{1}{2}×\text{AC × AD}\\=\frac{1}{2}×21×10.5\sqrt{3}\\=\frac{441\sqrt{3}}{4}\text{cm}^{2}$$

∴ Area of the segment = Area of sector formed by the arc – Area of ΔOAC

$$=\bigg(231-\frac{441\sqrt{3}}{4}\bigg)\text{cm}^{2}$$

**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.**

** $$\textbf{(Use}\space\pi=\textbf{3.14}\space\textbf{and}\sqrt{3}=\textbf{1.73})$$**

**Sol.** Since, chord AB subtends an angle 60° to the centre

∴ ∠AOB = 60°

Draw OD ⊥ AB and is bisect.

∵ ΔOAD ≅ OBD

∴ ∠AOD = ∠BOD = 30°

i.e., ∠AOB = 120°

Draw OD ⊥ AB it bisects, such that

AD = BD

∵ ΔAOD ≅ ΔBOD

⇒ ∠AOD = ∠BOD = 60°

Also, radius of circle, r = 12 cm

In right ΔOAD,

$$\text{sin 60\degree}=\frac{\text{AD}}{\text{AO}}\\\Rarr\space\frac{\sqrt{3}}{2}=\frac{\text{AD}}{12}\\\Rarr\space\text{AD = 6}\sqrt{2}=10.392\space\text{cm}\\\text{and cos 60° =}\frac{\text{OD}}{\text{AO}}\\\Rarr\space\frac{1}{2}=\frac{\text{OD}}{12}$$

⇒ OD = 6 cm

∴ Length of chord, AB = 2 AD = 2 × 10.392

= 20.784 cm

$$\text{Now, Area of ΔAOB =}\frac{1}{2}×\text{AB×DO}\\=\frac{1}{2}×20.784×6$$

= 62.352 cm^{2}

$$\text{Now, area of sector OACBO =}\frac{\theta\pi r^{2}}{360\degree}\\=120×\frac{3.14×(12)^{2}}{360}\\=\frac{3.14×144}{3}\\=\frac{452.16}{3}$$

= 150.72 cm^{2}

∴ Area of the corresponding segment

= Area of sector OACBO – Area of ΔAOB

= 150.72 – 62.352

= 88.368 cm^{2}

**8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). 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)**

**Sol.** Given,

side of a square = 15 m

∴ Area of square = (15)^{2} = 225 m

Also, length of rope = 5 m

∴ Readius of arc = 5 cm

(i) ∴ Area of the field, graze by the horse,

$$\text{A}_1=\frac{\theta\pi r^{2}}{360\degree}\\=90×\frac{3.14}{360\degree}×(5)^{2}$$

(∵ Angle between two adjacent sides of a square is 90°)

$$=\frac{3.14×25}{4}\\=\frac{78.5}{4}$$

= 19.625 cm^{2}

(ii) If length of rope = 10 m = r_{1} (say)

∴ Area of field, graze by the horse

$$\text{A}_2=\frac{\theta\pi r^{2}}{360\degree}\\=\frac{90\degree×3.14×(10)^{2}}{360\degree}\\=\frac{3.14×100}{4}\\=\frac{90\degree×3.14×(10)^{2}}{360\degree}$$

∴ Required increase in the grazing area

= A_{2} – A_{1}

= 78.5 – 19.625

= 58.875 cm^{2}

**9. A 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 figure. Find :**

**(i) the total length of the silver wire required.**

**(ii) the area of each sector of the brooch.**

**Sol.** Given, diameter of circle d = 35 mm

∴ Circumference of circle

$$=\pi d=\frac{22}{7}×35=110\space\text{mm}^{2}$$

Now, length of 5 diameter = 5 × 35 = 175 mm

(i) The total length of the silver wire required

= πd + 5d = 110 + 175 = 285 mm^{2}

(ii) Here, we see that the total circle is divide into 10 sector.

$$\therefore\space\text{Angle of each sector =}\frac{360}{10}=36\\\therefore\text{Angle of each sector of the brooch}\\=\frac{\theta\pi r^{2}}{360}=\frac{36}{360}×\frac{22}{7}\bigg(\frac{35}{2}\bigg)^{2}\\\bigg(\because r=\frac{d}{2}=\frac{35}{2}\text{mm}\bigg)\\=\frac{22}{70}×\frac{1225}{4}\\=\frac{26950}{70×4}=\frac{385}{4}\space\text{mm}^{2}$$

**10. An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.**

**Sol.** The central angle of an umbrella is 360°.

Since, umbrella has 8 ribs.

$$\therefore\space\text{Angle between two ribs =}\frac{360\degree}{8}=45\degree$$

∴ Area between two ribs = Area of one sector of an umbrella

$$=\frac{\theta\pi r^{2}}{360\degree}\space(\because\text{ r = 45, \text{given}})\\=45×\frac{22}{7}×\frac{(45)^{2}}{360\degree}\\=\frac{22}{7×8}(45)^{2}\\=\frac{22275}{28}\text{cm}^{2}$$

**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.**

**Sol.** Given, length of wiper blade= 25 cm = r (say)

Angle made by the blade, θ =115º.

∴ The area cleaned by one blade = Area of sector formed by blade

$$=\frac{\theta\pi r^{2}}{360}\\=115\degree×\frac{22}{7×360\degree}×(25)^{2}\\=\frac{23×22}{7×72}×625\\=\frac{23×11×625}{7×36}\\=\frac{158125}{252}\text{cm}^{2}$$

∴ Total area cleaned by both blade

=2 × the area cleaned by one blade

$$=\frac{2×158125}{252}\\=\frac{158125}{126}\text{cm}^{2}$$

**12. To warn ships for under water rocks, a lighthouse spreads a red coloured 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)**

**Sol.** Given, sector angle θ = 80°

and distance of radius r = 16.5 km

$$\therefore\space\text{Area of sector =}\frac{\theta\pi r^{2}}{360\degree}\\=\frac{80\degree×3.14×(16.5)^{2}}{360\degree}\\=\frac{2×3.14×272.25}{9}\\=\frac{1709.73}{9}$$

= 189.97 km^{2}

Which is the required area of the sea over which the ships are warned.

**13. A round table cover has six equal designs as shown in figure. 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}.**

$$\textbf{(Use}\space\sqrt{\textbf{3}} = \textbf{1.7)}$$

**Sol.** ∵ Central angle of a circle is 360Ä.

$$\therefore\space\text{Angle of each sector =}\frac{360\degree}{6}\\=60\degree$$

Now, we determine the area of one segment of a circle.

