Bihar Board Class 10th Maths Chapter 12 Areas Related to Circles Solution

 

BSEB Bihar Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.1

Question 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.
Solution:
Let r be the radius of the circle whose circumference is equal to the sum of the circumferences of the two circles of radii 19 cm and 9 cm.
∴ 2πr = 2π( 19) + 2π(9)
or r = 19 + 9 = 28
Hence, the radius of the 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 area equal to the sum of the areas of the two circles.
Solution:
Let r be the radius of the circle whose area is equal to the sum of the areas of the circles of radii 8 cm and 6 cm.
∴ π² = π(8)² + π(6)² or r² = 8² + 6²
or r² = 64 + 36 or r² = 100
or r = 100−−−√ = 10
Hence, the radius of the new circle is 10 cm.

Question 3.
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.

Solution:
The area of each of five scoring regions are as under :

Question 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?
Solution:
Distance covered by the car in 10 minutes
= (66×1000×100×1060) cm
= 66000006 = 1100000 cm
Circumference of each wheel of the car
= 2πr = (2×227×40)cm [∴ r = 802 cm = 40]
Number of revolutions in 10 minutes when the car is travelling at a speed of 66 km/h = 1100000 ÷ (2 x 225 x 40)
= 4375
Hence, each wheel of car makes 4375 revolutions in 10 minutes.

Question 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) it units
(C) 4 units
(D) 7 units
Solution:
(A) Because,
Here, 2πr = πr², where r is the radius.
So, r² – 2r = 0 or r(r – 2) = 0
or r = 0 or r = 2
But, r ≠ 0. r = 2 units

BSEB Bihar Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.2

Question 1.
Find the area of a sector of a circle with radius 6 cm, if angle of the sector is 60°.
Solution:
We know that the area A of a sector of angle θ in a circle of radius r is given by A = θ360∘ x πr².
Here, r = 6 cm, θ = 60°
∴ A = [60∘360∘×227×36] cm² = (1327) cm²
= 18.86 cm²

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Let r cm be the radius of the circle. Then,
Circumference = 22 cm means 2πr = 22
So, 2 x 227 x r = 22
or r = 72
So, radius of the circle is 72 cm.
Area of the quadrant of a circle

Question 3.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
Clearly, minute hand of a clock describes a circle of radius equal to its length, i.e., r = 14 cm.
Since the minute hand rotates through 6° in one minute, therefore area swept by the minute hand in one minute is the area of a sector of angle 6° in a circle of radius 14 cm.

Hence, the required area i.e., the area swept in 5 minutes

Question 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 n = 3.14)
Solution:
Here, r = 10 cm and θ = 90°.
(i) Area of the minor segment

Aliter. Area of the minor segment = Area of the sector – Area of right triangle

(ii) Area of the major sector

Question 5.
In a circle of a radius 21 cm, an arc subtends an 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.
Solution:
Here, r = 21 cm and θ = 60°
(i) Length of the arc, l
= θ180∘ x πr
= (60∘180∘×227×21) cm = 22 cm
(ii) Area of the sector, A
= θ360∘ x πr²
= (60∘360∘×227×21×21) cm² = 231 cm²
(iii) Area of the segment

Question 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).
Solution:
Here, r = 15 cm and θ = 60°.
∴ Area of the minor segment

Area of the major segment = Area of the circle – Area of the minor segment
= (3.14 x 225 – 20.4375)cm²
= (706.5 – 20.4375)cm²
= 686.0625 cm²

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:
Here, r = 12 cm and 0 = 120°

Area of the corresponding segment of the circle = Area of the minor segment (Shown shaded in the figure)
= Area of sector OAB – Area of ∆ OAB … (1)
Now, area of sector OAB
= 120∘360∘ x 3.14 x 12² cm² = 3.14 x 48 cm² … (2)
For area of ∆ OAB, draw OM ⊥ AB. So, AM = BM and ∠AOM = ∠BOM = 60°
Now, OMOA = cos 60°.
Therefore, OM = OA cos 60°
= 12 x 12 cm = 6 cm.
Also, AMOA = sin 60°.
So, AM = OA sin 60° = 12 x 3√2 cm = 63–√ cm
∴ AB = 2 x 63–√ cm = 123–√ cm
Hence, area of ∆ OAB
= 12AB x OM
= 12 x 123–√ x 6 cm² = 363–√ cm² … (3)
So, from (1), (2) and (3), area of the minor segment
= (3.14 x 48 – 363–√) cm²
= (3.14 x 48 -36 x 1.73) cm²
= 12(12.56 – 3 x 1.73) cm²
= 12(12.56 – 5.19) cm²
= 12 x 7.37 cm²
= 88.44 cm²

Question 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

(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 10 m long instead of 5 m. (Use π = 3.14)
Solution:
(i) The horse will graze over a quadrant of a circle with centre at the corner A of the field and radius AF = 5 m.
Then, the area of the quadrant of this circle

= π×524 m² = 3.14×254 m²
= 78.54m² = 19.625 m²

(ii) In the 2nd case, radius = 10 m.
The area of the quadrant of this circle

= π×1024 m² = 3.14×1004 m²
= 3144m² = 78.5 m²
∴ Increase in the grazing area = (78.5 – 19.625) m² = 58.875 m²

Question 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 the figure.

