# Exercise 12.1

**Unless stated otherwise, use Ï€ =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.**

**Answer:**

Let the radius of the third circle be R.

Circumference of the circle with radius R = 2Ï€R

Circumference of the circle with radius 19 cm = 2Ï€ Ã— 19 = 38Ï€ cm

Circumference of the circle with radius 9 cm = 2Ï€ Ã— 9 = 18Ï€ cm

Sum of the circumference of two circles = 38Ï€ + 18Ï€ = 56Ï€ cm

Circumference of the third circle = 2Ï€R = 56Ï€

â‡’ 2Ï€R = 56Ï€ cm

â‡’ R = 28 cm

The radius of the circle which has circumference equal to the sum of the circumferences of the 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 area equal to the sum of the areas of the two circles.**

**Answer:**

Let the radius of the third circle be R.

Area of the circle with radius R = Ï€R^{2}

Area of the circle with radius 8 cm = Ï€ Ã— 8^{2}Â = 64Ï€ cm^{2}

Area of the circle with radius 6 cm = Ï€ Ã— 6^{2}Â = 36Ï€ cm^{2}

Sum of the area of two circles = 64Ï€ cm^{2}Â + 36Ï€ cm^{2}Â = 100Ï€ cm^{2}

Area of the third circle = Ï€R^{2}Â = 100Ï€ cm^{2}

â‡’ Ï€R^{2Â }= 100Ï€ cm^{2}

â‡’ R^{2}Â = 100 cm^{2}

â‡’ R = 10 cm

Thus, the radius of the 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.**

**Answer:**

### Diameter of Gold region= 21 cm

RadiusÂ ofÂ goldÂ region= 21/2= 10.5 cm

Area of Gold Region = Ï€rÂ²

= Ï€(10.5)Â² =( 22/7)Ã—Â 110.25 = 346.5 cmÂ²

AreaÂ ofÂ GoldÂ Region=Â 345.5 cmÂ²

RadiusÂ ofÂ redÂ regionÂ = Radius for gold + red region= 10.5 + 10.5= 21 cm

Area of Red Region = Ï€Â²(21Â² – 10.5Â²)

[Area of a ring= Ï€ (RÂ²-rÂ²), where R= radius of outer ring & r= radius of inner ring]

= 22/7 (21Â² â€“ 10.5Â²)Â Â Â Â [ aÂ²-bÂ²= (a+b)(a-b)]

= 22/7 (21 + 10.5)(21 â€“ 10.5)

= (22/7 )x 31.5 x 10.5 = 1039.5 cmÂ²

AreaÂ ofÂ RedÂ RegionÂ = 1039.5 cmÂ²

RadiusÂ ofÂ blueÂ regionÂ = Radius of blue region = Now radius for gold + red+ blue region= 21+10.5= 31.5 cm

Area of Blue Region = Ï€(31.5Â² â€“ 21Â²)

= 22/7 (31.5Â² – 21Â²)

= 22/7 (31.5 +21)(31.5 – 21)

= (22/7 )x 52.5 x 10.5 = 1732.5 cmÂ²

AreaÂ ofÂ BlueÂ RegionÂ =1732.5 cmÂ²

Now,

RadiusÂ ofÂ blackÂ region= radius for gold + red+ blue + black region= 31.5+10.5= 42 cm

Area of Black Region = Ï€(42Â² â€“ 31.5Â²)

= 22/7 (42Â²-31.5Â² )

= 22/7 (42+31.5)(42-31.5)

= (22/7 )x 73.5 x 10.5 = 2425.5 cmÂ²

AreaÂ ofÂ BlackÂ RegionÂ =2425.5 cmÂ²

Now

RadiusÂ ofÂ whiteÂ region= radius for gold + red+ blue + black+ white region= 42+10.5= 52.5 cm

Area of White Region= Ï€(52.5Â² â€“ 42Â²)

= 22/7 (52.5Â²-42Â² )

= 22/7 (52.5+42)(52.5-42)

