**NCERT Solutions Class 10 Maths Chapter 12 Area Related to Circles** – Here are all the NCERT solutions for Class 10 Maths Chapter 12. This solution contains questions, answers, images, explanations of the complete Chapter 12 titled Area Related to Circles of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 12 Area Related to Circles. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 12 Area Related to Circles in one place.

## NCERT Solutions Class 10 Maths Chapter 12 Area related to circles

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Class | 10 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 12 |

Chapter Name |
Area related to circles |

### NCERT Solutions Class 10 Maths chapter 12 Area related to circles

Class 10, Maths chapter 12, Area related to circles solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Area related to circles

Q.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.

Ans :Radius \( \left(r_{1}\right)\) of the 1^{st}circle = 19cm Radius \( \left(r_{2}\right)\) of the 1st circle = 9 cm Let the radius of 3^{rd}circle be r. Circumference of 1^{st}circle = \( 2 \pi r_{1}=2 \pi(19)\) = \( 38 \pi\) Circumference of 2^{nd}= \( 2 \pi r_{1}=2 \pi(9)\) = \( 18 \pi\) Circumference of 3^{rd}circle =\( 2r\pi\) That given , Circumference of 3^{rd}circle = Circumference of 2^{nd}circle = Circumference of 1^{st}circle \( 2 \pi r=38\pi+18\pi=56 \pi\) \( r=\frac{56 \pi}{2 \pi}=28\) Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm.

Q.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.

Ans :Radius \( \left(r_{1}\right)\) of the 1st circle = 8 cm Radius \( \left(r_{2}\right)\) of the 1st circle = 6 cm Let the radius of 3rd circle be r. Area of 1ts circle = \( \pi r_{1}^{2}=\pi(8)^{2}=64 \pi\) Area of 1ts circle = \( \pi r_{1}^{2}=\pi(6)^{2}=36 \pi\) That given, Area of 3rd circle =Area of 2nd circle =Area of 1st circle \( \pi r^{2}=\pi r_{1}^{2}+\pi r_{2}^{2}\) \( \pi r^{2}=64 \pi+36 \pi\) \( \pi r^{2}=100 \pi\) \( r=\pm 10\) 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.

Q.3:Fig. 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.

Ans :Radius \( \left(r_{1}\right)\) of gold region (i.e. 1st circle) = \( \frac{21}{2}=10.5 \mathrm{cm}\) Given that each circle is 10.5 cm wider than the previous circle. Therefore . radius \( r_{2}\) of 2nd circle =10.5 + 10.5 =21 cm radius \( r_{3}\) of 2nd circle =21 + 10.5 =31.5 cm radius \( r_{4}\) of 2nd circle =31.5 + 10.5 =42 cm radius \( r_{5}\) of 2nd circle =42 + 10.5 =52.5 cm Area of gold region = Area of 1st circle =\( \pi r_{1}^{2}=\pi(10.5)^{2}=346.5 \mathrm{cm}^{2}\) Area of red region = Area of 2nd circle - Area of 1st circle =\( \pi r_{2}^{2}-\pi r_{1}^{2}\) =\( \pi(21)^{2}-\pi(10.5)^{2}\) =\( 441 \pi-110.25 \pi=330.75 \pi\) = \( 1039.5 \mathrm{cm}^{2}\) Area of blue region = Area of 3rd circle - Area of 2nd circle =\( \pi r_{3}^{2}-\pi r_{2}^{2}\) =\( \pi(31.5)^{2}-\pi(21)^{2}\) =\( 992.25 \pi-441 \pi=551.25 \pi\) = \( 1732.5\mathrm{cm}^{2}\) Area of black region = Area of 4th circle - Area of 3rd circle =\( \pi r_{3}^{2}-\pi r_{2}^{2}\) =\( \pi(42)^{2}-\pi(31.5)^{2}\) =\( 1764 \pi-992.25 \pi=771.75 \pi\) = \( 2425.5\mathrm{cm}^{2}\) Area of white region = Area of 5th circle - Area of 4th circle =\( \pi r_{3}^{2}-\pi r_{2}^{2}\) =\( \pi(52.5)^{2}-\pi(42)^{2}\) =\( 2756.25\pi-1764 \pi=992.25 \pi\) = \( 3118.5\mathrm{cm}^{2}\) Therefore, areas of gold, red, blue, black, and white regions are \( \ 346.5 \mathrm{cm}^{2}\) , \( 1039.5 \mathrm{cm}^{2}\) , \( 1732.5\mathrm{cm}^{2}\) , \( 2425.5\mathrm{cm}^{2}\) , \( \ 3118.5\mathrm{cm}^{2}\) respectively .

