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This chapter is an integral part of the mathematics syllabus and deals with calculating areas related to circles, such as the area of a sector, the area of a segment, and the length of an arc.
The chapter provides students with the fundamental concepts of circles, such as the formulae for calculating the circumference and area of a circle. The chapter also covers various practical applications of circles-related areas, making it an essential topic for students to learn.
The NCERT Solutions provided here have been designed to help students understand the concepts covered in this chapter clearly and concisely. These solutions have been prepared by experienced teachers and subject matter experts, considering the latest CBSE guidelines and syllabus.
The solutions provide step-by-step explanations to all the questions in the NCERT textbook, ensuring that students understand the concepts thoroughly. The solutions also provide tips and tricks to solve problems quickly and accurately.
We hope that these NCERT Solutions will help students prepare for the Class 10 board exams and also help them build a strong foundation for higher studies in mathematics.

Download PDF of NCERT Solution for Class 10th Maths Chapter 12 Areas Related To Circles

Answers of Math NCERT Solution for Class 10th Maths Chapter 12 Areas Related To Circles

Unit 12

Areas Related to Circles

Exercise 12.1 

Q1:

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:

Radius (r1) of 1st circle = 19 cm Radius (r2) or 2nd circle = 9 cm Let the radius of 3rd circle be r. 

Circumference of 1st circle = 2πr1 = 2π (19) = 38π Circumference of 2nd circle = 2πr2 = 2π (9) = 18π Circumference of 3rd circle = 2πr 

Given that,

Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle 2πr = 38π + 18π = 56π 

HCoBJsc UGcFcMuFMEwDlxw4b1ei9 lGxa DLKYpMC2qGDaBnOschE X9qkrrTc0q7M1oZuYTDNrwXOlQD32EH5w8UrD7L4wlZlE22q7FTnnE94d5nkLS4Ej5dou1Yuwvcby5 E

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

Q2:

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:

Radius (r1) of 1st circle = 8 cm Radius (r2) of 2nd circle = 6 cm Let the radius of 3rd circle be r. 

Area of 1st circle

Area of 2nd circle

Given that,

Area of 3rd circle = Area of 1st circle + Area of 2nd circle
ODQkHyvTnMtWSjxiejEdZccoV ISGY4bk9LS d6cbjeUxp9

However, the radius cannot be negative. Therefore, the radius of the circle having area equal to the sum of the areas of the two circles is 10 cm.

Q3:

Given figure depicts an archery target marked with its five scoring areas 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 are 10.5 cm wide. Find the area of each of the five scoring regions.

N 6yb631RR4SYJDBY8yuR1VBC88roCdtXTV3tdoTi2dRCN4ZgZ tDcA82FPHeP9j8

Answer:QL2m5Eb2MoF z1D4ReSHb3l9 2aVAN6hj RJUr

Radius (r1) of gold region (i.e., 1st circle)

Given that each circle is 10.5 cm wider than the previous circle. Therefore, radius (r2) of 2nd circle = 10.5 + 10.5

21 cm

Radius (r3) of 3rd circle = 21 + 10.5

= 31.5 cm

Radius (r4) of 4th circle = 31.5 + 10.5

= 42 cm

Radius (r5) of 5th circle = 42 + 10.5

= 52.5 cm

Area of gold region = Area of 1st circle   

Area of red region = Area of 2nd circle – Area of 1st circle
ZSHvLDPUN4hB pZq piO5 ttf0BGDQYjYt5prj9l1H2YxzN15b5

Area of blue region = Area of 3rd circle – Area of 2nd circle
673t6MrQiCEqQoTk6uwI11qlMywCpMuJu0

Area of black region = Area of 4th circle – Area of 3rd circle
8J ZhGr54iLuWNFwkKBYedVGAWpMBkqWpkAGZ1cJMokocl9I2pHrdVPlX0JgNjbd9kWnbt XSFuPzF89mL8XgPOHH1vERxtmpIOyLrzZAMIU9CBnzACL7S yTuNIt61k3m rDZE

Area of white region = Area of 5th circle – Area of 4th circle
1a896nFBx6MpoxNSNef 8qTIBD

Therefore, areas of gold, red, blue, black, and white regions are 346.5 cm2, 1039.5 cm2, 1732.5 cm2, 2425.5 cm2, and 3118.5 cm2 respectively.

Q4:

The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km? per hour?

Answer:

Diameter of the wheel of the car = 80 cm Radius (r) of the wheel of the car = 40 cm Circumference of wheel = 2πr 

= 2π (40) = 80π cm Speed of car = 66 km/hour 

chxhLzULOAuZ ArPciVCSm66YKm17c6a2I0S6AfjVnk oQKME

Distance travelled by the car in 10 minutes

= 110000 × 10 = 1100000 cm

Let the number of revolutions of the wheel of the car be n. n × Distance travelled in 1 revolution (i.e., circumference)

= Distance travelled in 10 minutes 

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

Q5:

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

  1. 2 units         (B) π units         (C)     4 units         (D)    7 units

Answer:

Let the radius of the circle be r.

Circumference of circle = 2πr Area of circle = πr

Given that, the circumference of the circle and the area of the circle are equal. This implies 2πr = πr2  

2 = r

Therefore, the radius of the circle is 2 units. Hence, the correct answer is A.

Exercise 12.2 

Q1:

Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
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Answer:
xnfebVjcrQWciZbVlXYWF jVyQecAcpXy6SbO4JA0MBZVE33eEBLkjWk1QEh6PgZhW1MkT NdUbYfKWtYvPwnClRp ABN1WYD 6o3Xm3h4nwzft8vd7 IbMQM7G8pTUq9twmAI8

Let OACB be a sector of the circle making 60° angle at centre O of the circle.

Area of sector of angle

Area of sector
2PwPAg0d1bO5d0iR3s17hOdpJv qYY0cEeoq BvMurtEbB6cG EvSUWG05epNp8ve6 1O4I5jneZX7xtd04HyqE1G2Lilk3RlkTBzmQVsl7 XvnZcOOkq4F13ptmKomQtw3LJcw

Therefore, the area of the sector of the circle making 60° at the centre of the circle is

Q2:

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

Answer:

bdMAQDIRH Qv icsNAecGKP jDlG0GrFkuDlmo mO32JOEl B60vwPJ9YhMCEAuURHNQu9z tjZN7auw Rpc8wABBnLBqrX 4PlOAHvlWvO6Oeg7ZhLrODu2 iTLdO NiHHKUo

Let the radius of the circle be r. 