= (22/7 )x 94.5 x 10.5 = 3118.5 cmÂ²

AreaÂ ofÂ whiteÂ RegionÂ =3118.5 cmÂ²

**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?**

**Answer:**

### The car travels in 10 minutes =66 / 6

= 11km

= 1100000cm

Circumference of the wheel = distance covered by the wheel in one revolution Thus, we have,

Circumference = 2 Ã— 22/7 Ã— 80/2 = 251.43cm

Thus, the number of revolutions covered by the wheel in 1100000cm = 1100000 / 251.43 â‰ˆ 4375

**5. Tick the correct answer in the following and justify your choice: If the perimeter and area of a circle are numerically equal, then the radius of the circle is:**

### (A) 2 units

### (B) Ï€ units

### (C) 4 units

### (D) 7 units

**Answer:**

### (A) Circumference = Area

### Correct option is A)

Perimeter of circle Â = 2Ï€r

Area of circle Â = Ï€rÂ²

According to the Question,

Perimeter of circle = Area of circle

2Ï€r = Ï€rÂ²

or, 2 = r

or, r = 2 units

# EXERCISE 12.2

**Unless stated otherwise, use Ï€ =22/7.**

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

**Answer:**

#### Area of the sector making angle Î¸

= (Î¸/360Â°)Ã—Ï€ r^{2}

Area of the sector making angle 60Â°

= (60Â°/360Â°)Ã—Ï€ r^{2}Â cm^{2}

= (1/6)Ã—6^{2}Ï€

= 36/6 Ï€ cm^{2}

= 6 Ã— 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 90Â°.

Circumference of the circle = 2Ï€r

= 22 cm

Radius of the circle = r

= 22/2Ï€ cm

= 7/2 cm

Area of the sector making angle 90Â°

= (90Â°/360Â°)Ã—Ï€ r^{2}Â cm^{2}

= (1/4)Ã—(7/2)^{2}Ï€

= (49/16) Ï€ cm^{2}

= (49/16) Ã— (22/7) cm^{2}

= 77/8 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.**

**Answer:**

#### We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360Â°

In 5 minutes, minute hand will rotate = 360Â°/ 60 x 5 = 30Â°

Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30Â° in a circle of 14 cm radius.

Area of sector of angle Î¸ = Î¸/360Â° . Ï€. r^{2}

Area of sector of 30Â° = 30Â°/360Â° . 22/7 . 14 .14

â†’ = 22/12 . 2 .14

#### = ((11).(14))/3

= 154/3 cm^{2}

= 51.33 cm^{2}

#### Therefore, the area swept by the minute hand in 5 minutes is 51.33 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 Ï€ = 3.14)**

**Answer:**

#### Radius of the circle = 10 cm

#### â€¨Major sector is making 360Â° âˆ’ 90Â° = 270Â°

#### â€¨Area of the sector making angle 270Â°

#### â€¨=(270Â°/360Â°) Ã— Ï€r^{2}Â cm^{2}

#### =(3/4) Ã— 10^{2}Ï€ = 75Ï€cm^{2}

#### =75 Ã— 3.14cm^{2}Â = 235.5cm^{2}

#### âˆ´ Area of the major sector = 235.5cm^{2}

#### â€¨Height of Î”AOB = OA = 10 cm

#### â€¨Base of Î”AOB = OB = 10 cm

#### â€¨Area of Î”AOB = 1/2 Ã— OA Ã— OB

#### â€¨= 1/2 Ã— 10 Ã— 10 = 50cm^{2}

#### â€¨Minor sector is making 90Â°

#### Area of the sector making angle 90Â°

#### â€¨= (90Â°/360Â°) Ã— Ï€r^{2}Â cm^{2}

#### = (1/4) Ã— 10^{2}Ï€ = 25Ï€cm^{2}

#### â€¨= 25 Ã— 3.14 cm^{2}

#### = 78.5 cm^{2}

#### â€¨Area of the minor segment = Area of the sector making angle 90Â° âˆ’ Area of Î”AOB

#### â€¨= 78.5 cm^{2}Â âˆ’ 50cm^{2}Â = 28.5cm^{2}

**5. In a circle of radius 21 cm, an arc subtends an angle of 60Â° 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

**Answer:**

#### In the mentioned figure,

O is the centre of circle,

AB is a chord

AXB is a major arc,

OA=OB= radius =21 cm

Arc AXB subtends an angle 60Â° at O.