Here, AC is a chord of the circle. Draw OD ⊥ AC it bisects, such that AB = BC.

∵ ΔAOD ≡ ΔCOD

∴ ∠COB = ∠AOB = 30°

In right ΔOCD,

$$\text{sin 30° =}\frac{\text{DC}}{\text{OC}}\\\Rarr\frac{1}{2}=\frac{\text{DC}}{28}$$

⇒ DC = 14 cm

$$\text{and cos 30° =}\frac{\text{OD}}{\text{OC}}\\\Rarr\frac{\sqrt{3}}{2}=\frac{\text{OD}}{28}$$

$$\Rarr\space\text{OD}=14\sqrt{3}\\=14×1.7=23.8\space\text{cm}$$

∴ AC = 2CD = 2 × 14 = 28 cm

$$\text{Area of ΔAOC =}\frac{1}{2}×\text{AC×OD}\\=\frac{1}{2}×28×23.8$$

= 333.2 cm^{2}

$$\text{Now, area of sector OABCO =}\frac{\theta\pi r^{2}}{360}\\=60\degree×\frac{22×(28)^{2}}{7×360}\\=\frac{22×4×28}{6}$$

= 410.67 cm^{2}

∴ Area of segment ABCA

= Area of sector OABCO – Area of ΔAOC

= 410.67 – 333.2 = 77.47 cm^{2}

∴ Area of six segment = 6 × 77.47

= 464.82 cm^{2}

Since, the cost of making the design at the rate of ₹ 0.35 per cm^{2}.

∴ Total cost = 464.82 × 0.35

= ₹ 162. 68

**14. Tick the correct answer in the following :**

**Area of a sector of angle p (in degrees) of a circle with radius R is**

$$\textbf{(a)}\space\frac{\textbf{p}}{\textbf{180}}×2\pi \textbf{R}\\\textbf{(b)\space}\frac{\textbf{p}}{\textbf{180}}×\pi \textbf{R}^{2}\\\textbf{(c)}\space\frac{\textbf{p}}{\textbf{360}}×2\pi\textbf{R}\\\textbf{(d)\space}\frac{\textbf{p}}{\textbf{720}}×2\pi\textbf{R}^{2}$$

$$\textbf{Sol.}\space(\text{d})\space\frac{p}{720}×2\pi\text{R}^{2}\\\text{Area of sector =}\frac{\theta\pi r^{2}}{360}=\frac{p\pi\text{R}^{2}}{360}$$

$$=\frac{p2\pi R^{2}}{720}$$

**Exercise 12.3**

$$\text{• Unless stated otherwise, use}\space\pi =\frac{22}{7}.$$

**1. Find the area of the shaded region in figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.**

**Sol.** Given, PQ = 24 cm, PR = 7 cm

We know any triangle drawn from diameter RQ to the circle is 90°.

Here, ∠RPQ = 90°

In right ΔRPQ,

RQ^{2} = PR^{2} + PQ^{2}

(By Pythagoras theorem)

⇒ RQ^{2} = 7^{2} + 24^{2}

⇒ RQ^{2} = 49 + 576

⇒ RQ^{2} = 625

⇒ RQ = 25 cm

(∵ Side can not be negative)

$$\therefore\space=\frac{1}{2}×\text{RP×PQ}\\=\frac{1}{2}×7×24=84\text{cm}^{2}\\\text{Area of semi circle =}\frac{\pi r^{2}}{2}\\=\frac{22}{7}×\frac{25}{2}×\frac{25}{2}×\frac{1}{2}\\=\frac{11×25×25}{7×2×2}\\=\frac{6875}{28}\text{cm}^{2}$$

∴ Area of the shaded region

= Area of the semi-circle – Area of right ΔROQ

$$=\frac{6875}{28}-84\\=\frac{6875-2352}{28}\\=\frac{4523}{28}\text{cm}^{2}$$

= 161.54 cm^{2}

Hence, the area of the shaded region = 161.54 cm^{2}.

**2. Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.**

**Sol.** Given, OB = 7 cm and OA = 14 cm

$$\text{Now, area of sector ODB =}\frac{\theta\pi r^{2}}{360\degree}\\=40\degree×\frac{22}{7}×\frac{7}{360\degree}\\=\frac{154}{9}\text{cm}^{2}\\\text{and area of sector OAC =}\frac{\theta\pi r^{2}}{360\degree}\\=40\degree×\frac{22}{7×360\degree}×(14)^{2}\\=\frac{22×28}{9}\\=\frac{616}{9}\text{cm}^{2}$$

Area of shaded part, ABCD

= Area of sector OAC – Area of sector OBD

$$=\frac{616}{9}-\frac{154}{9}\\=\frac{462}{9}=\frac{154}{3}\text{cm}^{2}$$

**3. Find the area of the shaded region in figure, if ABCD is a square of side 14 cm and APD and BPC are semi-circles.**

**Sol.** Given, side of square = 14 cm

Also, APD and BPC are semi-circles, therefore

$$\text{its radius are r =}\space\frac{14}{2}$$

=7cm.

Now, Area of semi-circle APD

$$=\text{Area of semi-circle BPC =}\frac{\pi r^{2}}{2}\\=\frac{22}{7×2}(7)^{2}=77\space\text{cm}^{2}$$

Now, Area of square ABCD = (side)^{2}

= (14)^{2} = 196 cm^{2}

∴ Area of shaded region = Area of square – [Area of semi-circle APD + Area of semi-circle BPC]

= 196 – (77+ 77)

= 42 cm^{2 }

**4. Find the area of the shaded region in figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.**

**Sol.** Given, radius of circle, r= 6 cm

Since, OAB is an equilateral triangle,

∴ ∠O = ∠A = ∠B = 60°

$$\therefore\space\text{Area of an equilateral ΔOAB =}\\\frac{\sqrt{3}}{4}\text{(side)}^{2}\\=\frac{\sqrt{3}}{4}(12)^{2}\\=36\sqrt{3}\space\text{cm}^{2}$$

∴ Area of circle outside the triangle

= Area of circle – Area of sector OPQO

$$=\pi r^{2}-\frac{60\degree}{360\degree}×\pi r^{2}\\=\pi r^{2}-\frac{\pi r^{2}}{6}=\frac{5}{6}\pi r^{2}\\=\frac{5}{6}\bigg[\frac{22}{7}(6)^{2}\bigg]\text{cm}^{2}$$