Circumference = 22 cm 

2πr = 22

52fIpnYadbfPtBep6SEqqW0AMnUfAZNFHPTgSYbmq6ukpk8KvX7ZwiqYdUk67urAZZEtIToeGWtYdQHgGrfY4VeJWqegbNkW08P40pmn

Quadrant of circle will subtend 90° angle at the centre of the circle.

Area of such quadrant of the circle
IRkNT9XkquhaF2S46ZDcsIfFwEiWx0YtFuhcf pCwU90rFgSkEXmn7dAcDHe21i5cUeNYD XaNIINz ZXyZ1yL

*Q3:

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 360°.

In 5 minutes, minute hand will rotate =

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

rlYQR4cFrPNzUhNP7hzbwmn3FzfDX uN9cGjaVL4VOspsn40vXQo0Xsxg8w2zS7xAg8kzaIjME VpsiRE7KGhFkLV0UYLbhgLqvzb5dE Hv9ySLf7ozl91f7mAb IOYXS4t2t38

Area of sector of angle θ =

YDbGahXZAXuuUrqNh6uHDAriDmnRa5UJ6AToFAOfsvkDAa7Adpchweg4UBzyZuEW3TrwSgOpC OC0P8ogWTFoFrD hO4 ecdBGODdkp3ecFIZVOmlhBesJIW 5bvBg2avYHb9r4

Area of sector of 30°

Therefore, the area swept by the minute hand in 5 minutes iscy E6GKC8Yk3pF6BP4dbxeB2SGmsgPoy d7xwif5NrH5yLOl rpVBI4xHeWhzEwXMZw26lzz 9yf Jt6u 9IWlH hyDQn2ItZ

Q4:

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:

  1.         Minor segment
  2.         Major sector [Use π = 3.14]

Answer:7vERB51RcaB4annx4xkvXU9ohO6V7d72hPUbf9gu2yLiENKvqCOKi5gAV233KUaTjSdAEvbgVbSlyWvaMqu03Yhy8UgXEWIqgDcu Fv gVB8RE

Let AB be the chord of the circle subtending 90° angle at centre O of the circle.

kND7Gyis 6OoI32f35m zr8fAYFSm6tQ8WMm9 N eme4SnFbFgduAXEj0pojGzlGIqYN3kqjMowytx cNROHKjjEkrEsmlCW4XD0V2UrIERfy9ev 0ie9VUbh08JPSY81v KfYA

Area of major sector OADB =

icdqIjrU Wn7qT8i3Nf7F1Q vheQnPdhLqboDooKDGmEtlqoAyJlfBKsD03Uvm2C3jB2ktTalGYp1X6WlTfy W3vbQ1XR8OLVKXMeepjeMrWJ092PIrZf0OPVadXI9znMJ15TqA

Area of minor sector OACB =

vg5WLd c7E1MbOESWIB zMOVG3ObDoE55R8482pHSuHJK8yVSjIX1XVU qn1dZGY3hi 6 RrL0CIWLVh5JNhlE7jbqCFlKPJo12hVR aXAc0JirNLXgs5hDpt7lrY7DWsZsGWCw

0M2mjabpDxefh8S3Y0ND XCeFI6oEATytI4a4o8gXcf2Mq3DKnZvciN2fsvQV64k3eg1RQlYGTimcpMFmQZCZmvLLoPrlIncyYIBkZMkq4oJD3vpWRo IRcftXB2v1ApAc97Mc0

Area of ΔOAB =

= 50 cm2

Area of minor segment ACB = Area of minor sector OACB – Area of ΔOAB = 78.5 – 50 = 28.5 cm2

Q5:

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

  1.         The length of the arc
  2.     Area of the sector formed by the arc
  3. Area of the segment forced by the corresponding chord

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Answer: 

Radius (r) of circle = 21 cm

Angle subtended by the given arc = 60°
BFSnoTAT14 YMaliHcugrY4uIFt8V9xisyVya UyyHqmXLjsLta ePFz 5v7tkvux7ih Sd4 q6lTaP2I28Yfaq2C1caJlJX6BDtp0YsbhYxEt9pZdx SpKcNLnYWz7 TUZnTyDV 3MadJO jAgTzAsNCfrGiwvPxHpjJ0GuJJZLOzlmcd1BG3oBr1W5 KHneEw7hBtTLTo 5i3LLFrKtA

Length of an arc of a sector of angle θ =    ZGcyTzk5MIYdZIg2UiUxy9N8Eayo0hdyyHEEKQNi6KPS7v qomIVBYqF0hkmrkFa YP aFECzrqskzRYS9l R5UkARhkc2kDh5k D58V3Hp Rrbm

Length of arc ACB =

= 22 cmZ1cXW6jmUjpU0sq3XB8NH1n9cmv8ZGIttFotHaMvlvjAsCiGk8loU3LOHfG7w7NPIAKLDNvZlN3OJA mZN SAy0FAxAa1T DZGNtDg o Lnsvh9Y3z8Jdn7JdiiNqonhZK7pabs
D9R0udZ25tDo

Area of sector OACB =
50ln4uCZeNQs1BWTsZEOYieYJBWGU7oIJIetc8ybS9HlsIJpvb9aMfbAgg 2dQ6snIjuTRg2BpcH1I ik20nRCOD5kdB6nJ8nETHuLaLZzjGjUOCmTvrQONrpvOVnl1pw4twQS4

In ΔOAB,

∠OAB = ∠OBA (As OA = OB)

∠OAB + ∠AOB + ∠OBA = 180° 2∠OAB + 60° = 180° 

∠OAB = 60°

Therefore, ΔOAB is an equilateral triangle.
1QLiuAdYotDnY8VYntQhxj2

Area of ΔOAB =
cRzmh01wkgEFdEw d28oFEGThHehEZi36OkGpej 0iz ycoExjVifv

Area of segment ACB = Area of sector OACB – Area of ΔOAB

Q6:

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.

[Use π = 3.14 and

Answer:

yJugU1BGKCidxL8XfWKWlWyLTTsAt1VleCn5WDPnxO

Radius (r) of circle = 15 cm
D9R0udZ25tDo

Area of sector OPRQ =
JlYL4hbj19onLq6g6A nNwG0mG1OrrikjBiczLnwFIb7AnvZ4oCLSxl5HLozZFoTZ LhwID2MuqYj3czugHDtDvU1hhDamT9mLtI1bn4eXUCX1obKTUgEhyFLDj83HMWm l9tzI

In ΔOPQ,

∠OPQ = ∠OQP (As OP = OQ)

∠OPQ + ∠OQP + ∠POQ = 180° 2 ∠OPQ = 120°

∠OPQ = 60°

ΔOPQ is an equilateral triangle.