i) Length of an arc AXB = 60/360 Ã— 2Ï€ Ã— r

= 1/6 Ã— 2 Ã— 22/7 Ã— 21

= 22cm

ii) Area of sector AOB = 60/360 Ã— Ï€ Ã— r^{2}

= 1/6 Ã— 22/7 Ã— (21)^{2}

= 231cm^{2}

(iii) Area of equilateral Î”AOB = âˆš3/4 Ã— (OA)^{2}Â = âˆš3/4 Ã— 212 = (441âˆš3)/4 cm^{2}

#### Area of the segment formed by the corresponding chord

= Area of the sector formed by the arc – Area of equilateral Î”AOB

= 231 cm^{2Â }– (441âˆš3)/4 cm^{2}

**6. A chord of a circle of radius 15 cm subtends an angle of 60Â° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use Ï€ = 3.14 and âˆš3 = 1.73)**

**Answer:**

#### Radius of the circle = 15 cm

Î”AOB is isosceles as two sides are equal.

âˆ´ âˆ A = âˆ B

Sum of all angles of triangle = 180Â°

âˆ A + âˆ B + âˆ C = 180Â°

â‡’ 2 âˆ A = 180Â° – 60Â°

â‡’ âˆ A = 120Â°/2

â‡’ âˆ A = 60Â°

Triangle is equilateral as âˆ A = âˆ B = âˆ C = 60Â°

âˆ´ OA = OB = AB = 15 cm

Area of equilateral Î”AOB

= âˆš3/4 Ã— (OA)^{2}Â = âˆš3/4 Ã— 15^{2}

= (225âˆš3)/4 cm^{2}Â = 97.3 cm^{2}

Angle subtend at the centre by minor segment = 60Â°

Area of Minor sector making angle 60Â°

= (60Â°/360Â°) Ã— Ï€ r^{2}Â cm^{2}

= (1/6) Ã— 15^{2}Â Ï€Â cm^{2}Â =Â 225/6 Ï€Â cm^{2}

=Â (225/6) Ã— 3.14 cm^{2Â }= 117.75Â cm^{2}

Area of the minor segment = Area of Minor sector – Area of equilateral Î”AOB

= 117.75Â cm^{2}Â – 97.3 cm^{2}Â = 20.4 cm^{2}

Angle made by Major sector = 360Â° – 60Â° = 300Â°

Area of the sector making angle 300Â°

= (300Â°/360Â°) Ã— Ï€ r^{2}Â cm^{2}

= (5/6) Ã— 15^{2}Â Ï€Â cm^{2}Â =Â 1125/6 Ï€Â cm^{2}

=Â (1125/6) Ã— 3.14 cm^{2}Â = 588.75Â cm^{2}

Area of major segment = Area of Minor sector + Area of equilateral Î”AOB

= 588.75Â cm^{2}Â + 97.3 cm^{2}Â = 686.05 cm^{2}

**7. A chord of a circle of radius 12 cm subtends anÂ angle of 120Â° at the centre. Find the area of theÂ corresponding segment of the circle.Â (Use Ï€ = 3.14 andÂ âˆš3Â = 1.73)**

**Answer:**

#### We have,

In a circle,

Radius

r = 12cm.