(∵ Central angle of a circle is 360° here we remove the area of 60° angle)

$$=\frac{110×36}{6×7}=\frac{660}{7}\text{cm}^{2}$$

∴ Total area of the shaded region

= Area of an equilateral ΔOAB + Area of circle outside the triangle

$$=\bigg(36\sqrt{3}+\frac{600}{7}\bigg)\text{cm}^{2}$$

**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 figure. Find the area of the remaining portion of the square.**

**Sol.** Given, side of a square is 4 cm.

Radius of quadrant r_{1} = 1 cm

$$\text{and radius of circle r}_2=\frac{2}{2}=\text{1 cm}\\\therefore\space\text{Area of one quadrant =}\frac{\pi r_1^{2}}{4}=\frac{22}{7×4}(1)^{2}\\=\frac{22}{28}\space\text{cm}^{2}\\\therefore\space\text{Area of four quadrant =}4\bigg(\frac{22}{28}\bigg)=\frac{22}{7}\text{cm}^{2}\\\text{Now, Area of circle =}\pi r_2^{2}=\frac{22}{7}(1)^{2}\\=\frac{22}{7}\text{cm}^{2}$$

and area of square ABCD = (side)^{2} = (4)^{2}

= 16 cm^{2}

∴ Area of shaded region = Area of square – (Area of four quadrant + Area of circle)

$$=16-\bigg(\frac{22}{7}+\frac{22}{7}\bigg)\\16-\frac{44}{7}\\=\frac{112-44}{7}\\=\frac{68}{7}\text{cm}^{2}\space\textbf{Ans.}$$

**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 figure. Find the area of the design.**

**Sol.** Given, radius of the circle = 32 cm

Let side of an equilateral ΔABC be a cm.

Let h be height of the triangle.

We know, in an equilateral triangle. Centroid and circumcentre are coincide.

$$\therefore\space\text{AO}=\frac{2}{3}\text{h cm}$$

Which is equal to the radius of circle.

$$\therefore\space\frac{2}{3}\text{h cm}=32$$

⇒ h = 48 cm

Here, we draw a perpendicular from vertex A to side BC. Which bisect them.

In right ΔABD,

AB^{2} = BD^{2} +AD^{2}

(by Pythagoras theorem)

$$\Rarr\space a^{2}=\bigg(\frac{a}{2}\bigg)^{2}+h^{2}\\\Rarr\space h^{2}=a^{2}-\frac{a^{2}}{4}=\frac{3a^{2}}{4}\\\Rarr\space(48)^{2}=\frac{3a^{2}}{4}\\\Rarr\space a^{2}=3072\\\Rarr a=\sqrt{3072}\\\therefore\space\text{Area of equilateral ΔABC =}\frac{\sqrt{3}}{4}(a)^{2}\\=\frac{\sqrt{3}}{4}×3072=768\sqrt{3}\space\text{cm}^{2}$$

∴ Area of shaded region = Area of circle – Area of ΔABC [∵ Area of circle = π(radius)^{2}]

$$\pi(32)^{2}-768\sqrt{3}\\=\frac{22}{7}(1024)-768\sqrt{3}\\=\bigg(\frac{22528}{7}-768\sqrt{3}\bigg)\text{cm}^{2}$$

**7. In figure, 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.**

**Sol.** Given, side of square is 14 cm.

i.e., AB = BC = CD = DA = 14 cm

$$\therefore\space\text{Radius of circle =}\frac{14}{2}=7\text{cm}\\\text{Area of quadrant of one circle =}\frac{\pi r^{2}}{4}\\=\frac{22}{7×4}(7)^{2}\\=\frac{154}{4}\space\text{cm}^{2}\\\therefore\space\text{Area of four quadrant of circle = 4}\bigg(\frac{154}{4}\bigg)$$

= 154 cm^{2}

Now, area of square = (side)^{2}

= (14)^{2}

= 196 cm^{2}

∴ Area of shaded region

= Area of square – Area of four quadrant

= 196 – 154

= 42 cm^{2}

**8. Figure depicts a racing track whose left and right ends are semi-circular.**

**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.**

**Sol.** Given, length of track = 106 m

width of track = 10 m

(i) ∴ The distance around the track along its inner edge

= PQ + RS + 2 circumference of semi-circle POR

$$= 106+106+2\bigg[\frac{2\pi\bigg(\frac{60}{2}\bigg)}{2}\bigg]\\= 212 + 60\pi\\= 212+60×\frac{22}{7}\\=212+\frac{1320}{7}\\=\frac{1484+1320}{7}\\=\frac{2804}{7}\text{m}$$

(ii) Area of the track = Area of outer track – Area of inner track

= (Area of ABCD + 2 × Area of semi-circle ATD) – (Area of PQRS + 2 × Area of semi-circle POR)

$$=\bigg[106×(60+20)+2×\frac{\pi}{2}\bigg(\frac{60+20}{2}\bigg)^{2}\bigg]-\\\bigg[106×60+2\frac{\pi}{2}\bigg(\frac{60}{2}\bigg)^{2}\bigg]\\=\bigg[106×80+\frac{22}{7}×(40)^{2}\bigg]\\-\bigg[106×60+\frac{22}{7}×(30)^{2}\bigg]\\=\bigg[106×80+\frac{22}{7}×40×40\bigg]\\-\bigg[106×60+\frac{22}{7}×30×30\bigg]$$

$$=\bigg(8480+\frac{35200}{7}\bigg)-\bigg(6360+\frac{19800}{7}\bigg)\\= 2120 + \frac{15400}{7}$$

= 2120 + 2200

= 4320 m^{2}

**9. In figure, 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.**

**Sol.** According to the given Q.

AB and CD are two diameters of the same circle with center O.

∴ OA = OB = OC = OD = radius of the circle R = 7 cm.

