Areas Related to Circles

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Area related to Circle Class 10 NCERT Solutions Maths Chapter 12 are made strictly in accordance with the CBSE Curriculum and the exam pattern. NCERT Solutions for Areas related to Circles Class 10 are curated by our team of subject experts at Toppr in a detailed manner. The NCERT textbook questions are answered in a way to provide you with a better understanding of the concepts in a systematic and step-by-step manner. It also contains the appropriate diagrams to explain the concepts to the students. As Class 10 exams are Board exams, these solutions will not only help the students in preparing for the board exams but also for the Olympiads. NCERT Solutions provided by Toppr are the best study material to excel in the exams. Also, the MCQs and long and short questions are all answered according to the weightage and the exam pattern. With the help of NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles you can also test your subject knowledge and analyze your shortcomings and work on them before the exams. These are the best resources designed after proper study and research and study to help the students in scoring good marks.

Table of Content

Exercise 12.1

Question 1

The radii of two circles are $19cm$ and $9cm$ respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.

Solution

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Let the radius of required circle $=rcm$

Radius of 1st circle $r_{1}=9cm$

Radius of 2nd circle $r_{2}=19cm$

As per the question

Circumference of the required circle $=$ Sum of circumference of two circles

Circumference of small circle $=2πr_{1}$

$=2π×9$

$=18π$

Circumference of small circle $=2πr_{2}$

$=2π×19$

$=38π$

Now,

Circumference of the required circle $=$ Sum of circumference of two circles

$2πr=18π+38π$

$2πr=58π$

$r=2π56π $

$r=28cm$

Hence, radius of new circle is 28 cm.

Question 2

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

Solution

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Let the radius of required circle is $r$ cm.

So, $πr_{2}=πr_{1}+πr_{2}$

$r_{2}=r_{1}+r_{2}$

$r_{2}=(8)_{2}+(6)_{2}$

$r_{2}=100$

$r=10cm$

So, $πr_{2}=πr_{1}+πr_{2}$

$r_{2}=r_{1}+r_{2}$

$r_{2}=(8)_{2}+(6)_{2}$

$r_{2}=100$

$r=10cm$

Question 3

Fif. 12.3 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.

Solution

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SOLUTION:

GIVEN:

Diameter of Gold region= 21 cm

$Radius of gold region=$ 21/2= 10.5 cm

Area of Gold Region = πr²

= π(10.5)² =( 22/7)× 110.25 = 346.5 cm²

$Area of Gold Region= 345.5$ cm²

$Radius of red region =$ Radius for gold + red region= 10.5 + 10.5= 21 cm

Area of Red Region = π²(21² - 10.5²)

[Area of a ring= π (R²-r²), where R= radius of outer ring & r= radius of inner ring]

= 22/7 (21² – 10.5²) [ a²-b²= (a+b)(a-b)]

= 22/7 (21 + 10.5)(21 – 10.5)

= (22/7 )x 31.5 x 10.5 = 1039.5 cm²

$Area of Red Region =$ 1039.5 cm²

$Radius of blue region =$ Radius of blue region = Now radius for gold + red+ blue region= 21+10.5= 31.5 cm

Area of Blue Region = π(31.5² – 21²)

= 22/7 (31.5² - 21²)

= 22/7 (31.5 +21)(31.5 - 21)

= (22/7 )x 52.5 x 10.5 = 1732.5 cm²

$Area of Blue Region =1732.5$ cm²

Now,

$Radius of black region=$ radius for gold + red+ blue + black region= 31.5+10.5= 42 cm

$Radius of black region=$ radius for gold + red+ blue + black region= 31.5+10.5= 42 cm

Area of Black Region = π(42² – 31.5²)

= 22/7 (42²-31.5² )

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

= (22/7 )x 73.5 x 10.5 = 2425.5 cm²

$Area of Black Region =2425.5$ cm²

Now

$Radius of white region=$ radius for gold + red+ blue + black+ white region= 42+10.5= 52.5 cm

$Radius of white region=$ radius for gold + red+ blue + black+ white region= 42+10.5= 52.5 cm

Area of White Region= π(52.5² – 42²)

= 22/7 (52.5²-42² )

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

= (22/7 )x 94.5 x 10.5 = 3118.5 cm²

$Area of white Region =3118.5$ cm²

Exercise 12.2

Question 1

Find the area of a sector of a circle with radius $6$cm if angle of the sector is $60_{o}$.