Î¸ = 120Â°

Area of segment APB = Area of sector OAPB – Area of Î”OAB

Area of sector OAPB

= Î¸/360Â° Ã— Ï€r^{2}

= 120Â°/360 Ã— 3.14 Ã— (12)^{2}

= 1/3 Ã— 3.14 Ã— 144

= 3.14 Ã— 4 Ã—12

= 3.14 Ã— 48

= 150.72 cm^{2}

Finding area of triangle Î”AOB

Area of Î”AOB = 12 Ã— Base Ã— Height

We draw

OMâŠ¥AB

âˆ´ âˆ OMB = âˆ OMA = 90Â°

In

Î”OMA and Î”OMB

âˆ OMA = âˆ OMB (Both 90Â°)

OM = OM( Both radius)

OA = OB (Common line)

âˆ´ Î”OMA â‰… Î”OMB (S.A.S. Congurency)

â‡’ âˆ AOM = âˆ BOM

Then,

BM = AM = 1/2 AB ……. (1)

Now,

âˆ AOM = âˆ BOM = 1/2 âˆ BOA

= 1/2 Ã— 1200

= 60Â°

In right angle triangle OMA

sin O= AM/AO

sin 60Â° = AM/12

âˆš3/2 = AM/12

AM = 6âˆš3

Again, In right angle triangle OMA

cos O = OM/AO

cos 60Â° = OM/AO

1/2 = OM/12

OM = 6cm.

Now, from equation (1) and we get,

AM = 1/2 AB

2AM = AB

AB = 2AM

AB = 2 Ã— 6âˆš3Â Â Â Â Â âˆ´ AM = 6âˆš3

AB = 12âˆš3 cm.

Now,

Area of Î”AOB = 1/2 Ã— Base Ã— Height

= 1/2 Ã— AB Ã— OM

= 1/2 Ã— 12âˆš3 Ã— 6

= 36âˆš3

= 36 Ã— 1.73

62.28 cm^{2}

Hence, the Area of segment of

APB = Area of sector OAPB âˆ’ Area of Î”OAB

= 150.72 âˆ’ 62.28

= 88.44 cm^{2}

Hence, this is the answer.

**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:**

**Answer:**

#### Side of square field = 15 m

Length of rope is the radius of the circle, r = 5 m

Since, the horse is tied at one end of square field; it will graze only quarter of the field with radius 5 m.

(i) Area of circle = Ï€ r^{2}Â = 3.14 Ã— 5^{2}Â = 78.5 m^{2}

Area of that part of the field in which the horse can graze = 1/4 of area of the circle

#### = 78.54 = 19.625 m^{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.

**Answer:**

#### Diameter =35mm

#### â€¨âˆ´ Circumference of circle

= 2 Ã— Ï€ Ã— r = 2 Ã— Ï€ Ã— 17.5 = 110 mm

i) The total length of the silver wire required

#### â€¨= Circumference of circle + (5 Ã— 35)

#### â€¨= 110 + 175 = 285mm

#### (ii) The area of each sector of the brooch

#### â€¨= Area of circle/10

#### â€¨= (Ï€ Ã—17.5 Ã— 17.5)/10 â€¨ = 385/4 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.**

**Answer:**

#### Here, area between two consecutive ribs = 1/8 Ã— Area of umbrella

Radius = r = 45cm

Area of umbrella = Ï€r^{2}

= 22/7 Ã— 45 Ã— 45 cm^{2}

= 22/7 Ã— 2025 cm^{2}

Required area = 1/8 Ã— Area of umbrella

= 1/8 Ã— 22/7 Ã— 2025 cm^{2}

= 22275/28 cm^{2}

= 795.54 cm^{2}

Hence, area between two consecutive ribs is 795.54 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 115Â° . Find the total area cleaned at each sweep of the blades.**

**Answer:**

#### Angle of the sector of circle made by wiper = 115Â°

Radius of wiper = 25 cm

Area of the sector made by wiper

= (115Â°/360Â°) Ã— Ï€ r^{2Â }cm^{2}

= 23/72 Ã— 22/7 Ã— 25^{2}

=Â 23/72 Ã— 22/7 Ã— 625 cm^{2}

= 158125/252 cm^{2}

Total area cleaned at each sweep of the blades

= 2 Ã—158125/252 cm^{2}

=Â 158125/126 = 1254.96 cm^{2}

**12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80Â° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.**

(Use Ï€ = 3.14)

**Answer:**

#### Let O bet the position of Lighthouse.

Distance over which light spread i.e. radius, r = 16.5 km

Angle made by the sector = 80Â°

Area of the sea over which the ships are warned = Area made by the sector.