Now, area of smaller circle whose diameter (OD = 7 cm) is

$$=\pi r^{2}=\pi\bigg(\frac{7}{2}\bigg)^{2}\\=\frac{22}{7}×\frac{49}{4}\\=\frac{77}{2}\space\text{cm}^{2}\\\text{Now, area of ΔABC =}\frac{1}{2}×\text{AB × OC}\\=\frac{1}{2}×2×\text{OA×OC}\\=\frac{1}{2}×2×7×7$$

(∵ OA = OC)

= 49 cm^{2}

$$\text{and area of semi-circle ABCA =}\frac{\pi r^{2}}{2}=\frac{22}{7×2}(7)^{2}$$

= 77 cm^{2}

∴ Area of segment BC and AC

= Area of semi-circle – Area of ΔABC

= 77 – 49 = 28 cm^{2}

∴ Area of total shaded region

= Area of small circle + Area of segment BC and AC

$$=\frac{77}{2}+28$$

= 38.5 + 28

= 66.5 cm^{2}

**10. The area of an equilateral ΔABC is 17320.5 cm2. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region.**

$$\textbf{(Use p = 3.14 and}\sqrt{\textbf{3}}=\textbf{1.73205})$$

**Sol.** Let side of an equilateral triangle be a.

$$\therefore\space\text{Area of an equilateral ΔABC =}\frac{\sqrt{3}}{4}(a)^{2}\\\Rarr\space 17320.5 =\frac{1.73205}{4}(a^{2})$$

⇒ a^{2} = 10000 × 4

⇒ a = 100 × 2 = 200 cm

Since, ΔABC is an equilateral triangle.

∴ ∠A = ∠B = ∠C = 60°

Here, AB = a = 200 cm

$$\therefore\space\text{Radius of circle =}\frac{200}{2}=100\text{cm}$$

[∵ Radius of circle = half the length of the side of the ΔABC]

$$\text{Area of sector in Ist circle =}\frac{\theta\pi r^{2}}{360}\\=\frac{60\degree×3.14×(100)^{2}}{360}$$

= 5233.33 cm2

∴ Area of three equal sectors = 3 × 5233.33

= 15700 cm^{2}

∴ Area of required shaded region

= Area of ΔABC – Area of three sectors

= 17320.5 – 15700

= 1620.5 cm^{2}

**11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.**

**Sol.** Given, radius of each circle, r = 7 cm

∴ Diameter of circle, d = 14 cm

(∵ Diameter = 2 × radius)

In a given figure, horizontal three circles touch each other.

∴ Length of side of the square

= 3 (Diameter of one circle)

= 3 × 14 = 42 cm

Now, area of one circle = πr^{2}

= π(7)^{2}

$$=\frac{22}{7}×(7)^{2}$$

= 154 cm2

Now, area of square ABCD = (side)^{2}

= (42)^{2}

= 1764 cm^{2}

∴ Area of the remaining portion of the handkershief

= Area of square – Area of nine circles

= 1764 – 9 × 154

= 1764 – 1386 = 378 cm^{2}

**12. If figure, OACB is a quadrant of a circle wth centre O and radius 3.5 cm. If OD = 2 cm, find the area of the**

**(i) quadrant OACB,**

**(ii) shaded region.**

**Sol.** Given, radius of quadrant, r = 3.5 cm = 7/2

**(i)** ∴ Area of quadrant OACB

$$=\frac{1}{4}×\text{Area of the circle}\\=\frac{\pi r^{2}}{4}=\frac{22}{7×4}\bigg(\frac{7}{2}\bigg)^{2}\\=\frac{1}{4}×\frac{22}{7}×\frac{7}{2}×\frac{7}{2}\\=\frac{17}{8}\text{cm}^{2}$$

$$\text{(ii) Now, area of ΔOBD =}\frac{1}{2}×\text{OB × OD}\\=\frac{1}{2}×3.5×2\\=\frac{1}{2}×\frac{7}{2}×2$$

(∵ Radius = OB = 3.5)

$$=\frac{7}{2}$$

∴ Area of shaded region

= Area of quadrant OBCA – Area of ΔOBD

$$=\frac{77}{8}-\frac{7}{2}\\=\frac{49}{8}\text{cm}^{2}\space\textbf{Ans.}$$

**13. In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)**

**Sol.** Since, OABC is a square.

$$\therefore\space\text{Diagonal of a square =}\sqrt{2}×\text{Side}\\=\sqrt{2}×20\space \text{cm}\\\lbrack\because\space\text{OA (side) = 20 cm}\rbrack\\\therefore\space\text{Radius of circle, r =}20\sqrt{2}\\\therefore\space\text{Area of quadrant =}\frac{\pi r^{2}}{4}\\=\frac{3.14 ×(20\sqrt{2})^{2}}{4}$$

= 628 cm^{2}

Now, area of square = (OA)^{2} = (20)^{2}

= 400 cm

∴ Area of shaded region = Area of quadrant– Area of square

= 628 – 400

= 228 cm^{2}

**14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If ∠AOB = 30°, find the area of the shaded** region.

**Sol.** Given, ∠COD = 30°, r_{1} = OC = 7 cm and r_{2} = OB = 21 cm

$$\therefore\space\text{Area of sector COD =}\frac{\theta\pi r_1^{2}}{360}\\=\frac{30}{360}×\frac{22}{7}×(7)^{2}\\=\frac{22×7}{12}=\frac{77}{6}\text{cm}^{2}\\\text{Area of sector OAB =}\frac{\theta\pi r_2^{2}}{360}\\=\frac{30}{360}×\frac{22}{7}×(21)^{2}\\=\frac{22}{12}×3×21\\=\frac{11×63}{6}$$

$$=\frac{693}{6}\text{cm}^{2}$$

∴ Area of shaded region = Area of sector OAB – Area of sector COD

$$=\frac{693}{6}-\frac{77}{6}\\=\frac{308}{3}\text{cm}^{2}$$

= 102.67 cm^{2} Ans.

**15. In figure, ABC is a quadrant of a circle radius 14 cm and a semi-circle is drawn with BC as diameter. Find the area of the shaded region.**

**Sol.** Given, radius of quadrant

r = 14 = AC = AB

In right ΔABC,

BC^{2} = AB^{2} + AC^{2}

(By Pythagoras theorem)

⇒ BC^{2} = (14)^{2} + (14)^{2}

$$\Rarr\space\text{BC}=14\sqrt{2}\\\text{Now, area of ΔABC =}\\\frac{1}{2}×\text{AC×AB}\\\frac{1}{2}×\text{14×14}$$

= 98 cm^{2}

and area of quadrant ABDC =

$$\frac{\pi r^{2}}{4}=\frac{22}{7×4}(14)^{2}$$

= 154 cm^{2}

$$\text{Now, area of semi-circle BCE =}\frac{\pi r^{2}}{2}\\=\frac{22}{7×2}\bigg(\frac{14\sqrt{2}}{2}\bigg)^{2}\\\bigg(\because r=\frac{\text{BC}}{2}\bigg)\\=\frac{22×14×14×2}{7×2×4}$$