Find :
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Solution:
(i) Silver wire used to make a brooch
= 2πr, where r = 352 mm
= 2 x 227 x 352 mm = 110 mm
Wire used in 5 diameters = 5 x 35 mm = 175 mm
∴ Total wire used = (110 + 175) mm = 285 mm.

(ii) The area of each sector of the brooch

Question 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.

Solution:
Area between two consecutive ribs
= 18 x πr², where r = 45 cm
= (18 x 227 x 45 x 45 ) cm²
= 2227528 cm² = 795.53 cm²

Question 11.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution:
Here, r = 25 cm and θ = 115°.
Total area cleaned at each sweep of the blades = 2 x Area of the sector
(having r = 25 cm and θ = 115°)

Question 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 are warned. (Use π = 3.14)
Solution:
Area of the sea over which the ships are warned

Question 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)
Solution:
Clearly from the figure,
Area of one design
= Area of the sector AOB – Area of ∆ AOB

Cost of making such designs at the rate of Rs 0.35 per cm²
= Rs (0.35 x 464.8)
= Rs 162.68

Question 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) p180 x 2πR
(B) p180 x πR²
(C) p360 x 2πR
(D) p720 x 2πR²
Solution:
We know that area A of a sector of angle 0 in a circle of radius r is given by
A = θ360∘ x πr²
But here, r = R and θ = p°.
∴ A = p∘360∘ x πR² = p720 x 2πR²
∴ (D) is the correct answer.

BSEB Bihar Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

Question 1.
Find the area of the shaded region in the figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

[Annual Paper (Delhi) 2009]
Solution:
Since ROQ is a diameter, therefore ∠RPQ = 90°.
RQ² = RP² + PQ²
or RQ² = 7² + 24² = 49 + 576 = 625
or RQ = 625−−−√ cm = 25 cm
∴ Radius r = 12 RQ = 252 cm

Question 2.
Find the area of the shaded region in the figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Solution:
Area of the shaded region = Area of sector AOC – Area of sector OBD

Question 3.
Find the area of the shaded A region in the figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Solution:
Area of the square ABCD = (14)² cm² = 196 cm²
Diameter of the semicircles = AD or BC = 14 cm
∴ Radius of each semicircle = 142 cm = 7 cm
∴ Area of the both semicircular regions
= 2 x 12πr² = πr²
= (227 x 49) cm² = 154 cm²
∴ Area of the shaded regions = Area of the square ABCD – Area of the both semicircular regions
= (196 – 154) cm² – 42 cm².

Question 4.
Find the area of the shaded region in the 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.
Solution:
Area of the circular portion

Area of the equilateral ∆ OAB
3√4(side)² = (3√4 x 144) cm²
= 363–√ cm²
Area of the shaded region = Area of circular portion + Area of the equilateral triangle
= (6607+363–√) cm²

Question 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 figure. Find the area of the remaining portion of the square.

Solution:
The area of the square ABCD = (side)²
= 4² cm²
= 16 cm²
The sum of the areas of the four quadrants at the four corners of the square
= The area of a circle of radius 1 cm
= 227 x 1² cm²
= 227 cm²
The area of the circle of diameter 2 cm, i.e., radius 1 cm²
= 227 x 1² cm²
= 227 cm²
Area of the remaining portion of the square = The area of the square ABCD – The sum of the areas of 4 quadrants at the four corners of the square – The area of the circle of diameter 2 cm
= (16 – 227 – 227) cm²
= (112−22−227) cm²
= 687 cm² – 9.71 cm²

Question 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 figure. Find the area of the design (shaded region).

Solution:
Let ABC be an equilateral triangle and let 0 be the circumcentre of the circumcircle of radius 32 cm.

Draw OM ⊥ BC.
Now, ∠BOM = 12 x 120° = 60°
So, from ∆ BOM, we have

Hence, area of ∆ BOC
= 12 BC x OM
= 12 x 323–√ x 16 cm²
Area of ∆ ABC = 3 x Area of ∆ BOC
= (3 x 12 x 323–√ x 16) cm²
= 7683–√ cm²
∴ Area of the design (i.e., shaded region) = Area of the circle – Area of ∆ ABC
= (225287−7683–√) cm²

Question 7.
In the figure, 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.