Area of ΔOPQ =
uoL2s RsF8kToMz6NPmL2UQfiKmPYQ6FEkmh44 bcRcSDHr XkEJnqcWRPiY7YjABF1wpQUvGJFXk3 O6Jm3eCB9 L1LzyjQZJdNMH sE3dsjLc7DAWCfYWF94wIbj TFAGrSXg

Area of segment PRQ = Area of sector OPRQ – Area of ΔOPQ

= 117.75 – 97.3125

= 20.4375 cm2

Area of major segment PSQ = Area of circle – Area of segment PRQ

zaYQN5Zslmv6eW67ix7wDs OcsZlXTUfmsqRITs7QJM60DevvEG4FBziaDfQRmUp

*Q7:

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.

[Use π = 3.14 and

Answer:k F1N8FOV3kFp2LxF9wFZ7mUNVpLp CSk1gLA WqjWY2YqDsOneAHPNOLIhJYTDCzZDd3mftuI4z2UKNr Y5bb1Y4FDjMTkKO

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST. SV = VT

In ΔOVS,
3Xk9fN6kdRH7CWf991MjTdQkMfaRkPbBje7cBMH2FMVb1bAHEWU6Df0DF
gEdlxxtxloaY8PMcsEPURFIj3JVlraSBQuHTKcV54mrSr8tE31HD87YnP6o1Q0SasrIAAkXHty8IZhE0I JmfnVi2NuQgNw1 dRmaMKd7 zahQ0gh8pz0uvHfC01aFqIXbb4Tms

Area of ΔOST =
sc 9VKrXA3C7eiz3bMYYP3GVFQDF7vGBESkYuHp9bSlKUbm1S0D67THn9XBG1wmyni5kkU8vs0 fVw V3zCiZNTioH7I5KfOPfQYg0qES9R3hhIbFAcoJJZjPP319hMqalu1UM
V6wvykeQyQRCOunM9AqXlVD0orbsDQRgIehRg4YmeYl2vDAgGFY7ChgSAJ3QjPvOlevif2l8Te Hy1LYis udu5CdqCrfdyQlVttphlkHXrQ1HS2h9oTyTwCWhF7M7Oz KKDuUs

Area of sector OSUT =

Area of segment SUT = Area of sector OSUT – Area of ΔOSThAIEtzSYEplyDv zAAhOVAOwOmBonfMCX4NNU7v4gD Vz25Qb4MeoJb s4cpeW2LMtNztBRyNHmP0fPgzlCmW9ppe6xoBbN2jRMTNceBXIPW Ck7ge21tsJYIgPp9sd929or0TQ

= 150.72 – 62.28  

= 88.44 cm2

Q8: 

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 the given figure). Find

  1. The area of that part of the field in which the horse can graze.
  2. The increase in the grazing area of the rope were 10 m long instead of 5 m. [Use À = 3.14]

Answer:6cV8MhcL2r7G3qbN7zxd5ESlKPuzuL2YN mTNeRnz5A0iijaXo89dwxeqarsFs2UHiv8UkRIbHjyFfWMCOR3AnaS5

From the figure, it can be observed that the horse can graze a sector of 90° in a circle of 5 m radius. Area that can be grazed by horse = Area of sector OACB

OSfTZME 5iXKXWN2o3ijY7fx7DxXE94zTTvnAI1A5Vqet

Area that can be grazed by the horse when length of rope is 10 m long

jtxgfLWTF5okf5U5Gcg32DLqz5qScKQF7p OQ8RicsZaNv1Vbv3Xn B0gc2zWlN9j9 JZPrntfGN8I

Increase in grazing area = (78.5 – 19.625) m2

= 58.875 m2

*Q9:

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.

  1.         The total length of the silver wire required.
  2.     The area of each sector of the brooch

TerQ3Ruee384n8rouRtTtHbQxeKy 2tr972b8vx02c1lD6CHDZuZVxIM4K FXVmlFH7Fp6YJZsugbi7x8ppsVrezx5hwvv3MmAZoxnRfygyYuOYSVRhE KCAS21lkrWN84I uCwzq9wKAaA Ilfx69gc1EDJGNvJxUjU8TwiJBcqEeFwFEKFicT2vzb46Q1fKiFcayAyawn2y5h9b 7eMxTBL4kvXyaHdqqQ9Q jySKUH2rmmXZcsf 5ZjIDzAlQMC kOetN1 D 1o

Answer: 

Total length of wire required will be the length of 5 diameters and the circumference of the brooch.

Radius of circle =

Circumference of brooch = 2πr

Gw7

= 110 mm

Length of wire required = 110 + 5 × 35 

= 110 + 175 = 285 mm

It can be observed from the figure that each of 10 sectors of the circle is subtending 36° at the centre of the circle.

8O jfJ9tOcPhd1Gkh6F6xCE30yKpstmauA KMHi0jYzE3HTU44bpuP0fiRZtQ6WwA6FJxVr5CHg c4t5dcxu5QnLfP8C8l4ayy9BwypSBzbHUmxdGjk9 TFf jm63zewEEMDm 4

RFKdYBC1mhUlSzfP1XLAU5lDtJm5dAbDJXgz8Eaxz25EseGCyWUDe h9c0tfJkfCTDx2w6fSfizJrbqu4YcGJUVzu3iyFwPWdtxl tv5oFZOYjqnD 82L LnXORX7rTSfq48JLc

Therefore, area of each sector =

Q10:

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.TerQ3Ruee384n8rouRtTtHbQxeKy 2tr972b8vx02c1lD6CHDZuZVxIM4K FXVmlFH7Fp6YJZsugbi7x8ppsVrezx5hwvv3MmAZoxnRfygyYuOYSVRhE KCAS21lkrWN84I uCw

RLGiNYx4yaYuZ76QxSyNH3EgRNyurmwwf TbHR7UKhPevR8tHTrqRQcciqvdBwronvKynHoHRYsQg X5M23

Answer:

There are 8 ribs in an umbrella. The area between two consecutive ribs is subtending    at the centre of the assumed flat circle.o4Ot ywngZyBlAHg8v4dIdSR E7wOTKUUXsSbMI0vA K0hNt7q56IMLAyckddZ1ww941f 0cDsoFb jnck oXp66

Area between two consecutive ribs of circle

nGLObsBIfLRVAIlg1XV2SCcxFQB4uLUClO1B oBl

Q11:

A car has two wipers which do not overlap. Each wiper has blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
TerQ3Ruee384n8rouRtTtHbQxeKy 2tr972b8vx02c1lD6CHDZuZVxIM4K FXVmlFH7Fp6YJZsugbi7x8ppsVrezx5hwvv3MmAZoxnRfygyYuOYSVRhE KCAS21lkrWN84I uCw

Answer:ULaTZQ4mKkOBWMj jEAYstXlQS9HQxHTq5zeqS8lhXjJrFJtzeoJ9wt2rcyZmVV6cpqoftK2fqS1EG6iU4R8t1kCqpKI4 IXcugs1JqABWx 82egCgR7wKc3vBv3vvCQjwA vQ8

It can be observed from the figure that each blade of wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.

