Areas Related To A Circle

Given radius of the circle (r)= 21 cm

Measure of angle (θ)= 120^o

Length of the arc =

=

= 44 cm

Area of the sector formed by the arc =

=

= 462 cm²

Given the rate of repair work per m² = Rs. 25

The radius of the outer circle (R₁) = 63 cm

radius of the inner circle (R₂) = 63 – 7

= 56 cm

(∵ the distance between the inner circle and outer circle is 7 cm)

Let the measured angle is 60^o

area of the repair road = () – ()

= () – (

= 436.15 m²

cost required to repair the road = 436.15 × 25

= Rs. 10,903.9445 (approximately)

Given radius of the circle (r) = 10 cm

Side of the hexagon (a) = 10 cm

we can divide the hexagon into 6 equilateral triangles as follows.

the side of an equilateral triangle will be 10 cm

we have to find the area of the shaded region as shown in fig. 13.11

area of the shaded region = area of circle - area of hexagon

= area of circle - area of 6 right equilateral triangles

= π r² – 6 ()

= 3.14 × 10² – 6 ( × 10²)

= 54.5 cm²

length of the minute hand (r) = 10 cm

(In clocks the length of minute hand will be radius of that clock dial)

The sector formed by the minute hand after 5 minutes will be as follows.

θ = = 30^o

Area of the sector formed by minute hand =

=

= 26.166 cm²

Radius of the sector (r) = 21 m

cost for constructing a meter wall = Rs. 25

cost of constructing a wall around the field = Rs. 1875

∴ Total length of wall = = 75 m

perimeter of the sector = 75 m

l + r + r = 75 m

l + 2(21) = 75 m

l = 33 m

∴ Arc length of CDB (L) = 33m

Area of the field =

=

= 346.5 m²

Cost of tiling 1 m² = Rs. 10

∴ Cost of tiling 346.45 m² = 10 × 346.5

= Rs. 3465

length of side of a square field (a) = 20 m

length of the rope (r₁) = 6 m

since it is square the center angle (θ) = 90^o

area of the gazing field =

=

= 28.26 m²

If the length of rope is increased by 2m

The radius of the gazing field (r₂) = 6m + 2m = 8m

area of the gazing field =

=

= 50.24 m²

increase in gazing field = 50.24 – 28.26

= 27.98 m²

radius of the circle (r) = 42 cm

angle of measure (θ) = 60^o

area of the segment =

=

= 923.16 cm²

since the center angle of measure is 60^o that is a equilateral triangle.

side of the triangle (a) = 42 cm

area of the equilateral triangle =

=

= 762.93 cm²

area of minor segment (BDC) = area of segment – area of triangle

= 923.16 – 762.93

= 160.23 cm²

Length of the chord = 10 cm

Angle of measure = 90^o

since triangle ABC is a right-angled triangle

AB, AC is same as the radius of the circle

BC² = AB² + AC²

10² = r² + r²

100 = 2 r²

50 = r²

r = √50

Area of segment =

=

= 39.25 cm²

Area of the triangle = AB× AC

= × (√50) × (√50)

= 25 cm²

Area of minor segment = Area of segment - Area of the triangle

= 39.25 – 25

= 14.25 cm²

Area of major segment = Area of circle - Area of minor segment

= π r² – 14.25

= 3.14 × (√50)² – 14.25

= 142.82 cm²

given length (l) = 12 cm

breadth (b) = 5 cm

we can notice from the fig. that the Hypotenuse of ADB is same as diameter of the circle.

from the triangle ABD

BD² = AD² + AB²

BD = √(5² + 12²)

(∵ AD is breadth, AB is length of the rectangle)

BD = √(25 + 144)

= √169

= 13 cm

∴ diameter of the circle = 13 cm

Radius of the circle (r) = 6.5 cm

Area of the shade region = Area of circle – area of rectangle

= π r² – b × h

= π × 6.5² – 5 × 12

= 72.732 cm²

side of a square park (a) = 80 cm

radius of the sectors (r) = 7 cm

we can notice that all the sectors can form a circle of radius 7 cm

so, the area of remaining part of park =

area of square – area of circle

= a² - π × r²

= 80² - π × 7²

= 6,240.06 cm²

Given side of a square (a) = 12 cm

Radius of each sector (r) = 12 cm

since it is square the angle (θ) = 90^o

let divide the square into 2 parts. (BAPC, DAQC)

Area of part – 1

The area of the sector BAPC =

= × 12² × 90

= 113.04 cm²

Area of square = a²

= 12²

= 144 cm²

Area of part 1 (APC) = Area of square - The area of the sector BAPC

= 144 – 113.04

= 30.96 cm²

∵ The shapes are identical

The area of part 2 (AQC) = 30.96 cm²

Area of the colored region = area of PART 1+ area of part 2

= 30.96 + 30.96

= 61.92 cm²

Cost for covering 1 m² silver foil = 100

Cost for covering 61.92 m² silver foil = 61.92 × 100

= 6192

Side of square plate (a) = 1 m = 100cm

radius of the segment (r) = 42 cm

we can notice that all the sectors can form a circle of radius 42 cm

the area of the remaining portion of the plate =

Area of square plate - area of the circle

= a² - π r²

= 100² - π × 42²

= 4,458.23 cm²

=0.4458 m²

given OA = 10.5 cm = radius (r)

OD = 6 CM

we can see that ODA is a right-angled triangle

area of triangle = OA× OD

= 10.5 × 6

= 31.5 cm²

we can see that segment OACB is of the circle.