= 22 × 7

= 154 cm^{2}

∴ Required shaded region = Area of semi-circle

BCE – (Area of quadrant ABCD – Area of ΔABC)

= 154 – (154 – 98)

= 98 cm^{2}

**16. Calculate the area of the designed region in figure, common between the two quadrants of circles of radius 8 cm each.**

**Sol.** Firstly, determine the area of the shaded region AEC.

$$\text{Now, area of sector BAEC =}\frac{\pi r^{2}}{4}\\=\frac{22}{7}×\frac{(8)^{2}}{4}$$

(∵ r = AB = 8 cm)

$$=\frac{352}{7}\text{cm}^{2}\\\text{Now, area of right ΔABC =}\\\frac{1}{2}×8×8\\=\frac{1}{2}×8×8$$

= 32 cm^{2}

∴ Area of shaded region AEC

= Area of sector BAEC – Area of ΔABC

$$=\frac{352}{7}-32\\=\frac{352-224}{7}\\\frac{128}{7}\text{cm}^{2}$$

∴ Rquired shaded region

= 2 × Area of shaded region AEC

$$= 2×\frac{128}{7}=\frac{256}{7}\text{cm}^{2}$$

**Selected NCERT Exemplar Problems**

**Exercise 12.1**

• Choose the correct answer from the given four options.

**1. If the sum of the areas of two circles with radius R _{1} and R_{2} is equal to the area of the circle with radius R, then**

**(a) R _{1} + R_{2}= R**

**(b) R _{1}^{2} + R_{2}^{2} = R^{2}**

**(c) R _{1} + R_{2} < R**

**(d) R _{1}^{2} + R_{2}^{2} < R^{2}**

**Sol.** (b) R_{1}^{2} + R_{2}^{2} = R^{2}

**Explanation:**

According to the given condition,

Area of circle = Area of Ist circle + Area of IInd circle

∴ πR^{2} = πR_{1}^{2} + πR_{2}^{2}

⇒ R^{2} = R_{1}^{2} + R_{2}^{2 }

**2. If the sum of the circumferences of two circles with radii R _{1} and R_{2} is equal to the circumfereence of a circle of radius R, then**

**(a) R _{1} + R_{2} = R**

**(b) R _{1} + R_{2} > R**

**(c) R _{1} + R_{2} = R**

**(d) Nothing definite can be said about the ****relation among R _{1}, R_{2} and R.**

**Sol.** (A) R_{1} + R_{2} = R

**Explanation:**

According to the given question,

Area of circumference of circle = Area of circumference of Ist circle + Area of circumference of IInd circle

∴ 2πR = 2πR_{1} + 2πR_{2}

⇒ R = R_{1 }+ R_{2}

**3. If the circumference of a circle and the perimeter of a square are equal, then**

**(a) Area of the circle = Area of the square**

**(b) Area of the circle > Area of the square**

**(c) Area of the circle < Area of the square**

**(d) Nothing definite can be said about the ****relation between the areas of the circle ****and square.**

**Sol.** (b) Area of the circle > Area of the square

**Explanation:**

According to the given condition,

Circumference of a circle = Perimeter of square

⇒ 2πr = 4a

(Where, r and a be radius of circle and side of square)

$$\Rarr\space\frac{22}{7}r=2a$$

⇒ 11r = 7a

$$\Rarr\space a=\frac{11}{7}r\\\Rarr\space r=\frac{7a}{11}\space\text{...(i)}$$

Now, area of circle, A_{1} = πr^{2}

$$=\pi\bigg(\frac{7a}{11}\bigg)^{2}\\=\frac{22}{11}×\frac{49 a^{2}}{121}$$

[From eq. (i)]

$$=\frac{14a^{2}}{11}\space\text{...(ii)}$$

and area of square, A_{2} = (a)^{2} ...(iii)

From eqs. (ii) and (iii),

$$\text{A}_1=\frac{14}{11}\text{A}_2$$

∴ A1 > A2

Hence, area of the circle > Area of the square.

**4. Area of the largest triangle that can be iscribed in a semi-circle of radius r units is**

**(a) r ^{2} sq units**

$$\textbf{(b)}\space\frac{\textbf{1}}{\textbf{2}}\textbf{r}^{2}\textbf{sq units}$$

**(c) 2r ^{2} sq units**

$$\textbf{(d)}\space\sqrt{\textbf{2}}\space \textbf{r}^{2}\textbf{sq units}$$

**Sol.** (a) r^{2} sq units

**Explanation:**

When we take a thord point on a circle and join by the end points of circle, then

(By property of circle)

∴ ΔABC is right angled triangle.

∴ Area of the largest ΔABC =

$$\frac{1}{2}×\text{AB × CD}\\=\frac{1}{2}×2r×r$$

= r^{2} sq units

**5. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is**

**(a) 22 : 7**

**(b) 14 : 11**

**(c) 7 : 22**

**(d) 11 : 14**

**Sol.** (b) 14 : 11

**Explanation:**

Let radius of circle be r and side of a square be a.

According to the given condition,

Perimeter of a circle = Perimeter of a square

∴ 2πr = 4a

$$\Rarr\space a=\frac{\pi r}{2}\space\text{...(i)}\\\text{Now,}\space\frac{\text{Area of circle}}{\text{Area of square}}\\=\frac{\pi r^{2}}{(a)^{2}}\\=\frac{\pi r^{2}}{\bigg(\frac{\pi r^{2}}{2}\bigg)}$$

[From eq. (i)]

$$=\frac{\pi r^{2}}{\pi^{2} r^{2}/4}=\frac{4}{\pi}\\=\frac{4}{22/7}=\frac{28}{22}\\=\frac{14}{11}$$

**6. If is proposed to build a singlecircular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be**

**(a) 10 m**

**(b) 15 m**

**(c) 20 m**

**(d) 24 m**

**Sol.** (a) 10 m

**Explanation:**

Area of first circular park, whose diameter is 16 m.

$$=\pi r^{2}=\pi\bigg(\frac{16}{2}\bigg)^{2}\\\bigg(\because r=\frac{d}{2}=\frac{16}{2}=8\space\text{m}\bigg)$$

^{2}

^{2}

^{2}= 64 π + 36 π

^{2}= 100π ⇒ R

^{2}= 100

**7. The area of the circle that can be inscribed in a square of side 6 cm is**

**(a) 36π cm ^{2}**

**(b) 18π cm ^{2}**

**(c) 12π cm ^{2}**

**(d) 9π cm ^{2}**

**Sol.** (d) 36π cm^{2}

**Explanation:**

Given, side of square = 6 m

∴ Diameter of a circle ‘d’ = side of square = 6 cm

$$\therefore\space\text{Radius of a circle =}\frac{d}{2}\\=\frac{6}{2}=3\text{cm}$$

∴ Area of circle = π(r)^{2}

= π(3)^{2}

= 9π cm^{2}

**8. The area of the square that can be inscribed in a circle of radius 8 cm is**

**(a) 256 cm ^{2}**

**(b) 128 cm ^{2}**

**$$\textbf{(c)\space 64}\sqrt{\textbf{2}}\space\textbf{cm}^{2}$$**

**(d) 64 cm ^{2}**

**Sol.** (b) 128 cm^{2}

**Explanation:**

Given, radius of circle, r = OC = 8 cm.