Area of such sector

zV6RG 3UNqrppglq5xQWG 4SCZ0vGrPTi7xvm8lsGA ypMRsVfFSPankx1jL6lfv9W1p8hVjI1NmU4VDCBoBAG4OwqNXVfqsxOKfBQg4T9M eAuIVh5GrI62Ww7U FMFEJQHTB8

Area swept by 2 blades
xmBZndDf633TWCaAtbTH7h2gjHaQat KD2PLJfgTFKg rJvg9PehZxQxc4kmFl1lmvYOk3y DvpFRUT5rF9uNQEBY1Z7MprOdJGEOaoRUwzO5Y

*Q12:

To warn ships for underwater 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 warned. [Use π = 3.14]

Answer:

It can be observed from the figure that the lighthouse spreads light across a sector of 80° in a circle of 16.5 km radius.

Area of sector OACB

Cb vxkzFXHHNh9JG XFSZJj7yF7LXH r3nfjhW4q3tkdFkRaQYEq5 a77QcdMmD3T qmnDPWkkt8dVOmnf9yLgz SKHfgRzYuYfw5Xl vZ72cnnvoyfaLmlqH2W0rNtojIRt3j4

Q13:

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 Rs.0.35 per cm2. [Use ]

JH6R 09dcUCpgBWsI4 v6XlDGY4ZI5h5wz95019T5qwImB17e B9VIJKZntThRAOku6s0f6Yl9qk6mxRKjFqu6WH1GE5DDcermjMqHZ2fqqS4F6FfNwDSpeq13k dLlvHM5FIDw

Answer:EnWT7r8WU4U3SMsEQ ynupyOFPdAlnJPqzwkwDuwghSaUkk0MWrlyn5O8hzKePGEW 4Wao65nDRIVNN4bu gmxV68aV7OGnfYFO4aO9c3JjtONQe7K3FnYCi

It can be observed that these designs are segments of the circle.

Consider segment APB. Chord AB is a side of the hexagon. Each chord will substitute at the centre of the circle. 

In ΔOAB,

∠OAB = ∠OBA (As OA = OB)

∠AOB = 60°

∠OAB + ∠OBA + ∠AOB = 180° 2∠OAB = 180° – 60° = 120°

∠OAB = 60°

Therefore, ΔOAB is an equilateral triangle.

Area of ΔOAB =

gM6GI MxOnaVeCG1MN0ZzdQsbWlNzYWcTWY9Cgekt8dS6zNnXMNV8xc9N4VGMXK59E9tJrLsSAINN4Z9 VrHshD2R3unCImafAktfVhcYCMz Mn2VX7VmD fBtDf

Area of sector OAPB

Area of segment APB = Area of sector OAPB – Area of ΔOAB

ZpD0SOnTMaaAhoV56hm5m Qx1 aj1iyLGwX4JZrDi45RdG7tySZsuEc3iZMEix iCnLbJxMmRjVbPlm7Ho8vxlR8ib4GUndEysxJ9S4cr0mhAn2cX4c7db upY56Q4Oe2EtIQI
T70LeR8Iz5oft3BwCVwBnIXHfuJyvReUS81YoebrAFZMcl tGPbW aA XQc2at4EGibPz8oNRsh3l8QTSd6qx9YBiBhpCihW0w A6y

Cost of making 1 cm2 designs = Rs 0.35

Cost of making 464.76 cm2 designs = Rs 162.68 Therefore, the cost of making such designs is Rs 162.68.

Q14:

Tick the correct answer in the following:

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

(A)         (B)         (C)         (D)

Answer:RmW ervpyQzbeOnt4wOM2aUNSDTBnF5pIjadWfpX4xrMfagyGl0TPTVcGhYEWrTaM3uO 2VczvObOwZv0QHHfrVNYbsvBEreQVItUu2 rSZzHgfZJedlACqgz4Ub8cX8khhBuTo

We know that area of sector of angle

Area of sector of angle

GW4hO99iuKTl3uRqViGMd7hs l0OYwMwzD6

Hence, (D) is the correct answer.

Q15:

Tick the correct answer in the following:

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

(A)         (B)         (C)         (D)

RmW ervpyQzbeOnt4wOM2aUNSDTBnF5pIjadWfpX4xrMfagyGl0TPTVcGhYEWrTaM3uO 2VczvObOwZv0QHHfrVNYbsvBEreQVItUu2 rSZzHgfZJedlACqgz4Ub8cX8khhBuTo

Answer:

We know that area of sector of angle

Area of sector of angle

Hence, (D) is the correct answer.

Exercise 12.3 

Q1:

Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°. 9E VEov1dfU4effkjKKjxQmeJEbwsc6gbjbXwo JnJ8OysWo1LJ3K7bHJV1GLOwJku3up1juK7R27F2s7MxpmxiiugApHLoge saS0LTwec2 OX7a6Qc1Q7Nlv oUA3tLtBcc

Answer:

Q9F3AYmzyidWnMBVIJ HYeHPm AHrlKuboh WP4dp3vk0Gpnm8xh 8NcmnQ2rPGusbviXi9RZ298dz26aJkXWDKR7rl6L LPN4DLejmHpeYRG 7IkGGn7 I2Rb4cWh N380LNpc

Radius of inner circle = 7 cm Radius of outer circle = 14 cm 

Area of shaded region = Area of sector OAFC – Area of sector OBED

=40°360°×π(14)2 – 40° 

*Q2:

Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

qiAOg8fTmiUN3g1BOJnn9dgbPHtN4YwwANgj nmBV6DleK22lIHG3 IOy7WfbbzDHQj 8ZWKuD BMwqKE gyQqUXBmwFbuHvYfjKe4a4BFVGk5TNPbJt6bUdX48 uSpnIsHulas

Answer:

It can be observed from the figure that the radius of each semicircle is 7 cm.