Q.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?

Ans :Diameter of the wheel of the car = 80 cm Radius (r) of the wheel of the car = 40 cm Circumference of wheel =\( 2 \pi r\) \( =2 \pi(40)=80 \pi \mathrm{cm}\) Speed of car = 66 km/hr \( =\frac{66 \times 100000}{60} \mathrm{cm} / \mathrm{min}\) \( =\frac{35000}{8}=4375\) Therefore , each wheel of the car will make 4375 revolutions.

Q.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

Ans :Let the radius of the circle be r. Circumference of circle = \( 2 \pi r\) Area of circle =\( \mathrm\pi r^{2}\) Given that, the circumference of the circle and the area of the circle equal. This implies \( 2 \pi r\) = \( \mathrm\pi r^{2}\) r=2 Therefore, the radius of the circle is 2 units. Hence, the correct answer is A.

Q.1:Find the area of a sector of a circle with radius 6 cm if angle of the sector is \( 60^{\circ}\).

Ans :Area of sector of angle \( \theta=\frac{\theta}{360^{\circ}} \times \pi r^{2}\) Area of sector OACB =\( \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(6)^{2}\) \( =\frac{1}{6} \times \frac{22}{7} \times 6 \times 6=\frac{132}{7} \mathrm{cm}^{2}\) Therefore , the area of sector of the circle making \( 60^{\circ}\) at the centre of the circle is \(\frac{132}{7} \mathrm{cm}^{2}\)

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

Ans :Let the radius of the circle be r. Circumference = 22 cm \( 2 \pi r\) = 22 \( r=\frac{22}{2 \pi}=\frac{11}{\pi}\) quadrant of circle will subtend \( 60^{\circ}\) at the centre of the circle . Area of such quadrant of the circle \( =\frac{90^{\circ}}{360^{\circ}} \times \pi \times r^{2}\) \( =\frac{1}{4 \pi} \times \pi \times\left(\frac{11}{ }\right)^{2}\) \( =\frac{121}{4 \pi}=\frac{121 \times 7}{4 \times 22}\) \( =\frac{77}{8} \mathrm{cm}^{2}\)

Q.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 Fig. Find the area of the design.

Ans :Radius (r) of circle 32 cm AD is the median of triangle ABC. \( \mathrm{AO}=\frac{2}{3} \mathrm{AD}=32\) AD = 48 cm In triangle ABD, =\( \mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}\) =\( \mathrm{AB}^{2}=(48)^{2}+\left(\frac{\mathrm{AB}}{2}\right)^{2}\) =\( \frac{3 \mathrm{AB}^{2}}{4}=(48)^{2}\) =\( \mathrm{AB}=\frac{48 \times 2}{\sqrt{3}}=\frac{96}{\sqrt{3}}\) =\( 32 \sqrt{3} \mathrm{cm}\) Area of equilateral a triangle , \( \frac{\sqrt{3}}{4} \times 32 \times 32 \times 3=96 \times 8 \times \sqrt{3}\) \( 768 \sqrt{3} \mathrm{cm}^{2}\) Area of circle = \( \pi r^{2}\) =\( \frac{22}{7} \times(32)^{2}\) =\( \frac{22}{7} \times 1024\) =\( \frac{22528}{7} \mathrm{cm}^{2}\) Area of design = area of circle - Area of triangle ABC =\( \left(\frac{22528}{7}-768 \sqrt{3}\right) \mathrm{cm}^{2}\)

Q.7:In Fig., 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.

Ans :Area of each of the 4 sectors is equal to each and is a sector of \( 90^{\circ}\)in a circle of 7 crn radius. Area of each sector = \( \frac{90^{\circ}}{360^{\circ}} \times \pi(7)^{2}\) =\( \frac{1}{4} \times \frac{22}{7} \times 7 \times 7\) =\( \frac{77}{2} \mathrm{cm}^{2}\) Area of Square of ABCD =\( (\text { Side })^{2}=(14)^{2}=196 \mathrm{cm}^{2}\) Area Of shaded portion = Area Of square ABCD - 4 x Area Of each sector = \( 196-4 \times \frac{77}{2}=196-154\) =\( 42 \mathrm {cm}^{2}\) Therefore , the area of shaded portion is \( 42 \mathrm{cm}^{2}\).

Q.8:Fig. 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.