∴ Diameter of the circle = AC = 2 × OC

= 2 × 8 = 16 cm

Which is equal to the diagonal of a square.

Let side of square be x.

In right ΔABC,

AC^{2} = AB^{2} + BC^{2}

(∵ By Pythagoras theorem)

⇒ (16)^{2} = x^{2} + x^{2}

⇒ 256 = 2x^{2}

⇒ x^{2} = 128

∴ Area of square = x^{2}

= 238 cm^{2}

**9. The diameter of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is**

**(a) 56 cm**

**(b) 42 cm**

**(c) 58 cm**

**(d) 16 cm**

**Sol.** (a) 56 cm

**Explanation:**

Circumference of first circle

= 2πr = πd_{1} = 36π cm

(Given, d_{1} = 36 cm)

Circumference of second circle = πd_{2} = 20π cm

(Given, d_{2} = 20 cm)

Accoding to the given condition,

Circumference of circle = Circumference of first circle + Circumference of second circle

⇒ πD = 36π + 20π

(Where D be diameter of a circle)

⇒ D = 56 cm

Hence, required diameter of a circle is 56 cm.

**10. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is**

**(a) 31 cm**

**(b) 25 cm**

**(c) 62 cm**

**(d) 50 cm**

**Sol.** (d) 50 cm

**Explanation:**

Let r_{1} = 24 cm and r_{2} = 7 cm

∴ Area of first circle = πr_{1}^{2} = π(24)^{2}

= 576π cm^{2}

Area of second circle = πr_{2}^{2} = π(7)^{2}

= 49π cm^{2}

According to the given condition,

Area of circle = Area of first circle

+ Area of second circle

∴ πR^{2} = 576π + 49π

⇒ R^{2} = 625 ⇒ R = 25 cm

∴ Diameter of a circle = 2R = 2 × 25 = 50 cm

**Exercise 12.2**

• *State whether the following statements are yes or no justify your answers.*

**1. In the area of the circle inscribed in a square of side a cm, πa ^{2} cm^{2}. Give reason your answer.**

**Sol.** No. Let ABCD be a square of side a.

∴ Diameter of circle = Side of square = a

$$\therefore\space\text{Radius of circle =}\frac{a}{2}$$

$$\therefore\space\text{Area of circle = π (Radius)}^{2}\\=\pi\bigg(\frac{a}{2}\bigg)=\frac{\pi a^{2}}{4}\\\text{Hence, area of the circle is}\space\frac{\pi a^{2}}{4}\text{cm}^{2}.$$

**2. Will it be true to say that the perimeter of a square circumscribing a circle of radius a cm is 8a cm? Give reason for your answer.**

**Sol.** Yes. Given, radius of circle r= a cm

∴ Diameter of circle, d = 2 × Radius = 2a cm

∴ Side of a square = Diameter of circle = 2a cm

∴ Perimeter of square = 4 × (Side)

= 4 × 2a

= 8a cm

**3. In figure, a square is inscribed in a circle of diameter d and another square is circumscribing the circle. Is the area of the outer square four times the area of the inner square? Give reason for your answer.**

**Sol.** No. Given diameter of circle is d.

∴ Diagonal of inner square = Diameter of circular = d

Let side of inner square EFGH is x.

∴ In right ΔEFG,

EG^{2} = EF^{2} +FG^{2}

(By Pythagoras theorem)

⇒ d^{2} = x^{2} + x^{2}

⇒ d^{2} = 2x^{2}

$$\Rarr\space x^{2}=\frac{d^{2}}{2}\\\therefore\space\text{Area of in square EFGH}\\ = (\text{Side})^{2}=x^{2}=\frac{d}{2}$$

But side of the outer square ABCD = Diameter of circle = d

∴ Area of outer square = d^{2}

Hence, area of outer square is not equal to four times the area of the inner square.

**4. Is it true to say that area of segment of a circle is less than the area of its corresponding sector? Why?**

**Sol.** No. It is true only in the case of minor segment. But in case of major segment area is always greater than the area of sector.

**5. Is it true that the distance travelled by a circular wheel of diameter d cm in one revolution is 2nd cm? Why?**

**Sol.** No. Because the distance travelled by the wheel is π d.

i.e., n(2r) = 2πr = circumference of wheel

(∵ d = 2r)

**6. In covering a distance s metre, a circular wheel of radius metre makes**

$$\frac{\text{s}}{2\pi r}\space\text{revolution.}$$

Is this statement true? Why?

**Sol.** Yes. The distance covered in one revolution is 2πr. i.e., its circumference.

**7. The numerical value of the area of a circle is greater than the numerical value of its circumference. Is this statement true? Why?**

**Sol.** No. If 0 < r < 2, then numerical value of circumference is greater ran numerical value of area of circle. And if r > 2, area is greater than circumference.

**8. If the length of an arc of a circle of radius r is equal to that of an arc of a circle of radius 2r, then the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle. Is this statement false? Why?**

**Sol.** Yes. It is true for arcs of the same circle. But in different circle, it is possible.

**9. The area of two sectors of two different circles with equal corresponding arc lengths are equal. Is this statement true? Why?**

**Sol.** No. It is true for arcs of the same circle. But in different circle, it is not possible.

**10. The areas of two sectors of two different circles are equal. Is it necessary that their corresponding arc lengths are equal? why?**

**Sol.** No. It is true for arcs of the same circle. But in different circle, it is not possible.

**11. Is the area of the largest circle that can be drawn inside a rectangle of length a cm and breadh b cm (a > b) is πb ^{2} cm? Why.**

**Sol.** No. The area of the largest circle that can be drawn inside is rectangle is

$$\pi\bigg(\frac{b}{2}\bigg)^{2}\text{cm.}$$

$$\text{Where}\space\bigg(\frac{b}{2}\bigg)\space\text{is the radius of the circle}$$

and it is possible when rectangle become a square.