Solution

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Area of sector$=360_{o}θ ×πr_{2}$ where $θ=60_{o}$

$r=6$cm

$⇒360_{o}60_{o} ×π×6×6=61 ×722 ×36=722 ×6$

$=7132 cm_{2}$.

Question 2

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

Solution

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$wehaveC=22cm⇒2πr=22cm⇒2×722 ×r=22∴r=3.5cmNow,areaofquadrantofacircle=4πr_{2} =722 ×4(3.5)_{2} =2822×12.25 =9.625cm_{2}. $

Question 3

The length of the minute hand of a clock is 14cm. find the area swept by the minute hand in 15 min.

Solution

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Minute hand completes full circle degree in $60$ minutes.

Angle swept by minute in $60$ minutes $=360°$

Angle swept by the minute hand in $15$ minutes $=60360° ×15=90°$

Therefore,

$θ=90°$

Length of minute hand $=r=14cm$

Area swept by minute hand in $15$ minutes $=$ Area of sector

As we know that area of sector is given as-

$A=360°θ ×πr_{2}$

Therefor,

Area swept by the minute hand in $15$ minutes $=360°90° ×(722 ×(14)_{2})=41 ×616=154cm_{2}$

Hence the area swept by the minute hand in $15$ minutes is $154cm_{2}$.

Hence the correct answer is $154cm_{2}$.

Question 5

In a circle of radius $21cm$, an arc subtends and 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

(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

Solution

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Question 6

A chord of a circle of radius $15cm$ 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 $3 =1.73$)

(Use $π=3.14$ and $3 =1.73$)

Solution

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In the mentioned figure,

$O$ is the centre of circle,

$AB$ is a chord

$AXB$ is a major arc,

$OA=OB=$ radius $=$ $15$ cm

Arc $AXB$ subtends an angle $60_{o}$ at $O.$

Area of sector $AOB=36060 ×π×r_{2}$

$=117.75cm_{2}$

Area of minor segment (Area of Shaded region) $=$ Area of sector $AOB−$ Area of $△$ $AOB$

By trigonometry,

$AC=15sin30$

$OC=15cos30$

And, $AB=2AC$

$∴$ $OC=15cos30=1523 =15×21.73 =12.975$ cm

$=(3.14×15×15)−20.4375$

$=686.0625cm_{2}$

$O$ is the centre of circle,

$AB$ is a chord

$AXB$ is a major arc,

$OA=OB=$ radius $=$ $15$ cm

Arc $AXB$ subtends an angle $60_{o}$ at $O.$

Area of sector $AOB=36060 ×π×r_{2}$

$=36060 ×3.14×(15)_{2}$

$=117.75cm_{2}$

Area of minor segment (Area of Shaded region) $=$ Area of sector $AOB−$ Area of $△$ $AOB$

By trigonometry,

$AC=15sin30$

$OC=15cos30$

And, $AB=2AC$

$∴$ $AB=2×15sin30=15$ cm

$∴$ $OC=15cos30=1523 =15×21.73 =12.975$ cm

$∴$ Area of $△AOB=0.5×15×12.975=97.3125cm_{2}$

$∴$ Area of minor segment (Area of Shaded region) $=117.75−97.3125=20.4375$ $cm_{2}$

Area of major segment $=$ Area of circle $−$ Area of minor segment

$∴$ Area of minor segment (Area of Shaded region) $=117.75−97.3125=20.4375$ $cm_{2}$