Ans :Distance around the track along its inner edge =AB + arc BEC + CD + arc DFA =\( 106+\frac{1}{2} \times 2 \pi r+106+\frac{1}{2} \times 2 \pi r\) =\( 212+\frac{1}{2} \times 2 \times \frac{22}{7} \times 30+\frac{1}{2} \times 2 \times \frac{22}{7} \times 30\) =\( 212+2 \times \frac{22}{7} \times 30\) =\( 212+\frac{1320}{7}\) =\( \frac{1484+1320}{7}=\frac{2804}{7} \mathrm{m}\) 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) =\( 106 \times 80-106 \times 60+\frac{1}{2} \times \frac{22}{7} \times(40)^{2}-\frac{1}{2} \times \frac{22}{7} \times(30)^{2}+\frac{1}{2} \times \frac{22}{7} \times(40)^{2}-\frac{1}{2} \times \frac{22}{7} \times(30)^{2}\) =\( 106(80-60)+\frac{22}{7} \times(40)^{2}-\frac{22}{7} \times(30)^{2}\) =\( 106(20)+\frac{22}{7}\left[(40)^{2}-(30)^{2}\right]\) =\( 2120+\frac{22}{7}(40-30)(40+30)\) =\( 2120+\left(\frac{22}{7}\right)(10)(70)\) =2120 + 2200 = \( 4320 \mathrm{m}^{2}\) Therefore , the area of shaded region is \( 4320 \mathrm{m}^{2}\)

Q.9:In Fig., 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.

Ans :Radius \( r_{1}\) of larger circle = 7cm Radius \( r_{2}\) of smaller circle =\( \frac{7}{2} \mathrm{cm}\) Area of smaller circle = \( \pi r_{1}^{2}\) =\( \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) = \( \frac{77}{2} \mathrm{cm}^{2}\) Area of semi-circle AECFB of larger circle = \( \frac{1}{2} \pi r_{2}^{2}\) = \( \frac{1}{2} \times \frac{22}{7} \times(7)^{2}\) = \( 77 \mathrm{cm}^{2}\) Area of triangle ABC =\( \frac{1}{2} \times \mathrm{AB} \times \mathrm{OC}\) =\( \frac{1}{2} \times 14 \times 7=49 \mathrm{cm}^{2}\) Area of the shaded region =Area of smaller circle + Area of semi-circle AECFB - Area of triangle ABC =\( \frac{77}{2}+77-49\) = \( 28+\frac{77}{2}=28+38.5=66.5 \mathrm{cm}^{2}\)

Q.10:The area of an equilateral triangle ABC is 17320.5 \( \mathrm{cm}^{2}\) . 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 Fig.). Find the area of the shaded region.

Ans :Let the side of the equilateral triangle be a, Area of equilateral triangle = 17320.5 \( \mathrm{cm}^{2}\) \( \frac{\sqrt{3}}{4}(a)^{2}=17320.5\) \( \frac{1.73205}{4} a^{2}=17320.5\) \( a^{2}=4 \times 10000\) A = 200 cm Each sector of measure \( 60^{\circ}\) Area of sector ADEF = \( \frac{60^{\circ}}{360^{\circ}} \times \pi \times r^{2}\) =\( \frac{1}{6} \times \pi \times(100)^{2}\) =\( \frac{3.14 \times 10000}{6}\) =\( \frac{15700}{3} \mathrm{cm}^{2}\) Area Of shaded region Area Of equilateral triangle - 3 x Area Of each sector \( 17320.5-3 \times \frac{15700}{3}\) \( 17320.5-15700=1620.5 \mathrm{cm}^{2}\)

Q.11:On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig). Find the area of the remaining portion of the handskter chief.

Ans :From the figure, it can be observed that the side of the squ«e is 42 cm. Area of square = \( (\text { Side })^{2}=(42)^{2}=1764 \mathrm{cm}^{2}\) Area of each circle =\( 9 \times 154=1386 \mathrm{cm}^{2}\) Area of the remaining portion of the = 1764 -1386 = 378 \( \mathrm{cm}^{2}\)

Q.12:In Fig. 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.

Ans :(i) Since OACB is a quadrant, it will subtend \( 90^{\circ}\) angle at O. Area of quadrant OACB = \( \frac{90^{\circ}}{360^{\circ}} \times \pi r^{2}\) \( =\frac{1}{4} \times \frac{22}{7} \times(3.5)^{2}=\frac{1}{4} \times \frac{22}{7} \times\left(\frac{7}{2}\right)^{2}\) \( \frac{11 \times 7 \times 7}{2 \times 7 \times 2 \times 2}=\frac{77}{8} \mathrm{cm}^{2}\) (ii) area of triangle = \( \frac{1}{2} \times \mathrm{OB} \times \mathrm{OD}\) \( \frac{1}{2} \times 3.5 \times 2\) \( =\frac{1}{2} \times \frac{7}{2} \times 2\) \( \frac{7}{2} \mathrm{cm}^{2}\) Area of the shaded region = Area of OACE - Area of triangle OBO \( =\frac{77}{8}-\frac{7}{2}\) \( \frac{77-28}{8}\) \( \frac{49}{8} \mathrm{cm}^{2}\)

Q.13:In Fig. a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.