**12. Circumference of two circles are equal. Is it necessary that their areas be equal? Why?**

**Sol.** Yes. If circumference of two circles are equal, then their corresponding radii are equal. So, their areas will be equal.

**13. Areas of two circles are equal. Is it necessary that their circumferences are equal? Why?**

**Sol.** Yes. If areas of two circles are equal, then their corresponding radii are equal. So, their circumference will be equal.

**14. Is it true to say that area of a square inscribed in a circle of diameter p cm in p ^{2} cm? Why ?**

**Sol.** No. When the square is inscribed in the circle the diameter of a circle is equal to diagonal of a square but not the side of the square.

**Exercise 12.3**

**1. The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so on to keep a speed of 66 kmh ^{–1}?**

**Sol.** Given, radius of wheel, r = 35 cm

∴ Length of wheel = Circumference of the wheel

= 2πr

$$= 2×\frac{22}{7}×35$$

= 220 cm

But speed of the wheel = 66 kmh^{–1}

$$=\frac{66×100}{60}\space\text{m/min}^{\normalsize-1}$$

= 110000 cm min^{–1}

∴ Number of revolutions in per minute

$$=\frac{110000}{220}$$

= 500 revolution

**2. A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20 m × 16 m. Find the area of the field in which the cow can graze.**

**Sol.** Let ABCD be a rectangular field of dimension 20 m × 16 m. Suppose a cow is tied at a point A. Let length of rope be AE = 14 m = r (say).

∴ Area of the field in which the cow graze

= Area of sector AFEG

$$=\frac{90}{360}×\pi r^{2}\\=\frac{90}{360}×\pi(14)^{2}$$

(∵ The angle between two adjacent sides of a rectangle is 90°)

$$=\frac{1}{4}×\frac{22}{7}×196$$

= 154 m^{2}

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

**Sol.** Length and breadth of a circular ends is 38 cm and 10 cm.

∴ Area of rectangle ACDF = Length × Breadh

= 38 × 10 = 380 cm^{2}

Both ends of flower bed a semi-circle.

$$\therefore\space\text{Radius of semi-circle =}\frac{\text{DF}}{2}\\=\frac{10}{2}=5\space\text{cm}\\\therefore\space\text{Area of one semi-circle =}\frac{\pi r^{2}}{2}\\=\frac{\pi}{2}(5)^{2}\\=\frac{25\pi}{2}\text{cm}^{2}\\\therefore\text{Area of two semi-circle =}2×\frac{25}{2}\pi$$

= 25 π cm^{2}

∴ Total area of flower bed = Area of rectangle ACDF + Area of two semi-circle

= (380 + 25π) cm^{2}

**4. In figure, AB is a diameter of the circle, AC = 6 cm and BC = 8 cm. Find the area of the shaded region. (Use π = 3.14)**

**Sol.** Given, AC = 6 cm, BC = 8 cm

We know, the triangle formed from the diameter of a circle to the circle is right angled triangle.

∴ ∠C = 90°

In right DACB, use Pythagoras theorem

∴ AB^{2} = AC^{2} + CB^{2}

⇒ AB^{2} = 6^{2} + 8^{2}

= 36 + 64

⇒ AB^{2} = 100

⇒ AB = 10

(∵ Side can not be negative)

$$\therefore\space\text{Area of ΔABC =}\frac{1}{2}×\text{BC×AC}\\\frac{1}{2}×8×6$$

= 24 cm^{2}

Here, diameter of circle, AB = 10 cm

$$\therefore\space\text{Radius of circle, r =}\frac{10}{2}=\text{5 cm}$$

∴ Area of circle = πr^{2} = 3.14 × (5)^{2}

= 3.14 × 25 = 78.5 cm^{2}

∴ Area of the shaded region = Area of circle – Area of ΔABC

= 78.5 – 24

= 54.5 cm^{2}

**5. Find the area of the shaded field shown in figure.**

∴ Radius of semi-circle DFE

r = 6 – 4 = 2 m

Now, area of rectangle ABCD = BC × AB

= 8 × 4 = 32 m^{2}

And area of semi-circle DFE

$$=\frac{\pi r^{2}}{2}=\frac{\pi}{2}(2)^{2}=2\pi m^{2}$$

∴ Area of shaded region = Area of rectangle ABCD + Area of semi-circles DFE

= (32 + 2π) m^{2}

**6. Find the area of the shaded region in figure.**

**Sol.** Join GH and FE.

Here, breadh of the rectangular BC = 12 m

∴ Length of inner rectangle EFGH

= 26 – (5 + 5) = 16 m

∴ Breadth of the inner rectangle EFGH

= 12 – ( 4 + 4) = 4 cm

Which is equal to the diameter of the semi-circle EJF, d = 4 cm

∴ Radius of semi-circle EJF, r = 2 m

∴ Area of two semi-circles EJF, HIG

$$=\bigg(\frac{\pi r^{2}}{2}\bigg)\\=2×\pi\frac{(2)^{2}}{2}=4\pi\space\text{m}$$

Now, area of inner rectangle EFGH = EH × FG

= 16 × 4 = 64 m^{2}

Now, area of outer rectangle ABCD

= 26 × 12 = 312 m^{2}

∴ Area of shaded region

=Area of outer rectangle – (Area of two semi-circles + Area of inner rectangle)

= 312 – (64 + 4π)

= (248 – 4π) m^{2}

**7. Find the area of the shaded region in figure, where are drawn with centres A, B, C and D intersect in pairs at mid-point P, Q, R and S of the sides AB, BC, CD and DA, respectively of a square ABCD. (Use π = 3.14)**

**Sol.** Given, side of a square BC = 12 cm

Since, Q is a mid-point of BC

$$\therefore\space\text{Radius = BQ}=\frac{12}{2}=\text{6\space cm}\\\text{Now, area of quadrant BPQ =}\frac{\pi r^{2}}{4}\\=\frac{3.14×(6)^{2}}{4}\\=\frac{113.04}{4}\text{cm}^{2}\\\text{area of four quadrants =}\frac{4×113.04}{4}$$

= 113.04 cm^{2}

Now, area of square ABCD = (12)^{2}

= 144 cm^{2}

∴ Area of the shaded region = Area of square – Area of four quadrants

= 144 – 113.04

= 30.96 cm^{2}

**8. In figure areas are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm. To intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of the shaded region. (Use π = 3.14)**

**Sol.** Since, ABC is an equilateral triangle,

∴ ∠A = ∠B = ∠C = 60

and AB = BC = AC = 10 cm

∴ E, F and D are mid-points of the sides.

