SEBA Class 10 Mathematics Chapter 12 Areas Related to Circles

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SEBA Class 10 Mathematics Chapter 12 Areas Related to Circles

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Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 10 Mathematics Chapter 12 Areas Related to Circles Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 10 Mathematics Chapter 12 Areas Related to Circles Solutions for All Subject, You can practice these here.

Areas Related to Circles

Chapter – 12

Exercise 12.1

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 (r1) of 1st circle = 19 cm

Radius (r2) or 2nd circle = 9 cm

Let the radius of the 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

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Given that,

Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle

2πr = 38π + 18π = 56π

Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given 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 an area equal to the sum of the areas of the two circles.

Ans: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 = πr12  π(8)2 = 64π

Area of 2nd circle  πr12 =  π(8)2  π(6)2 = 36π

Given that, Area of 3rd circle = Area of 1st circle + Area of 2nd circle 

πr12  = πr12  + πr12 

πr12 = 64πr12 + 36 π 

πr2  = 100π

r = 100

r = + 10

3. 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 is 10.5 cm wide. Find the area of each of the five scoring regions[𝑈𝑠𝑒 𝜋 = 22 7 ].

Ans: 

Radius (r1) of good region (i.e., 1st circle = 21/ = 10.5 cm 

Given that circle is 10.5 cm wider than the previous circle 

Therefor , radius (r2) of 2nd circle = 10.5 + 10.5 

21cm

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

= 42cm

Area of good region = Area of 1st  circle = πr12 = π(10.5)2 = 346.5cm2

Area red region = Area of 2nd circle – area of 1 st circle 

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 is travelling at a speed of 66 km per hour? 

Ans:  Diameter of the car = 80 

Radius (r) of the wheel of the car = 40cm

Circumference of wheel = 2πr

= 2π (40) = 80π cm

Speed of car = 66 km/hour

= 110000 cm/min

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

N = 1100000 x 7 / 80 x 22

= 1100000 x 7/ 80 x 22

= 35000/8 =  4375

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

5. Thick the correct answer in the following and justify your choice : if the perimeter area of a circle are  numerically equally , then the radius o the circle is

(a) 2 units 

(b) n nuits 

(c ) 4 unites 

(d) 7 uniter 

Ans: Let the radius of the circle  be r

The implies  2 Пr = Пr2

2 = r

Given that the Circumference of the circle are equal This implies 2П = Пr2

This implies 2Пr = nr2  

2 = r 

Therefore , the radius of the circles is 2 units. Hence , the correct answer is (A).

Exercise 12.2

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

Ans: 

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

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 

2Пr = 22 

r = 22/ 2П = 11/ П 

= 1/ 4П = (11/22 )2

Quadrant of will subtend 90 o angles at the circle.

Area  of such quadrant of the circle = 90o/360o x X r2

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

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

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

Therefore , the area swept by the minute hand in minute in will be the area of a sector of sectors of 30o in a circle of 14 cm radius

Area of sector of 30o =  θ/360o ✖ Пr2

Area of sector of 30= 30o /360ox 22/7 x 14 x 14 

= 22/12 x 2 x 14 

= 11x 14 / 3 

= 154/3 cm2

Therefore , the areas swept by the hand in 5 minutes is 154/3 cm2

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

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

Ans: Length of the arc 60/360 x 2 x 22/7 x 21 cm = ⅙ x 2 x 22x 3cm2=22cm

(ii) Area of the sector formed by the arc 

Ans: The area  of the sectors formed by the arc corresponding to sector angle 60o

= 60/360x πr2=⅙ x 22/7 x (21)2 cm2 = ⅙ 22/7 x 21 x 21 cm = 231cm2

(iii) Area of the segment forced by the corresponding chord 

Ans: 

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. [𝑈𝑠𝑒 𝜋 = 3.14 𝑎𝑛𝑑 √3 = 1.73]

Ans: 

Radius (r) of circle = 15 cm

Area of sector OPRQ =  60/360o x πr2  

= 117.75cm2

In ∆OPQ,

∠OPQ = ∠OQP (As OP = OQ)

∠OPQ + ∠OQP + ∠POQ = 180°

2 ∠OPQ = 120°

∠OPQ = 60°

∆OPQ is an equilateral triangle. 