Ans :In \( \triangle \mathrm{OAB}\), \( \mathrm{OB}^{2}=\mathrm{OA}^{2}+\mathrm{AB}^{2}\) \( =(20)^{2}+(20)^{2}\) \( \mathrm{OB}=20 \sqrt{2}\) Radius of circle = \( 20 \sqrt{2} \mathrm{cm}\) Area of quadrant OPBQ = \( \frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(20 \sqrt{2})^{2}\) \( \frac{1}{4} \times 3.14 \times 800\) \( 628 \mathrm{cm}^{2}\) Area of OABC = \( (\text { Side })^{2}=(20)^{2}=400 \mathrm{cm}^{2}\) Area of shaded region = Area of quadrant OPBQ - Area of OABC =\( (628-400) \mathrm{cm}^{2}\) = \( 228 \mathrm{cm}^{2}\)

Q.14:AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig.). If angle AOB = \( 30^{\circ}\) find the area of the shaded region.

Ans :Area of the shaded region = Area of sector OAEB = Area of sector OCFD =\( \frac{30^{\circ}}{360^{\circ}} \times \pi \times(21)^{2}-\frac{30^{\circ}}{360^{\circ}} \times \pi \times(7)^{2}\) =\( \frac{1}{12} \times \pi\left[(21)^{2}-(7)^{2}\right]\) =\( \frac{1}{12} \times \frac{22}{7} \times[(21-7)(21+7)]\) =\( \frac{22 \times 14 \times 28}{12 \times 7}\) =\( \frac{308}{3} \mathrm{cm}^{2}\)

Q.15:In Fig. 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.

Ans :As ABC is a quadrant of the circle,angle BAC will be of measure \(\ 90^{\ 0}\) . In triangle ABC, \( \mathrm{BC}^{2}=\mathrm{AC}^{2}+\mathrm{AB}^{2}\) =\( (14)^{2}+(14)^{2}\) =\( B C=14 \sqrt{2}\) Radius \( \ r_{1}\) Of semi-circle drawn on \( \mathrm{BC}=\frac{14 \sqrt{2}}{2}=7 \sqrt{2} \mathrm{cm}\) Area of triangle ABC =\( \frac{1}{2} \times \mathrm{AB} \times \mathrm{AC}\) = \( \frac{1}{2} \times 14 \times 14\) =\( 98 \mathrm{cm}^{2}\) Area of Sector ABCD = \( =\frac{90^{\circ}}{360^{\circ}} \times \pi ^{2}\) =\( \frac{1}{4} \times \frac{22}{7} \times 14 \times 14\) =\( 154 \mathrm{cm}^{2}\) Area of semi-circle drawn on BC = \( =\frac{1}{2} \times \pi \times r_{1}^{2}=\frac{1}{2} \times \frac{22}{7} \times(7 \sqrt{2})^{2}\) \( \frac{1}{2} \times \frac{22}{7} \times 98=154 \mathrm{cm}^{2}\) Area of shaded region = Area of semi-circle - (Area of sector ABCD - Area of triangle ABC) = 154 - (154-98) =\( 98 \mathrm{cm}^{2}\)

Q.16:Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.

Ans :The designed is the common region between two sectors BAEC and DAFC. Area of sector \( \mathrm{BAEC}=\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(8)^{2}\) =\( \frac{1}{4} \times \frac{22}{7} \times 64\) = \( \frac{22 \times 16}{7}\) = \( \frac{352}{7} \mathrm{cm}^{2}\) Area of triangle BAC = \( \frac{1}{2} \times \mathrm{BA} \times \mathrm{BC}\) =\( \frac{1}{2} \times 8 \times 8=32 \mathrm{cm}^{2}\) Area of the designed portion = 2 x (Area of segrnent AEC) = 2 x (Area of sector BAEC - Area of triangle BAC) =\( 2 \times\left(\frac{352}{7}-32\right)\) =\( 2\left(\frac{352-224}{7}\right)\) =\( \frac{2 \times 128}{7}\) =\( \frac{256}{7} \mathrm{cm}^{2}\)

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