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

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

PSEB 10th Class 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:

Radius of first circle (r1) = 19 cm
Radius of second circle (r2) = 9 cm
Let radius of third circle be R cm
According to condition
circumference of first circle + circumference of second circle = circumference of third circle
2πr1 + 2πr2 = 2πR
2π (r1 + r2] = 2πR
19 + 9 = R
∴ R = 28
∴ Radius of third circle (R) = 28 cm.

Question 2.
The Radii of two circles are 8 cm and 6 cm respectively. Find radius of circle which is having area equal to sum of the area of two circles.
Solution:
Radius of first circle (r1) = 8 cm
Radius of second circle (r2) = 6 cm
Let radius of third circle be R cm
According to question
Area of third circle = Area of first circle + Area of second circle
πR2 = πr12 + πr22
πR2 = π[r12 + r22]
R2 = (8)2 + (6)2
R = √64+36 = √100
R = 10 cm
∴ Radius of required circle (R) = 10 cm.

Question 3.
Fig. depicts an archery target marLed with its five scoring areas from the centre ‘utwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score ¡s 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

  

Hence, area of gold ring; red ring; blue ring : black ring; white ring are 3465 cm2 ; 1039.5 cm2; 1732.5 cm2; 24255 cm2 ; 3118.5 cm2 respectively.

Question 4.
The wheels of a ca are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a spel of 66 km per hour?
Solution:

Question 5.
Tick the correct answer in the following and justify your choice : If the perimeter and area of a clrde are numerically equal, then the radius of the circle Is
(A) 2 units
(B) π units
(C) 4 units
(D) n units
Solution:
Perimeter of circle = Area of circle
2πR = πR2
2R = R2
⇒ R = 2
∴ Correct option A is (R) = 2 unit.

Ex 12.2

Question 1.
Find the area of sector of a circle with radius 6 cm, if angle of the sector is 60°.
Solution:
Radius of sector of circle (R) = 6 cm

 

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:
Length of minute hand of clock = Radius of circle (R) = 14 cm

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.
Solution:
Radius of circle (R) = 10 cm
Central angle (θ) = 90°

Question 5.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(j) 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:
(i) Radius of circle (R) 21 cm

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:
Radius of circle = (R) = 15 cm
Central angle (θ) = 60°

In ∆OAB, central angle θ = 60°
OA = OB = 15 cm
∴ ∠A = ∠B
Now, ∠A + ∠B + ∠O = 180°
2∠A ÷ 600 = 180°
∠A = 60°
∴ ∠A = ∠B = 60
∴ ∠OAB is equilateral.
[ Area of minor segment] = [Area of minor sector] – [Area of equilateral triangle]

Area of minor segment = 20.43 cm2.
Area of major segment = Area of circle – Area of major segment.
= πR2 – 20.43
= 3.14 × 15 × 15 – 20.43
= 706.5 – 20.43
= 686.07 cm2
Area of major segment = 686.07 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 it π = 3.14 and √3 = 1.73).
Solution:
Radius of circle (R) = 12 cm
Central angle (θ) = 120°

In ∆OAM, from O draw, angle bisector of ∠AOB as well as perpendicular bisector OM of AB.
∴ AM = MB = 1/2 AB
In ∆OMA,
∠AOM + ∠OMA + ∠OAM = 180°
LOAM = 30°
Similarly, ∠OAM = 30° = ∠OBM

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 fig.). 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:
Side of square = 15 m
(i) Length of Peg = Radius of rope (R) = 5 m
Centrral angle (θ) = 90° [Each angle of square]

 

∴ Increase in grazing area = Area of sector OCD – Area of sector OAB
= 78.5 – 19.625 = 58.875 m2
Increase of grazing area = 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 fig. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

∴ Area of each brooch = 96.25 m2.

Question 10.
An umbrella has 8 ribs which are equally spaced (see fig). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Area of sector = 795.53 cm2
∴ Area between two consecutive ribs of the umbrella = 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:
Length of blade (R) = 25 cm
Sector angle (θ) = 115°
Wiper moves in form of sector.

Question 12.
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 800 to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use x = 3.14)
Solution:
Sector angle (θ) = 80°
Radius of sector (R) = 16.5 km

Question 13.
A round table cover has six equal designs as shown in fig. 1f the radius of the cover is 28 cm, find the cost of making the designs at the rate of 0.35 per cm2. (Use √3 = 1.7)
Solution:
Number of equal designs = 6
Radius of designs (R) = 28 cm
Each design is in the shape of sector central angle (θ) = 6 = 60°

Since central angle is 60° and OA = OB
∴ ∆OAB is equilateral triangle having side 28 cm.