Area of each semi-circlexmvcfZ3hIj4URazzapOLqK9a8 Xv wbJLUqnHjn9PJfO1Q

o7sI8gRbJ5MAMSHxtSyag3

Area of square ABCD = (Side)2 = (14)2 = 196 cm2 Area of the shaded region 

= Area of square ABCD – Area of semi-circle APD – Area of semi-circle BPC

= 196 – 77 – 77 = 196 – 154 = 42 cm

Q3:    Find the area of the shaded region in the given 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. 

6ZchPx 3YWB8zdGLC8m7lTd1PCrAdpK rCdSMFBhqKw3qXAL 7FNciv TWQMErsua1PaetD06UwJBHu9
QpYAA

Answer:

We know that each interior angle of an equilateral triangle is of measure 60°.K3FHdno6I67sinMjNCWekjRta0sVN40cr1DI0UVYRIZcXXfgU7UJb2TGvtv9VsJAnj2VQwYKahkO4l0GDFkQrl7vyXjbyll1SdpX29qrGSpXIu88IO38SsZINiJ8hzdI nt5EDc

Area of sector OCDE

fCyURPmZJKLQdbSB4e7ZNUf fKICNN331KrPn7mmJCmtdASxvRZGBxxEqW11PPLtx27j9AhuLQoGX8fSjBB3k45rI EvoBKNmMuj2458rPUziCsJr26ZSfarddh1DfsYpWR7iA

Area of 

Area of circle

Area of shaded region = Area of ΔOAB + Area of circle – Area of sector OCDE

M0rZA0VGKEYEnpPX4kli0NP95F 8yImWmW9SIaojvU 6YPyc3pVXJ zh2zJsK4s0EcwM8VoEQTgyrDXrf2eTmDz OJmBc Uy MzYfU9BV3QcWtvRK5OeVFPZCLNTBhN6ybwlbT8

*Q4:

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 the given figure. Find the area of the remaining portion of the

square.

isWEHTvkMAbCHwuR3IV7 7bh0H8IapC2EwJ6jLNczr95GCxrEY75t0nzBMxqRxSrbuEm8ZrVrk7 f4 frjp3OMAy1pOs767Zuz2F 4ccj Z504sAwEq6vWWQjmo0ErmSL0oEFoc

Answer:
TT8Fi0ARJND68UGwpgR nS3OE2zT1YyegzqCjcWyKJPmbM0 5FtsiHMgOruwm79xlhloe2ubasoq1VFt JPLCkN3GIpvTEJd1NAwRESApUkhZHHw 7w aKpFMQ F AF64Pa nQ

Each quadrant is a sector of 90° in a circle of 1 cm radius. 

Area of each quadrant

Area of square = (Side)2 = (4)2 = 16 cm2 Area of circle = πr2 = π (1)2

RfziU5a Xj5PVhwqBuF8oHrTynVScNBxylOk8voD4y8C1M2qOLI8YNhGlMt4KnKtJrjQpvzWrziMms1hAuzJDkvNSY0fC9KcTf8KYIucvCkf3rdNvVD0ygtYAn68mFQxR Aycg

Area of the shaded region = Area of square – Area of circle – 4 × Area of quadrantaTic1BID5egOwV0r27OHNo5j3 1rJk oVVj7 JL PIetmUBbyHANQs qOdfl

Q5:

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the given figure. Find the area of the design (Shaded region).

Answer:X2GV3B60KbU6sgX1jvXQZv11IRuRCt rNEwfwTyWODaXfCPp0Dm6C 7hmRgo7Utkrm9jC3GOXwF6oiXEz8f97JQhamico UaGOXhjRg Q9NgKn tcUYiCG8Mg IX6IvU9ZdsY4

28wHFmQwNV2H7U2 SaFyGc7TPB8ggGYjqMcHwjxi5H6SXBMo

Radius (r) of circle = 32 cm AD is the median of 0B3jxS7TsfJZu VOViuiZ 8GsgRWxf8MbFXHBLwiBKx2w EkkRuMp3MxdYjYlueebDiHkUPkoQwau It0LWxyS9eRj62MAIABC.

AD = 48 cm 

In ΔABD,

AB2 = AD2 + BD210T515IW
o0u33liuSvL98G P8b z

Area of equilateral triangle,

hvzr9moxtw65m6TV8WudMgplLGOXn8WXgXBZD8irpBPddxya2lZvWq61xeUp5AgDDnaKbRdFFVaIaJQMEPUgwNrsh6hR2kp0TeXFk4jiF6F1rHqWx2sVSpBEfOO 6SYifY2W80Q

Area of circle = πr2QwihQFJaoEVHnR3PunCz9ECekFMyxMMswav1M7o6KS3nue8qXcVJIqMIh4 K yZy3toZL Gvh6oI6v0P6M7Y9Ok4p6 QsFnf50zm Ct 2JCmkNtX0zRfpwXxNOhVWVYV7vP 5jM

Area of design = Area of circle – Area of ΔABC

*Q6:

In the given figure, ABCD is a square of side 14 cm. With centres 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. WAartZleA EGHQfjkDVz0HpmoTAVL8TIla0pMP2VwkyIvZGjCLfLw2tziZgipe2eoE7rXfSouHZRPJ7YV6Zu9AkEWDLR ybUpdHwFUkXFk5MKuDicr2giq9WGGP dok87CSPsSE

Answer:

Z0OJ7zDQg4KqsdbSlFVYlxTf4frjwbA3t18SPsss1O7iagGcthkr 0ZjESxA8sdL70 MZYIfEO KZMDDKFmPWyBheIczLYfp5i hB9YOc4 ZO9KcyNFO5PJtNhKkhcMu6hU4lDM

Area of each of the 4 sectors is equal to each other and is a sector of 90° in a circle of 7 cm radius.

Area of each sectorOuhgF NDyCUywiEp tXpi88JPydHwv3KmLeUr IUvfsCynrdIxz4O0swV 70FlSYKm xcevZ8nQchY1es6mDDrDdsLYXL1mHwiR0Xa3 8qSCLYzVpQ9O jyYSbdg3wUIV5bV t0

Area of square ABCD = (Side)2 = (14)2 = 196 cm

Area of shaded portion = Area of square ABCD – 4 × Area of each sectorfOYuPJvJoezSDkWEDwUXT1aTFJS2Wtr3vImgAtk5hpEr08riPC8xAwlkiepiGfu4JZ SarO8CCg 7o6CKzj46pMJfWoLtOy6jViBhPD0LkQmFGaGypT

Therefore, the area of the shaded portion is 42 cm2.