Area of sector = (80Â°/360Â°) Ã— Ï€ r^{2}Â km^{2}

= 2/9 Ã— 3.14 Ã— (16.5)^{2}Â km^{2}

=Â 189.97 km^{2}

**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 â‚¹ 0.35 per cm**^{2}Â . (Use âˆš3 = 1.7)

^{2}Â . (Use âˆš3 = 1.7)

**Answer:**

#### r = 28 cm, Î¸ = 360Â°/6 = 60Â°

Area OAMB = Î¸/360Â° Ã— Ï€r^{2}Â = 60Â°/360Â° Ã— 22/7 Ã— 28 Ã— 28

= 1232/3 cm^{2}Â = 410.67 cm^{2}

Now, in Î”ONA and Î”ONB

i) âˆ ONA = âˆ ONB [each 90Â°]

ii) OA = OB [Radii of common circle]

iii) ON = ON [common]

âˆ´ Î”ONA â‰… Î”ONB [By RHS congruency]

Hence, AN = NB = 1/2 AB

& âˆ AON = âˆ BON = 1/2 âˆ AOB = 60Â°/2 = 30Â°

Now in Î”ONA, cos 30Â° = ON/OA â‡’ âˆš3/2 = ON/28 â‡’ ON = 14âˆš3cm

& sin 30Â° = AN/OA â‡’ 1/2 = AN/28 â‡’ AN = 14cm

& 2AN = 14 Ã— 2 = 28cm = AB

âˆ´Â arÎ”AOB = 1/2 Ã— AB Ã— ON = 1/2 Ã— 28 Ã— 14âˆš3 = 196âˆš3 = 196 Ã— 1.7 = 333.2 cm^{2}

âˆ´ Area of one design = 410.67 âˆ’ 333.2 = 77.47 cm^{2}

âˆ´ Area of six design = 6 Ã— 77.47 = 464.82 cm^{2}

Therefore, Cost of making the designs = Rs. (464.82 Ã— 0.35)

= Rs.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

(A) p/180 Ã— 2Ï€R

(B) p/180 Ã— Ï€ R^{2}

(C) p/360 Ã— 2Ï€R

(D) p/720 Ã— 2Ï€R^{2}

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

**Answer:**

Area of a sector of angle p

= p/360 Ã— Ï€ R^{2}

= p/360 Ã— 2/2 Ã— Ï€ R^{2}

=Â 2p/720 Ã— 2Ï€R^{2}

Hence, Option (D) is correct.

# EXERCISE 12.3

**Unless stated otherwise, use Ï€ =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.**

**Answer:**

#### PQ = 24 cm and PR = 7 cm

#### â€¨âˆ P = 90Â° (Angle in the semi-circle)

#### â€¨âˆ´ QR is hypotenuse of the circle = Diameter of the circle.

#### â€¨By Pythagoras theorem,

#### QR^{2}Â = PR^{2}Â + PQ^{2}

#### â€¨â‡’ QR^{2}Â = 7^{2}Â + 24^{2}

#### â‡’ QR^{2}Â = 49 +576

#### â‡’ QR^{2}Â = 625

#### â‡’ QR = 25cm

#### â€¨âˆ´ Radius of the circle = 25/2cm

#### â€¨Area of the semicircle = (Ï€R^{2})/2

#### = (22/7 Ã— 25/2 Ã— 25/2)/2) cm^{2}

#### = 13750/56 cm^{2}Â = 245.54 cm^{2}

#### â€¨Area of the Î”PQR = 1/2 Ã— PR Ã— PQ

#### â€¨= 1/2 Ã— 7 Ã— 24 cm^{2}

#### = 84 cm^{2}

#### â€¨Area of the shaded region

#### = Area of the semicircle – Area of the Î”PQR

#### = 245.54 cm^{2}Â âˆ’ 84 cm^{2}Â = 161.54cm^{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 âˆ AOC = 40Â°.**

**Answer:**

#### Given that,

Radius of the small circle, OB = 7 cm

Radius of second circle, OA = 14 cm

and âˆ AOC = 40Â°

We know that, the area of a sector that subtends an angle Î¸ at the centre of the circle is Î¸/360Â° Ã— Ï€r^{2}

where, Î¸ is in degrees.