Solution:
The area of the square ABCD = (side)² = 142 cm²
= 196 cm².
The sum of the areas of the four quadrants at the four corners of the square
= The area of a circle of radius 142 cm = 7 cm²
= π(7)² cm² =(227 x 49) cm² = 154 cm²
Area of the shaded region = The area of the square ABCD – The sum of the areas of four quadrants at the four corners of the square
= (196 – 154) cm² = 42 cm²

Question 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.
Solution:
We have OB = O’C = 30 m
and, AB = CD = 10 m
OA = O’D = (30 + 10) m = 40 m

(i) the distance around the track along its inner edge.
= BC + EH + 2 x circumference of the semicircle of radius OB = 30 m
= ( 106 + 106 + 2 x 12 x 2π(30) ) m
= ( 212 + 2 x 227 x 3o) m
= ( 212 + 13207 )m = (1484+13207) m
= 28047 m
m = 400.57 m

(ii) Area of the track = Area of the shaded region = Area of rectangle ABCD + Area of rectangle EFGH + 2(Area of the semicircle of radius 40 m – Area of the semicircle with radius 30 m)

Question 9.
In the 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.

Solution:
Area of the sector OCB = (90∘360∘×227×7×7) cm²
= 772 cm²
Area of ∆ OCB = 12 x OC x OB
= 12 x 7 x 7 cm²
= 492 cm²
The area of the segment BPC = Area of the sector OCB – Area of the A OCB
= (772 – 492) cm²
= 282 cm²
= 14 cm²
Similarly, the area of the segment AQC
= 14 cm²
Also,the area of the circle with DO as diameter
= (227 x 72 x 72) cm² = 772 cm²
Hence,the total area of the shaded region
= (14 + 14 + 772) cm²
= (28+28+772) cm²
= 1332 cm²
= 66.5 cm²

Question 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 figure). Find the area of the shaded region.

(Use π = 3.14 and 3–√ = 1.73205)
Solution:
Let each side of the triangle be a cm. Then,

Thus, radius of each circle is 2002 cm = 100 cm.
Now, area of shaded region
= Area of ∆ ABC – 3 x (Area of a sector of angle 60° in a circle of radius 100 cm)
= [17320.5 – 3(60∘360∘ x 3.14 x 100 x 100)] cm²
= (17320.5 – 15700) cm² = 1620.5 cm²

Question 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.

Solution:
Side of the square ABCD
= AB – 3 x diameter of circular design
= 3 x (2 x 7) cm – 42 cm
∴ Area of the square ABCD
= (42 x 42) cm² = 1764 cm²
Area of one circular Resign
= πr² = (227 x 7 x 7) cm²
= 154 cm²
∴ Area of 9 such designs
= (9 x 154) cm² = 1386 cm²
∴ Area of the remaining portion of the handkerchief
= Area of the square ABCD – Area of 9 circular designs
= (1764 – 1386) cm² – 378 cm²

Question 12.
In the 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.

Solution:

Hence, area of the shaded region = Area of quadrant – Area of ∆ BOD
= (778 – 72) cm² = (77−288) cm²
= 498 cm² = 6.125 cm²

Question 13.
In the 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)

Solution:
Radius of the quadrant = OB = OA2+AB2−−−−−−−−−√
= 202+202−−−−−−−−√ cm
= 201+1−−−−√ cm
= 202–√ cm
∴ Area of quadrant OPBQ
= 14πr²
= 14 x 314 x (202–√)² cm²
= ( 14 x 3.14 x 80o) cm² = 628 cm²
Area of the square OABC = (20) cm² = 400 cm²
Hence, area of the shaded region = Area of quadrant – Area of square OABC
= (628 – 400) cm² = 228 cm².

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If ∠AOB = 30°, find the area of the shaded region.

Solution:
Let A1 and A2 be the areas of sector OAB and OCD respectively. Then,
A1 = Area of a sector of angle 30° in a circle of radius 21 cm

A2 = Area of a sector of angle 30° in a circle of radius 7 cm

Question 15.
In the 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.

[Annual Paper (Foreign) 2008]
Solution:
Here the radius of the quadrant with A as the centre is 14 cm.
Then, the area of the quadrant ABMC

Area of the segment BMC of the circle = Area of the quadrant ABMC – Area of ∆ BAC
= (154 – 98) cm²
= 56 cm²
Now, since AC = AB = 14 cm and ∠BAC = 90°, therefore by Pythagoras Theorem,
BC = AC2+AB2−−−−−−−−−√
= 142+142−−−−−−−−√
= 142–√ cm
∴ Radius of the semicircle BNC = 12 x 142–√ cm = 72–√ cm
∴ Area of the semicircle BNC
= (72–√ )² cm²
= (12×227×98) cm² = 154 cm²
Hence, the area of the shaded region
= the area of the region between two arcs BMC and BNC = The area of semicircle BNC – The area of the segment of the circle BMC
= (154 – 56) cm² = 98 cm²

Question 16.
Calculate the area of the designed region in figure common between the two quadrants of circles of radius 8 cm each.

Solution:
Here, 8 cm is the radius of each of the quadrants ABMD and BNDC.
Sum of their areas

Area of the square ABCD = (8 x 8) cm² = 64 cm²
Area of the designed region = Area of the shaded region
= Sum of the areas of quadrants – Area of the square ABCD
= (7047−64) cm² = (704−4487) cm²
= 2567 cm²
= 36.57 cm²