Q7:

The given figure depicts a racing track whose left and right ends are semicircular.kchLDPnWmVkKqvriVwD0bPz6qBqemLhqeg4 Q1uAHuaezlybWLAXQ2NtIABAoG6A GOxrFNO0GATZwbu1reWWqE4Nk 668CxWgY1kzz4

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:

  1.         The distance around the track along its inner edge
  2.     The area of the track
1cPpW8lRdz76fM vemJ9nP WWKEFHq2SUgI 1jL mXdusfrWUzd7AMIJb sMTXsoOeseKwgLp

Answer: 
qSFXNdDIruCFC2Ot4GJkpkunnrR6EnPFXH4hV3xXZYU2dyF3edTL1WuoqZmddc3Ys6RmlQXJkuiFR9mNiceHx4HRn2Eb7oI JUwRJ1Ilr igTkHIrzwubPsP OO4zkeSaQvqvv0

Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA

kuOJnjLnsUDijFnaZ9 nwLj0XN3oMsGAh1uCmmnWMk1xFg6kAZCSKlB1I9Z3Xk679xmAcvc8x

Area of the track = (Area of GHIJ – Area of ABCD) + (Area of semi-circle HKI – Area of semi-circle BEC) + (Area of semi-circle GLJ – Area of semi-circle

AFD)

Therefore, the area of the track is 4320 m2.

*Q8:

In the given 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.  

OLSPwDsOdrGLnXwnu3ti5m0ELJWVf9pvY2ME0vyvS4E8k Yia4n1469GZhZs2yrNJWw3RKX0WKc F288TfWPDj2VgZydqjJ Ng9GNW6FnTsZxbRX6m74Cb6Glj9d2 p9YTK7te4

Answer:YxU5oxAEigDL97g5jJ3lUGuBOxIZc7m MxnY9lwX2MeTE4py4jRX8P1v LgkItMeKCixbtW9V2xwSVGJR6JDMtXvnaW4xySCK4Sqmo6C

Radius (r1) of larger circle = 7 cm

Radius (r2) of smaller circle  nggy0GsMz0xRUnFdGpwBy93h yGd9HDfNHVav1fXODljMK RmH6e5nWQVVcdJJvRIzRsZ FEiiRG5B5TsJUA5aE8KuY3FuP8MY3eGgYuC K4milrAiBgnehqwHyNOTzQb5yadKY

Area of smaller circle

Area of semicircle AECFB of larger circle

o7sI8gRbJ5MAMSHxtSyag3

7BRqkH c C KP5 utBjgyUeAAIU4xpChsXBFib7 AgtLco30MzjT58pKBsbm0oJiS58Puag1LCUBWOMqm9z hFFAn9LkjW0Tnd7JdlJ El36ERC 73aFJLox7 tstKBhqwJpe3c

Area ofDc2cUxyLzOYGvJjCBvwD26fOek9BSkz169oX8IOReg2teq gZz8iEsS9KzNsIzAY7ItAOdY5i9C1gMqw0qn omnAcoI0vX pVZI3UWojhgmX8G6iUf 7RjdE9c3bK3NwnUKkOV0

Area of the shaded region

= Area of smaller circle + Area of semi-circle AECFB – Area of ΔABCnVOOu8FKNLdx0prrRWGdtbE996eB0ZI8La vPOttnnJFGT6HITSK Eg2AT97So Mq53u UQovk3jeAQbKKSDl0mtzh3qrNWUwZrRfuGmKqK4XeECL5IPXd72WIsKXeCWRwBRhKQ

*Q9:

The area of an equilateral triangle ABC 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 the given figure). Find the area of shaded region. [Use π = 3.14 and

APcYNEr5PJLAskO2Si1FQPzkeALQtoCF85FAWng1ewAHs0TgmsddADYg4jHtPcoGbPMOPszh7N0pu2Kiwm2zKM9K1S6tz6d7cpy0A3gR3KL3C0OQLJGNkyndbAPWOpyiZD4bcTs

Answer:

Let the side of the equilateral triangle be a. 

Area of equilateral triangle = 17320.5 cm2

EUpDYt6vWnKUPu7HYZ wgdyU0AUrYHh gts9J5OU UX8PHfe4qSrTyIM830f4YOEW7c19TInIpZFI 4N6fW9Hz0JaqEQrmBEPKNKvMBkcgaHLdM4dIfTJ9O4Wiwc0dFURkd9s s

Each sector is of measure 60°.

Area of sector ADEF 4exJ2 79QPhOhLabx

Area of shaded region = Area of equilateral triangle – 3 × Area of each sector

5SKnY sczZxsDa2gpiXr2YEATMUSHRMYMcs9sjA lXcsSygmuHTpVj7jiXn6efJZME51KukR15dUnkYlPWHjbfVPPQRzGf0GlLtniuEyEUNQkIoUd1WR6BDQYlQN 7WNYKvQwAw

Q10:

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

Q6onKRIFqQ3Bh2GnWdSRCGNWST4OOTpya8hpWVJ mIT8VF50VzB PVeaaU2gysS3 Xinid0PSe0BPZCJnM0hav38ylkYGzd CTl7RxX33Tz6M bofUlkFeGyuM J K8O3Se6L00

Answer:
0Gq8yFFNGehRMmSoblPWkrEqnCUuHw4qMlkeZCTOapW00r2tJvf2noGbZTcE7i0oz43ksF4JgpdgdvdWudLV2XuPlGI7MpxOPbqwI PlhBwMjmFH9f vZy17ENdvHZs9VROkaaE

From the figure, it can be observed that the side of the square is 42 cm. 