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

= 117.75 – 97,3125

= 20,4375cm2

= 3.14 x 225 – 20,4375

= 706.5 – 20,4375

= 686.0625 cm2

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]

Ans: 

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

SV = VT 

In Δ OVS ,

OV/OS = cos60o 

OV/12 = ½ 

OV = 6cm 

SV/SO = sin 60o = √3/2 

SV/12 =  √3/2 

SV = 6√3cm

ST  = 2SV = 2 X 6√3 cm

Area of ∆OST = ½ x ST X OV

= ½ X 12√3 X6

= 36 √3 = 36 X 1.73  = 62.28 cm2

Area of segment  = area of sector sout – area of ∆ OST

= 150.72 – 32,28

= 88.44cm2

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 the given figure). Find [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 of the rope were 10 m long instead of 5 m. 

Ans: From the figure, it can be observed that the horse can graze a sector of 90° in a circle of 5 m radius.

(i) Area that can be grazed by horse = Area of sector OACB 

=  90o/360oπr2  

= ¼ x 3.14 x (5)2 

= 19.625m2

Area that can be grazed by the horse = Area of sector OABC 

= 90O/360O πr2  

¼ X 3.14 X (5) 2

= 19.625m2

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

= 90o / 360ox π x (10)

= ¼ x 3.14 x 100

= 78.875m2

(ii)  Increase in grazing area = (78.5 − 19.625) m2

= 58.875 m2

9. A brooch is made with silver wire in the form of a circle with a 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.

Ans: Diameter of the circle = 35mm

35/2 Then radius r = mm

The total length of wire required = The length of wire required for

making 5 diameters + the length of wire required for making the perimeter of the circle.

5x35mm + 2π×35/2mm2

= (175 + 22 /7 x 35)]mm -x 35 mm = 285mm

(ii) Here r = 35/2mm

The sector angle of each brooch = 360/10 degrees . i.e θ = 36o Therefore , the area of each brooch

11. 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.

Ans: Here, r = 45cm

θ = sector angle between two consecutive ribs. 

= 360o/8 = 450 [: There are 8 sector of same size]

Therefore, the area between two consecutive ribs of the umbrella 

= The area of one sector 

= 45/360 xπr= ⅛ x 22/7 x (45)2cm2

= 11/28 x 45 x cm2 = 22275/28 cm2

12. 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]

Ans: Here, r = 16.5km =-km 33/2 

 θ = 80° 

The area of the sea over which the ship are warned 

= 820/360 x πr2 2/9 x 3 .14 x  (93/2)2 km2 

= 1.57 x 121 km2 = 189.97

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 Rs.0.35 per cm2. [𝑈𝑠𝑒 √3 = 1.7]

Ans: Here r = 285cm 

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) 9/180 x 2π R 

(b) P/180 X πR2

(c) p/360 x 2π R 

(d) P/720 x 2πR 

Ans: (d) P/720 x 2πR 

Exercise 12.3

1. Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. [𝑈𝑠𝑒 𝜋 =22/7]

Ans: 

In the figure くRPQ = 90

[Angle subtended by a diameter on the circumference]

Therefore ߡRPQ is the right angled at P

RP = 7cm 

PQ = 24cm 

Then by pythagoras There we have.

QR2 = 25cm 

∴ The radius of the circle = 25/2cm 

Now the area of the shaded region 

= The area of the semicircle – the area of ΔRPQ

2. 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. [𝑈𝑠𝑒 𝜋 =22/7]

Ans: 

Here, the radius of the smaller circle = 7cm 

Area of the sector  OBD (The sector angle is 40o)

= 40/360 x π x (14)2 cm2

= 1/9 x 22/7 x 196 cm2 = 1/9 x 22 x 28 cm2 = 616/9 cm2

The area of shaded region 

= the area of the sector OCA – The area of the sector OBD

= (616/9 – 154 / 9) cm2 = 462/9 cm2

3. 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. [𝑈𝑠𝑒 𝜋 = 22 7] 

Ans: 

The area of square ABCD = (14)2 cm 2 = 196cm2

The sum of the areas semicircle APD} [∴ The area of the two semi – circle are equal ]

The area of the shaded region 

= The area of square ABCD=- The sum of the areas of the semicircle APD and BPC 

= 196cm2 – 154cm2 = 42cm2 

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

Ans: 

Radius of arc (R) = 6cm

Side of equilateral triangle OAB = 12cm

Central angle of sector of circle = Area of circle – Area of sector 

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 the given figure. Find the area of the remaining portion of the square. [𝑈𝑠𝑒 𝜋 =22/7].