âˆ´ Area of minor sector OBD = 40/360 Ã— 22/7 Ã— 7 Ã— 7Â Â Â Â [âˆµ Ï€ = 22/7]

= 1/9 Ã— 22 Ã— 7

= 17.11 cm^{2}

Also, area of minor sector OAC = 40/360 Ã— 22/7 Ã— 14 Ã— 14

= 1/9 Ã— 22 Ã— 2 Ã— 14

= 68.4 cm^{2}

Now, area of the shaded region = Area of sector OAC âˆ’ Area of sector OBD

= 68.4 âˆ’ 17.1

= 51.3 cm^{2}

Hence, the area of the shaded region is 51.3 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 semicircles.**

**Answer:**

#### Area of the shaded region = Area of Square ABCD âˆ’ (Area of semicircle APD + Area of semicircle BPC)

#### = (14)^{2}Â âˆ’ [1/2 Ï€ (14/2)^{2}Â + 1/2 Ï€ (14/2)^{2}]

#### = (14)^{2}Â âˆ’ 22/7 (7)^{2}

#### = 196 âˆ’ 154

#### = 42 cm^{2}

Hence, the area of the shaded region is 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.**

**Answer:**

#### Area of Shaded region = Area of equilateral triangle ABO + Area of Major sector

#### Area of Equilateral Triangle ABO = âˆš3/4 Ã— a^{2}Â = âˆš3/4 Ã— 12 Ã— 12 cm^{2}

#### = 62.352 cm^{2}

#### Area of major sector = Î¸/360 Ã— Ï€ Ã— r^{2}

= 300/360 Ã— Ï€ Ã— 6 Ã— 6 cm^{2}

= 94.2 cm^{2}

#### Area of Shaded region = 62.352 + 94.2 = 156.55 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 figure.**

**Answer:**

#### Side of the square = 4 cm

Radius of the circle = 1 cm

Four quadrant of a circle are cut from corner and one circle of radius are cut from middle.

Area of square = (side)^{2}Â = 4^{2}Â = 16 cm^{2}

Area of the quadrant = (Ï€ R^{2})/4 cm^{2}Â = (22/7 Ã— 12)/4 = 11/14 cm^{2}

âˆ´ Total area of the 4 quadrants = 4 Ã— (11/14) cm^{2}Â = 22/7 cm^{2}

Area of the circle = Ï€ R^{2}Â cm^{2}Â = (22/7 Ã— 12) = 22/7 cm^{2}

Area of the shaded region = Area of square – (Area of the 4 quadrants + Area of the circle)

= 16 cm^{2}Â – (22/7 + 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 figure. Find the area of the design (shaded region).**

**Answer:**

#### cos Î¸ = AD/OA

cos 30Â° = âˆš3/2

AD = âˆš3/2 Ã— 32

AD = 16âˆš3cm

AB = 32âˆš3cm

shaded region = Area of circle – Area of triangle

= Ï€(32)^{2}Â âˆ’ 3 Ã— 1/2 Ã— 32âˆš3 Ã— 16

= ((22578/7) âˆ’ 768âˆš3)cm^{2}

**7. In figure ABCD is a square of side 14 cm. With centers A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.**

**Answer:**

#### Side of square = 14 cm

Four quadrants are included in the four sides of the square.