Area of square = (Side)2 = (42)2 = 1764 cm

Area of each circle = πr2

Area of 9 circles = 9 × 154 = 1386 cm2

Area of the remaining portion of the handkerchief = 1764 – 1386 

= 378 cm2

Q11:

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

  1. Quadrant OACB
  2. Shaded region6ZchPx 3YWB8zdGLC8m7lTd1PCrAdpK rCdSMFBhqKw3qXAL 7FNciv TWQMErsua1PaetD06UwJBHu9
N56a82sE79rL0g554shTwzM LIKvF4tSKeMSc kme0QTDvYZe5Ar1u1PZGwcu7VRt8ZFX d D6wY03B8EV5CDgu72qVMaLKHd d79kVsniRVoOdYijGMa6fMsI6wn84FieI u3Y

Answer:ncm5n0GqizD8Cf0ROe7i9wt7KSUodxNU4zVch wlWCTVZ3sdUxqzjFjRbHs JSED6FcY7QrKN zd0WBgRhMBrZ9YJmCGJmyYAfuW QNAVCCdpE

  1. Since OACB is a quadrant, it will subtend 90° angle at O.

    Area of quadrant OACB EVdJWgfeJbGH9aA1cQ5ROv w PJUmeEji Gd2gwKkEgFiO7rqnxeixgNasL2BLnIgPk4wR7aUkBGpoONBv9MzbQwLxMunzdk30hv8PXRUkQbxBZEK 7q MXF4muLCgveLNroUVA

  1. Area of ΔOBD
4mIgqWZw163 CVB1nuPDnd9t MQ0b45F4MIhcuJ8nRXElRGIdRrRdgBUUua71YNMLACERO42QA 1 JtdAEG 20KPLitqZhknSAP3m6EEewl3kbjXy7VG4lrZewLZKAguf3VKjng

Area of the shaded region = Area of quadrant OACB – Area of ΔOBDl8hNIa4sSCO4b59aCrPubEZyKRBIhCWH4lbBARsoECcFru3jNwMuM3a8ka Sz7hli1VQGNKoliOVlru5SBEz289THTFJBt1E 9oOyIn7JQt weBVw3J6CNbmuG14eXX6OMDt9gU

Q12:

In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. [Use tPYhIDJOcpzj53hRn080wpZsyxs0Lljgo3mZhFJBI4dp 0Uh9Ictuead9 sSRSj9Fuwc41JxdD wKjoiVqzWof 3pJU6hqtBVXEQspejk553zP9JRn3OusPDW1KF4QGE5 q3c3k3.14]BLS227ZTtIrEKrzSTFDs7sefJM2iKiv6WCevf3XZOrFXEaeoJekaqt0FkoxWlNOWx3yG2n5w9cbvJ0EUqxF7UocDSFZR4y gvR4FwHFWEVIPe I0lRMb9 yHIyDa7vsTbXXcKAQ

Answer:

puPxfljainj2SGkIzUrOt Vr4tSb2tc8gHwhX11euNorg0B3uL0S BOeWwwDGvyrRtiuzd

In ΔOAB,

OB2 = OA2 + AB2

= (20)2 + (20)2

Radius (r) of circle

Area of quadrant OPBQ

Area of OABC = (Side)2 = (20)2 = 400 cm

Area of shaded region = Area of quadrant OPBQ – Area of OABC

= (628 – 400) cm

= 228 cm2

Q13:

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see the given figure). If ∠AOB = 30°, find the area of the shaded region. 6ZchPx 3YWB8zdGLC8m7lTd1PCrAdpK rCdSMFBhqKw3qXAL 7FNciv TWQMErsua1PaetD06UwJBHu9

WQarU5KE2AOObrCxp9jEwns

Answer:jlhz UswUfJughHkqRUB8dTbMowi4mcQea7M0Nrh1L s8l15KndTZb3j8Ss4tnhW NTKlo2I kJ9O9WspMQfyA7NDt7F9Gcwv2tNXei1gDYtk0pEhlNV4hHe9yRcRkzLfDnklmQ

Area of the shaded region = Area of sector OAEB – Area of sector OCFDKMUB5jN0MsksFPcKCi3ngaUoFKm04vwiv3GQ3FN0pXrHnV7DV94Twm rUNhq QsPWIm96x0IU4PH3S9bImkOlT RjNTVX 5XvKORIISzbqNxmGGWa4ky6AOIzW54NUfpef iBX4

Q14:

In the given 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. yfcR k ySFr4LvxTNMMO3f xW FrTrCICGOyRDsSJ 5s2 G6GOR6axYfZOFt 4Miq15oazOhThbqH

Answer:

pHzXMy bfq82rp6mzSU2quGBmMbfl00NL7OZgN36SzYv R2TWG0 lognGGX9mA4duaMs9yWiHk Gz0b2BEyEMNXzF9kahQb8z5AEq Eqe7ypK5KntYrDUcI3IagN lskNws3R8Y

As ABC is a quadrant of the circle, ∠BAC will be of measure 90 º.

 In ΔABC,

BC2 = AC2 + AB

= (14)2 + (14)2yq GogPyNV1KmYplgds4xVzncLDkyXSLEC FGE fLDVWfpfaoeZ2iDeVqV6p5tWXaeZTgKDB2ky jyLn2TUbbKCkmHXZ55X9GsiUDSyw6shvWkdd96TQmaBDRTD3L4kkeEF5UII

Radius (r1) of semi-circle drawn on 

Area of v2PJqplLz8z6SYsr6usuFT5GOitcFMA4dxItNInq2uZIBEY

uni CN9EU7Ul4s1KHEelMte 6S5cL3 95bbTxLtDP01WK1rmURms6Tz 6wNnl zLm r9lLAfYJnzenbzZvqmeLhsDN1sTbpUydPq5Sf5FnpTqZBaaTRcOWCBNJkcO85WCtdfglY
x3 vz hZFPTxMA1VFHt3K3R93iS5L8rJEgUDZlVL1CYPFdsbC0DfPj0fo K3D7q lSx eHI1lz1mgjJC1Kv1DlWMNfjAq0 fVW6gD3JmH34pCLUz uhlB5dZ1NvUup0YCpfyQ1s

Area of sectork8DVuqeUGhql7z JrSQPczzD7VIKDSEndryf49lKLCkULmZPxMb12f9iLI3MdyDmHOOzIDGvEHxq4TUB QeRMIbUV1hxrqNkDq7v1FTuQUFErt8qVZ mL9rutrMCYSma5CXuKck

ylPaOzuRIHIWfA5GXs0PUBlf3SVTLh1RW5FkjJJEXt67V0MQSuZN2EqNtDjk3GY0uOt4POjYCyNVER6P2w eGZma0C24JsNadeFYXk eDNvr0bcmnJJIKu6955 wSeeyeV0NsHQ

= 154 – (154vgV54dujTplCVqSDUkhaLBFYuaREw AT6N04G7fJLmUVwBUEpUTrNoKhndbi 1XZJyRon s4O vTS8FV44f5UJ0Dkd yBiRtG9rEXStiXrBf4k4O VoeG1pGPyShDkustNY9FHg