Area of major segment $=$ Area of circle $−$ Area of minor segment

$=(3.14×15×15)−20.4375$

$=686.0625cm_{2}$

Question 7

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 $3 =1.73$

Solution

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Given that:-

Radius of circle $(r)=12cm$

$θ=120°$

To find:-

Area of segment $APB=?$

Solution:-

Area of sector $OAPB(A_{1})=360°θ ×πr_{2}$

$⇒A_{1}=360120 ×3.14×12_{2}=150.72cm_{2}$

Let M be the point on $AB$ such that $AB⊥OM$

$∴∠OMA=∠OMB=90°$

Now in $△OMA$ and $△OMB$,

$∠OMA=∠OMB[each90°]$

$OA=OB[∵OA=OB=r]$

$OM=OM[common]$

By R.H.S. congruency,

$△OMA≅△OMB$

Now by CPCT,

$∠AOM=∠BOM$

$AM=BM$

Therefore,

$∠AOM=∠BOM=21 ∠AOB=60°$

$AM=BM=21 AB.....(1)$

Now in right angled $△AOM$

$sin60=OAAM [∵sinθ=hypotenusePerpndicular ]$

$⇒sin60=12AM $

$⇒AM=63 cm$

$cos60=OAOM [∵cosθ=hypotenusebase ]$

$⇒cos60=12OM $

$⇒OM=6cm$

Now, from $eq_{n}(1)$, we have

$AB=2AM=123 cm$

Now, area of $△AOB(A_{2})$ will be-

$A_{2}=21 ×AB×OM=21 ×123 ×6=363 cm_{2}=62.28cm$

$∴$ Area of segment $APB=A_{1}−A_{2}=150.72−62.28=88.44cm_{2}$

Hence the aea of the corresponding segment of the circle is $88.44cm_{2}$.

Question 8

A horse is tied to a peg at one corner of a square shaped grass field of side $15m$ by means of a $5m$ long rope. 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 $10m$ long instead of $5m$. (Use $π=3.14$)

(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 $10m$ long instead of $5m$. (Use $π=3.14$)

Solution

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Side of square$=15m$

Length of rope$=5m=$radius

The area available for horse to graze is nothing but "Area of Quadrant of a circle'

$∴$ Area of Quadrant $=4π×r_{2} $$=43.14×5×5 $

$=19.625m_{2}$

If the length of rope is increased to $10m$ then the new radius ,$=10m$

$∴$ Area of new quadrant $=43.14×10×10 $

$=78.5m_{2}$

$∴$ Increase in grazing area $=78.5−19.625=58.875m_{2}$

Length of rope$=5m=$radius

The area available for horse to graze is nothing but "Area of Quadrant of a circle'

$∴$ Area of Quadrant $=4π×r_{2} $$=43.14×5×5 $

$=19.625m_{2}$

If the length of rope is increased to $10m$ then the new radius ,$=10m$

$∴$ Area of new quadrant $=43.14×10×10 $

$=78.5m_{2}$

$∴$ Increase in grazing area $=78.5−19.625=58.875m_{2}$

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Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability

Question 1. What do you understand by a sector with reference to a circle?.

Answer. The sector of the circle refers to that part of the circular region that is enclosed by the two radii and the corresponding arc..

Question 2. What do you understand by a segment with reference to a circle?.

Answer. The segment of the circle refers to that part of the circular region that is enclosed between a chord and the corresponding arc. We can categorize the segment of the circle as a minor segment and a major segment. The major segment covers a larger area corresponding to the minor segment..

Related Chapters

- Chapter 1 : Real NumbersChapter 2 : PolynomialsChapter 3 : Pair of Linear Equations in Two VariablesChapter 4 : Quadratic EquationsChapter 5 : Arithmetic ProgressionChapter 6 : TrianglesChapter 7 : Coordinate GeometryChapter 8 : Introduction to TrigonometryChapter 9 : Some Applications of TrigonometryChapter 10 : CirclesChapter 11 : ConstructionsChapter 13 : Surface Areas and VolumesChapter 14 : StatisticsChapter 15 : Probability