Ans: 

Side of square = 4cm = 1cm 

Diameter of circle (r) = 1 cm

Radius of circle (R) 1cm 

Area of square = (side)

= (4)2 = 16cm2

Area of circle = πR2 =1 X 22/7 X 1X 1

= 22/7 = 3.14 cm2

Required area = area of square – area of 4 quadrant – area of circle = (16 – 3.14 – 3.14)cm2 = 9.27cm2 Required area = 9.72cm2

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 the given figure. Find the area of the design (Shaded region). [𝑈𝑠𝑒 𝜋 =227] 

Ans: 

Radius (r) of circle = 32 cm

AD = 48 cm

AO = ⅔ AD = 32 

In ∆ABD,

AB2 = AD2 + BD2

7. 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. [𝑈𝑠𝑒 𝜋 =22/7]

Ans: 

Side of square ABCD = 14cm 

Radius of circle (R) = 7cm 

Sector angle (θ) = 90°

Area of four square  =  (side)

= 14 x 14 = 196 cm2

Required shaded area = area of square – of quadrant 

= 196 – 154 = 42cm2

8. The given 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

[𝑈𝑠𝑒 𝜋 =22/7] 

Ans: 

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

Area of the track = (Area of GHIJ − Area of ABCD) + (Area of semicircle HKI − Area 

of semicircle BEC) + (Area of semicircle GLJ − Area of semi-circle AFD)

Therefore , the area pf the track is 4320m2

9. 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. [𝑈𝑠𝑒 𝜋 = 22/7] 

Ans: 

Radius (r1) of larger circle = 7 cm

Radius (r2) of smaller circle = 7/2cm 

Area of smaller circle =  Пr12

= 22/7 x 7/2 x 7/ 2 

= 77/ 2 cm

Area of semicircle AECFB of larger circle  = ½ Пr12

= ½ x 22/7 x (7)

= 77 cm

Area of ΔABC = ½ X AB X OC 

= ½ X 14 X 7  = 49 CM2

Area of smaller circle + Area of semi- circle AECFB –  Area of   ΔABC 

= 77/2 + 77/2 = 28 + 38.5 = 66.5 cm2

10. 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 √3 = 1.73205]

Ans: 

Let the side of the equilateral triangle be a.

Area of equilateral triangle = 17320.5 cm2

Area of sector ADEF = 60o/360ox П x r2

= ⅙ x П x (100)2

= 3.14 x 10000/6

= 15700/3 cm2

Area of shaded region =  area of equilateral triangle – 3 x Area of each sector = 

17320.5 .3 x 15700/3

= 17320.5 – 152700 = 1620.5 cm2

11. 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. [𝑈𝑠𝑒 𝜋 = 22 7 ] 

Ans: 

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

Area of square = (Side)2 = (42)2 = 1764 cm2

Area of each circle = πr2 = 22/7 x (7)2 = 154 cm2

Area of 9 circle = 9 x 154 = 1386 cm2

Area of the remaining potion of the handkerchief = 1764 – 1386 – 378 cm2

12. 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

(i) Quadrant OACB

(ii) Shaded region[𝑈𝑠𝑒 𝜋 =/227]

Ans:  

 Since OACB is a quadrant, it will subtend 90° angle at O. Area of quadrant OACB = 90o /360o x πr2

(ii) Area of  ΔOBD = ½ x OB X OD 

= ½ X 3.5 x 2

= ½ x 7/2 x 2 

= 7/2 cm2

Area of the shaded region  = Area of quadrant OACB – area of   Δ OBD 

= 77/8 – 7/2

= 77-28/8 

= 49/8 cm 2

13. 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 𝜋 = 3.14]

Ans:  

Side of square ABCD =  20cm 

く Aoc = 90o

AB = OA 

∴ In Δ OAB 

OB2 =  OA2 + AB

Area of square OABC = (side)2 =(20)

Area of square = 400

Radius of  quadrant (R) = 20 √2 

Sector angle (θ) = 90o

∴ Area of sector 

14. 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. [𝑈𝑠𝑒 𝜋 = 22 7 ]

Ans: 

Radius of sector OBA  (R) = 21 cm

Radius of sector ODC (r) = 7cm 

Sector angle (θ) = 30o

Area of bigger (OAB) 

15. 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. [𝑈𝑠𝑒 𝜋 =22/7]

Ans:  Radius of quadrant ACPB (r ) = 14cm

Sector angle (θ) = 90o

AB = AC 7cm 

16. Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each. [𝑈𝑠𝑒 𝜋 = 22 7 ] 

Ans: 

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

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