= 98 cm2

*Q15:

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

1cPpW8lRdz76fM vemJ9nP WWKEFHq2SUgI 1jL mXdusfrWUzd7AMIJb sMTXsoOeseKwgLp

6QpcsYLrVPHnhcWEGPPjR2ZRUn uOl7gVEZ1i XcYA VuMBfIp56x2BzI7l5ghdzIJ efmXEoYzYuMx7yuxfD T0A1M8rE2 o MQMGzVWD4Afrqr GSXl5jqJpjwf8cHTF7v04s

Answer:

dxUY8P lHS5aT Dlr5HS22QhhgpHkSqCw KWWrE0

The designed area is the common region between two sectors BAEC and DAFC. ymnyBJR SFcuMF5gGqYMR8USncyyyYapt2jlRTqfFuT9iUGNMeN5NOQ5w1hJYpbqKyGKnXByPaRUqOITziARnoIgv24epw0Aw63tATm1CB12MqtVBKf gACdj6omqzUkjuj9pRg

Area of sector

Area of ΔBAC
CcYqxnDb5V2crQl9hClEB 6ESQ5D9HFlIzydNBpCPuOhqwrnePQpKS5gDTSHS2j nVAKG4XNrSUPNzjFiVhUC2WgCTjfF7ybqDu13uQB6YgBfvERJewsl JfzaxnmCBXut5 vE

Area of the designed portion = 2 × (Area of segment AEC)

= 2 × (Area of sector BAEC – Area of ΔBAC)

YlCK pLqOmQ8VSt91zJ2T0mbjEmrN4bGEzJAIjclvWQvecNoZA8h1ggtONpXZsua e8BImMfZpu3W3bY6QUx6Jt3GAxycnajLB8cAaQqXTLO1K L3PckuGdim cjiZ3AfpdJCnE

Q16:

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

6QpcsYLrVPHnhcWEGPPjR2ZRUn uOl7gVEZ1i XcYA VuMBfIp56x2BzI7l5ghdzIJ efmXEoYzYuMx7yuxfD T0A1M8rE2 o MQMGzVWD4Afrqr GSXl5jqJpjwf8cHTF7v04s

Answer:dxUY8P lHS5aT Dlr5HS22QhhgpHkSqCw KWWrE0

The designed area is the common region between two sectors BAEC and DAFC. 

Area of sector

ymnyBJR SFcuMF5gGqYMR8USncyyyYapt2jlRTqfFuT9iUGNMeN5NOQ5w1hJYpbqKyGKnXByPaRUqOITziARnoIgv24epw0Aw63tATm1CB12MqtVBKf gACdj6omqzUkjuj9pRg

    Area of ΔBAC

Area of the designed portion = 2 × (Area of segment AEC)CcYqxnDb5V2crQl9hClEB 6ESQ5D9HFlIzydNBpCPuOhqwrnePQpKS5gDTSHS2j nVAKG4XNrSUPNzjFiVhUC2WgCTjfF7ybqDu13uQB6YgBfvERJewsl JfzaxnmCBXut5 vE

= 2 × (Area of sector BAEC – Area of ΔBAC)

=2×3527-32cm2

=2567cm2

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

The 12th Chapter of NCERT Solutions for Class 10 Maths covers the concepts of perimeter (circumference) and area of a circle and applies this knowledge in finding the areas of two special ‘parts’ of a circular region known as sector and segment.

Areas Related to Circles is a part of Mensuration, and this Unit holds a total weightage of 10 marks in the final exam. In the final examination, one question is sometimes asked from this chapter.

List of Exercises in class 10 Maths Chapter 12

Exercise 12.1 Solutions (5 Solved Questions)

Exercise 12.2 Solutions (14 Solved Questions)

Exercise 12.3 Solutions (16 Solved Questions)

NCERT solutions for Class 10 Maths Chapter 12 is about parts of circle, their measurements and areas of plane figures. SWC’s subject experts have prepared solutions for each question adhering to the CBSE syllabus.

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

ExerciseTopic
12.1Introduction
12.2Perimeter and Area of a Circle
12.3Areas of Sector and Segment of a Circle
12.4Areas of Combinations of Plane Figures
12.5Summary

Key Features of NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles

  • NCERT Solutions help students strengthen their concepts on circle related areas.
  • Questions are explained using diagrams which make learning more interactive and comprehensive.
  • Easy and understandable language used in NCERT solutions.
  • Step by step solving approach helps students to clear their basics.
  • Helps students to solve complex problems at their own pace.

Conclusion

The NCERT Solutions for Class 10 Maths Chapter 12, Areas Related to Circles, provides a comprehensive and detailed explanation of the fundamental concepts of circles and their applications in calculating the areas related to circles. The chapter covers various topics, such as the formulae for calculating the circumference and area of a circle, the area of a sector, the area of a segment, and the length of an arc.
Experts have carefully crafted these solutions to provide step-by-step explanations of all the questions in the NCERT textbook. The solutions also offer helpful tips and tricks to help students solve problems quickly and accurately.
Using these solutions, students can strengthen their understanding of the subject, improve their problem-solving skills, and prepare effectively for the Class 10 board exams. These solutions are an excellent resource for students who wish to excel in mathematics and build a strong foundation for higher studies.
Overall, the NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles are indispensable for students looking to score well in mathematics and enhance their analytical and mathematical abilities.

What is the importance of studying NCERT Solutions for Class 10 Maths Chapter 12?

The solutions of NCERT Class 10 Maths Chapter 12 helps students in procuring a summary of the question paper pattern, including a variety of questions, such as mostly repeating questions, concise and long answer type questions, multiple-choice questions, marks etc. The numerous you solve, the more assurance you will get towards your success.

Mention concepts that are important from an exam perspective in NCERT Solutions for Class 10 Maths Chapter 12?

The important from an exam perspective in NCERT Solutions for Class 10 Maths Chapter 12 Introduction of area of circle, perimeter and area of a circle, areas of sector and segment of a circle, areas of combinations of plane figures and lastly explains the summary of the whole chapter.

How many exercises are there in NCERT Solutions for Class 10 Maths Chapter 12?

There are 3 exercises in NCERT Solutions for Class 10 Maths Chapter 12. First exercise has 5 questions, second exercise has 14 exercises and last and final exercise has 16 questions based on perimeter and area of a circle, areas of sector and segment of a circle and areas of combinations of plane figure.


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