âˆ´ Radius of the circles = 14/2 cm = 7 cm

Area of the square ABCD = 14^{2Â }= 196 cm^{2}

Area of the quadrant = (Ï€ R2)/4 cm^{2}Â = (22/7 Ã— 72)/4 cm^{2}

= 77/2 cm^{2}

Total area of the quadrant = 4 Ã— 77/2 cm^{2Â }= 154 cm^{2}

Area of the shaded region = Area of the square ABCD – Area of the quadrant

= 196 cm^{2}Â – 154 cm^{2}

= 42 cm^{2}

**8. Figure 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.

**Answer:**

#### (i) â€¨The distance around the track along its inner edge

#### â€¨= EG + FH + 2 Ã— (circumference of the semicircle of radius OE = 30 cm)

#### â€¨= 106 + 106 + 2(1/2 Ã— 2Ï€ Ã— 30) = 212 + 60Ï€

#### â€¨= 212 + 60 x 22/7 = (212+1320/7)m = (1484 + 1320)/7m = 2804/7m = (400)4/7m

#### â€¨(ii) â€¨Area of the track = Area of the shaded region

#### = Area of the rectangle AEGC + Area of the rectangle BFHD + 2 + (Area of the semicircle of radius 40 m âˆ’ Area of the semicircle with radius 30 mâˆ’)

#### = [(10 Ã— 106) + (10 Ã— 106)] + 2{1/2 Ã— 22/7 Ã— (40)^{2}Â âˆ’ 1/2 Ã— 22/7 Ã— (30)^{2}}

#### â€¨= 1060 + 1060 + 22/7 [(40)^{2}Â âˆ’ (30)^{2}]

#### = 2120 + 22/7 Ã— 700 = 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.**

**Answer:**

#### OD is diameter of smaller circle.

#### â€¨â€¨Area of the smaller circle

#### = Ï€ Ã— 3.5^{2}Â = 38.49 cm^{2}

#### Area of Î”ABC = 0.5 Ã— (14)(7) = 49 cm^{2}

#### Area of semi-circle = 0.5Ï€(7)(7) = 76.97 cm^{2}

#### Area of shaded portion in semi-circle = 76.97 âˆ’ 49 = 27.97 cm^{2}

#### Total required area = 38.49 + 27.97 = 66.46 cm^{2}

**10. The area of an equilateral triangle ABC is 17320.5Â cm**^{2}. With each vertex of the triangle as centre, aÂ 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. (Use Ï€ = 3.14 andÂ âˆš3Â = 1.73205)

^{2}. With each vertex of the triangle as centre, aÂ 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. (Use Ï€ = 3.14 andÂ âˆš3Â = 1.73205)

**Answer:**

#### Let the side of the equilateral triangle beÂ *a*.

Area of equilateral triangle = 17320.5 cm^{2}

âˆš3/4 (*a*)^{2}Â = 17320.5

#### 1.73205/4Â *a*^{2}Â = 17320.5

*a*^{2}Â = 4 x 10000

*a*Â = 200 cm

#### Each sector is of measure 60Â°.

Area of sector ADEF = 60Â°/360Â° x Ï€ xÂ *r*^{2}

#### = 1/6 x Ï€ x (100)^{2}

#### = (3.14 x 10000)/6

#### = 15700/3 cm^{2}

#### Area of shaded region = Area of equilateral triangle âˆ’ 3 Ã— Area of each sector

#### = 17320.5 – 3 x 15700/3

= 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.**

**Answer:**

#### Side of square = 3 Ã— diameter of 1 circle

= 3 Ã— 14 = 42cm

Area of remaining portion = Area of square – 9 Ã— area of 1 circle

= (42 Ã— 42) – 9 Ã— 22/7 Ã— 7 Ã— 7

= 1764 âˆ’ 1386

= 378 cm^{2}

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

#### (i) quadrant OACB

#### (ii) shaded region

**Answer:**

#### (i) Area of quadrant OACB = Area of Circle/4 = (Ï€ Ã— r^{2})/4 = 1/4 Ã— 22/7 Ã— 3.5 Ã— 3.5

#### = 9.625cm^{2}

#### (ii) Area of the shaded region = Area of Quadrant âˆ’ Area of â–³BDO

= 9.625 âˆ’ (1/2 Ã— 3.5 Ã— 2)