Area of one shaded designed portion = area of segment = Area of the sector OAB – area of ∆OAB

Area of one shaded designed portion = 77.46
Area of six designed portions = 6 [Area of one designed]
= 6 [77.46] = 464.76 cm2
Cost of making 1 cm2 = ₹ 0.35
Cost of making of 464.76 cm2 = 464.76 × 0.35 = ₹ 162.68.

Question 14.
Tick the correct answer in the following:
Area of a sector of angle p° of a circle with radius R is

Ex 12.3

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

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

Question 3.
Find the area of the shaded region in fig., if ABCD is a square of side 14 cm and APD and BPC are semi circles.

Question 4.
Find the area of the shaded region in fig., where a circular arc of radius 6 cm has been drawn iith vertex O of an equilateral triangle OAB of side 12 cm as centre.

Shaded Area = Area of equilateral triangle OAB + Area of major sector of circle
= 62.28 + 94.28 = 62.28 cm2
Shaded Area = Area of equilateral triangle OAR + Area of major sectoç of circle
= 62.28 + 94.28 = 156.56
Shaded Area = 156.56 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 fig. Find the area of the remaining portion of the square.
Solution:
Side of square = 4 cm
Radius of each semi circle cut out (r) = 1 cm

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 fig. Find the area of the design (shaded region).

  

Question 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.

Question 8.
Fig. depicts a racing track whose left and right ends are semicircular. The distahce between the two inner parallel line segments is 60 m and they are each 106 m long. 1f the track is 10 m wide, find
(i) the distance around the track along its inner edge.
(ii) the area of the track.

(i) Here AB = DC = 106 m
AF = BE = CG = HD = 10m
Diameter of inner semicircle (APD and BRC) =60 m
∴ Radius of inner semicircle (APD (r) = 30 m
Radius of outer semicircle (R) = r + 10 = 30 + 10 = 40 m
Distance around the track along inner edge = AB + circumference of semi circle BRC + CD + circumference of semi circle DPA
= 2 AB + 2 [circumference of semi circle BRC]
= 2 (106) + 2((2πr/2))
= 212 + 2πr
= 212 + 2 × 22/7 × 30
= 212 + 60×22/7
= 212 + 188.57 = 400.57 m.
∴ Distance around the track along its inner edge = 400.57 m

(ii) Area of track = Area of rectangle ABEF + Area of region BEMGCRB + Area of rectangle CGHD + area of region.
= 2 Area of rectangle ABCD + 2 Area of region (II)
= 2 (AB × AF) + 2
[Area of semi circle with Radius 60 cm – Area of semi circle with radius 30 cm]
= 2 [106 × 10] + 2 [latex]\frac{\pi \mathrm{R}^{2}}{2}-\frac{\pi r^{2}}{2}[/latex]
= 2 × 1060 + 2π/2 [R2 – r2]
= 2120 + 22/7 (402 – 302)
= 2120 + 22/7 [1600 – 900]
= 2120 + 22/7 [700]
= 2120 + 2200 = 4320 m2
Area of track = 4320 m2

Question 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.

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 of the length of the side of the triangle (see Fig.). Find the area of the shaded region.
(Use n = 3.14 and ,√3 = 1.73205)

Solution:
Area of equilateral triangle ABC = 17320.5 cm2

Area of three sector = 3 × 15700/3 cm2
∴ Required shaded Area = Area of triangle – Area of three sectors
= 17320.5 – 15700 = 1620.5 cm2
∴ Hence, Required shaded Area = 1620.5 cm2

Question 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 handkerchief.

Question 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.

Solution:

∴ Shaded Area = Area of quadrant OACB – Area of ∆ODB
= 9.625 – 3.5 = 6.125 cm2
∴ Hence, Shaded Area = 6.125 cm2.

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

Solution:
Side of square ABCO = 20 cm
∠AOC = 90°
AB = OA

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

Area of smaller sector (ODC) = 12.83 cm2
Now, Shaded Area = Area of bigger sector OAB – Area of smaller sector OCD
= 115.5 – 12.83 = 102.66
Hence, Shaded Area = 102.66 cm2.

Question 15.
In fig., ABC is a quadrant of a circle of radius 14 cm and a semi circle is drawn with BC as diameter. Find the area of the shaded region.

Solution:
Radius of quadrant ACPB (r) = 14 cm
Sector angle (θ) = 90°
AB = AC = 14 cm

Required Area = Area of semi circle – [Area of sector – Area of ∆BAC]
= 154 – [154 – 98]
= (154 – 56) cm2 = 98 cm2
Hence, Shaded Area = 98 cm2.

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

Area of sector = 50.28 cm2
Area of ∆ABD = 1/2 × AB × AD
= 1/2 × 8 × 8
= 32 cm2

∴ Area of segment DMBPD = Area of sector ∆BPD – Area of ∆ABD
= 50.28 – 32 = 18.28 cm2
Hence, Shaded area = 2 area of segment DMBPD = 2 (18.28) = 36.56 cm2