∴ AE = EC = CD = BD = BF = FA = 5 cm

$$\text{Now, area of sector CDE =}\frac{\theta\pi r^{2}}{360}\\=\frac{60×3.14}{360}(5)^{2}\\\frac{3.14×25}{6}\\=\frac{78.5}{6}$$

= 13.0833 cm^{2}

∴ Area of shaded region

= 3(Area of sector CDE)

= 3 × 13.0833

= 39.25 cm^{2}

**9. A piece of wire 20 cm long is bent into the form of an arc of a circle subtending an angle of 60° at the centre. Find the radius of the circle.**

**Sol.** AB is an arc of circle.

Here, θ = 60°

$$\therefore\space\text{Length of arc =}\frac{\theta}{360}×2\pi r\\20=\frac{60\degree}{360}×2\pi r\\\Rarr\space\frac{20×6}{2\pi}=r\\\Rarr\space r=\frac{60}{\pi}\text{cm}\\\text{Hence, radius of circle is}\frac{60}{\pi}\text{cm.}$$

**Exercise 12.3**

**1. The area of a circular playground is 22176 m**

^{2}. Find the cost of fencing this ground at the rate of ₹ 50 per m.**Sol.**Given,

^{2}

^{2}= 22176

^{2})

^{2}= 1008 × 7

^{2}= 7056

**2. The diameters of front and rear wheels of a tractor are 80 cm and 2 m, respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions.**

**Sol.** Given, diameter of front wheels, d_{1} = 80 cm

and diameter of rear wheels, d_{2} = 2 m = 200 cm

∴ Its radius r_{1} = 40 cm, r_{2} = 100 cm

∴ Circumference of the front wheel

$$=2\pi r_1=\frac{2×22}{7}×40\\=\frac{1760}{7}$$

∴ The total distance covered by the wheel

$$=\frac{1400×7}{1760}=\frac{9800}{1760}$$

= 5.56 cm

**3. In figure ABCD is a trapezium with AB || DC. AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.**

**Sol.** Given, AB = 18 cm, DC = 32 cm and height, y = 14 cm

Given, arc of radii = 7 cm

Since, AB || DC

∴ ∠A + ∠D = 180°

∴ Area of sector A and D points

$$=\frac{\theta×\pi r^{2}}{360}\\=\frac{180\degree}{360\degree}×\frac{22}{7}×(7)^{2}$$

= 11 × 7 = 77 cm^{2}

Similarly, area of sector B and C = 77 cm

$$\text{Now, area of trapezium =}\\\frac{1}{2}(\text{AB+DC})×h\\=\frac{1}{2}(18+32)×14\\=\frac{50}{2}×14=350\space\text{cm}^{2}$$

∴ Area of shaded region

= Area of trapezium – (Area of sector A and D + Area of sector B and C)

= 350 – (77 + 77)

= 196 cm^{2}

**4. On a square cardboard sheet of area 784 cm2, four congreuent 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 the circular plates.**

**Sol.** Since, area of square = 784

⇒ (side)^{2} = (28)^{2}

⇒ side = 28 cm

Since, all four are congruent circular plates.

∴ Diameter of each circular plate = 14 cm

∴ Radius of each circular plate = 7 cm

Now, area of one circular plate = πr^{2}

$$=\frac{22}{7}(7)^{2}$$

= 154 cm^{2}

∴ Area of four circular plates = 4 × 154

= 616 cm^{2}

∴ Area of the square sheet not covered by the circular plates = 784 – 616 = 168 cm^{2}

**5. An archery target has three regions formed by the concentric circles are shown in figure. If the diameters of the concentric circles are in the ratio 1 : 2 : 3, then find the ratio of the areas of three regions.**

**Sol.** Let the diameters of concentric circles are k, 2k and 3k.

$$\therefore\space\text{Radius of concentric circles are}\\\space\frac{k}{2},\space\text{and}\space\frac{3k}{2}.\\\therefore\space\text{Area of inner circle, A}_1=\\\pi\bigg(\frac{k}{2}\bigg)^{2}=\frac{k^{2}\pi}{4}\\\therefore\space\text{Area of middle region,}\\ A_2=\pi(k)^{2}-\frac{k^{2}\pi}{4}=\frac{3k^{2}\pi}{4}$$

[∵ Area of ring = π(R^{2} – r^{2}) where ‘R’ is the radius of outer ring and ‘r’ is the radius of inner ring.

and area outer region, A_{3} =

$$\pi\bigg(\frac{3k}{2}\bigg)^{2}-\pi k^{2}\\=\frac{9\pi k^{2}}{4}-\pi k^{2}\\=\frac{5\pi k^{2}}{4}$$

∴ Required ratio = A_{1} : A_{2} : A_{3}

$$=\frac{k^{2}\pi}{4}:\frac{3k^{2}\pi}{4}:\frac{5\pi k^{2}}{4}$$

= 1 : 3 : 5

**6. Find the area of the shaded region given in figure.**

**Sol.** Join JK, KL, LM and MJ.

Their are four equally semi-circle at point E, F, G, H

And LMJK formed a square.

∴ FH = 14 – (3 + 3) = 8 cm

∴ Side of square should be 4 cm and radius of semi-circle of both ends are 2 cm each.

∴ Area of square JKLM = (4)^{2} = 16 cm^{2}

$$\text{Area semi-circle HJM =}\frac{\pi r^{2}}{2}\\=\frac{3.14×(2)^{2}}{2}$$

= 6.28 cm^{2}

∴ Area of four semi-circle = 4 × 6.28 = 25.12 cm^{2}

∴ Area of shaded region = Area of square ABCD – [Area of four semi-circle + Area of square JKLM]

= 196 – [25.12 + 16]

= 196 – 41.12

= 154.88 cm^{2 }

## NCERT Solutions for Class 10 Mathematics Chapter 12 Free PDF Download

Please Click on Free PDF Download link to Download the NCERT Solutions for Class 10 Mathematics Chapter 12 Areas Related to Circles