= 6.125 cm^{2}

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

**Answer:**

#### Side of square = OA = AB = 20 cm

Radius of the quadrant = OB

OAB is right angled triangle

By Pythagoras theorem in Î”OAB ,

OB^{2}Â = AB^{2}Â + OA^{2}

â‡’ OB^{2Â }= 20^{2}Â + 20^{2}

â‡’ OB^{2Â }= 400Â + 400

â‡’ OB^{2}Â = 800

â‡’ OB = 20âˆš2 cm

Area of the quadrant = (Ï€R2)/4 cm^{2}Â = 3.14/4 Ã— (20âˆš2)^{2}Â cm^{2}Â = 628 cm^{2}

Area of the square = 20 Ã— 20 = 400Â cm^{2}

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

= 628 – 400 cm^{2}Â = 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 Fig. ). If âˆ AOB = 30Â°, find the area of the shaded region.**

**Answer:**

#### Radius of the larger sector, R=21 cm

#### â€¨Radius of the smaller sector, r=7 cm

#### Angle subtended by sectors of both concentric circles = 30Â°

#### â€¨Area of the sector making angle Î¸

#### â€¨= (Î¸/360Â°) Ã— Ï€r^{2}

#### Area of the larger sector

#### = (30Â°/360Â°) Ã— Ï€R^{2}cm^{2}

= 1/12 Ã— 22/7 Ã— 21^{2}Â cm^{2}

= 1/12 Ã— 22/7 Ã— 21 Ã— 21 cm^{2}

= 231/2 cm^{2}

#### â€¨Area of the smaller circle

#### = (30Â°/360Â°) Ã— Ï€r^{2}Â cm^{2}

= 1/12 Ã— 22/7 Ã— 7^{2}Â cm^{2}

= 1/12 Ã— 22/7 Ã— 7 Ã— 7 cm^{2}

= 77/6 cm^{2}

#### Area of the shaded region = Area of the larger sector âˆ’ Area of the smaller sector

#### â€¨= (231/2 âˆ’ 77/6) cm^{2}

= 616/6 cm^{2}

= 308/3 cm^{2}

= (102) 2/3 cm^{2}

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

**Answer:**

#### Radius of the the quadrant ABC of circle = 14 cm

AB = AC = 14 cm

BC is diameter of semicircle.

ABC is right angled triangle.

By Pythagoras theorem in Î”ABC,

BC^{2}Â = AB^{2}Â + AC^{2}

â‡’ BC^{2}Â = 14^{2}Â + 14^{2}

â‡’ BC = 14âˆš2 cm

Radius of semicircle = 14âˆš2/2 cm = 7âˆš2 cm

Area of Î”ABC = 1/2 Ã— 14 Ã— 14 = 98 cm^{2}

Area of quadrant = 1/4 Ã— 22/7 Ã— 14 Ã— 14 = 154 cm^{2}

Area of the semicircle = 1/2 Ã— 22/7 Ã— 7âˆš2 Ã— 7âˆš2 = 154 cm^{2}

Area of the shaded region = Area of the semicircle + Area of Î”ABC – Area of quadrant

= 154 + 98 – 154 cm^{2}Â = 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.**

**Answer:**

#### AB = BC = CD = AD = 8 cm

#### Area of Î”ABC = Area of Î”ADC = 1/2 Ã— 8 Ã— 8 = 32 cm^{2}

Area of quadrant AECB = Area of quadrant AFCD = 1/4 Ã— 22/7 Ã— 8^{2}

= 352/7 cm^{2}

Area of shaded region = (Area of quadrant AECB – Area of Î”ABC) +(Area of quadrant AFCD – Area of Î”ADC)

= (352/7 – 32) + (352/7 -32) cm^{2}

= 2 Ã— (352/7 -32) cm^{2}

=Â 256/7 cm^{2}

# EXERCISE 12.3