so, the area of OACB = × r²

= × 10.5²

= 86.54625 cm²

area of blue colored region = area of OACB - area of triangle

= 86.54625 – 31.5

= 55.04625 cm²

Area of the circular park = 616 m²

width of the track = 3.5 m

Area of the circular park = 616 m²

π × r² = 616

r² = 196

r = √196

r = 14 m

∴ the radius of the circular park (r₁) = 14 m

the radius of the park with track (r₂) = 14 + 3.5

= 17.5 m

Circumference of the track = 2π (r₂)

= 2× (17.5⁾

= 110 m

Cost of fencing per meter = Rs. 5

Cost of fencing 110 m = 110 × 5

= 550

Speed of the cycle = 140RPM

Diameter of the wheel = 60 cm

No. of revolution in 2 hours = 120 × 140

= 16,800

Distance covered in 2 hours = Revolutions × Diameter

= 16,800 × 60

= 1008000 cm

= 10080 m

= 10.8 km

Given radius = 20cm

if we connect the chord to the center point that forms a segment.

the centre measure angle (θ) =90^o

we can observe a right-angled triangle OAB

Area of segment =

=

= 314 cm²

Area of the triangle = OA× OB

= × (20) × (20)

= 200 cm²

Area of minor segment = Area of segment - Area of the triangle

= 314 – 200

= 114 cm²

from the figure we can notice all these

1. OAB is an equilateral triangle

(∵ All the sides are equal)

2. OAPB is a minor segment with Radius 14 cm, the angle (θ) between them is 60^o

3. ABQ is a semi-circle with centre as M radius 7 cm

Area of the segment =

=

= 102.62 CM²

Area of the triangle = × OA²

= × 14²

= 84.870 CM²

Area of the semi-circle =

Area of the segment - Area of the triangle

= 102.62 - 84.870

= 17.75 cm²

Area of the semi-circle = × r²

= × 7²

= 76.96 cm²

area enclosed by two arcs =

Area of the semi-circle - Area of the semi-circle

= 76.96 – 17.75

= 59.20 cm² (approximately)

side of a square garden ABCD (a) = 70 m

From the fig. in the below we can see that

1. OAXD is a segment

2. OAD is an right-angled triangle.

3. AXD is a minor segment

OA, OD are the radii of the segment

from the triangle OAD

AD² = OA² + OD²

70² = r² + r²

2r² = 4900

r² = 2450

r = √2450

Area of minor segment = Area of segment - Area of triangle

= - × OA²

= - ×

= 1924.22 – 1225

=699.22 cm²

= 700 cm²

since AXD, BYC are same

∴ Area of lawn = 2 × area of minor segment

= 2 × 700

= 1400 cm²

side of a square (a) = 20 cm

we can see the semi-circle in the square having 10 cm as radius (r)

area of the colored region =

area of 4 semi-circles - area of the square

= 4 × ( × r²) - a²

= 2 (π × 10²) - 20²

= 228 cm²

radius of the circle = 30 cm

Area of the circle = π r²

= 3.14 × 30²

= 2826 cm²

PQR is an equilateral triangle

∠ QPR = 60^O

∠ QOR = 2 × ∠ QPR = 120^o

In OQR triangle,

let OS is perpendicular to QR

In triangle OQR as OQ = OR

∠ QOS = ∠ QPR = 60^O

In triangle OSQ ∠S = 90^o

sin 60^o =

=

QS = 15 √3 cm

Now,

QR = 2 QS = 2 (15 √3) = 30√3 cm

Area of equilateral triangle = × QR²

= × (30√3)²

= 1167.75 CM²

Area of design = Area of circle - area of triangle PQR

= 2826 – 1167.75

= 1658.25 CM²

Side of square (r) = 42 cm

In triangle RAB,

RA = RB = 5 cm

In minor sector RQCS angle is 90^o

Area of the shaded region = Area of minor sector RQCS + Area of RAB

= + × RA × RB

= + × 5 × 5

= 1384.74 + 12.5

= 1372.24 cm²

side of square (a) = 8 cm

from the fig. we can see that of circle is placed at four parts of the square. if we join them a circle is formed with radius (r₁) 2 cm

Another circle is placed at centre of the square with radius (r₂) 2 cm

Area of colored region = Area of square - Area of circles at 4 corners + Area of circle at centre

= a² – (4 × ( × r₁²) + (π r₂²))

= a² – (π r₁² + π r₂²)

= 8² – (3.14 (2² + 2²))

= 38.88 cm²

from the given figure

PQR is a right-angle triangle

PQ = 8 cm

QR = 6 cm

PR² = PQ² + QR²

PR² = 8² + 6²

PR = √100 = 10 cm

Radius of the incircle (r) = = = = 2 CM

area of coloured region = Area of triangle - area of circle

= × PQ × QR - π r²

= × 8 × 6 - π 2²

= 24 – 12.566

= 11.433 cm²

we know that area of minor sector =

So, option(d) is the correct answer.

we know that when arc length (l) and radius (r) of sector is given

Area of the sector is = rl

So, option(a) is the correct answer.

Given radius (r)= 9 cm

∵ They are perpendicular to each other θ = 90^o

area of the minor sector =

=

=

= 63.585 cm²

So, option(b) is the correct answer.

Given radius (r)= 21 cm

angle(θ) = 120^o

area of the minor sector =

=

=

= 461.58 cm²

= 462 cm²

So, option(a) is the correct answer.

Area of the circle = πr²

circumference of the circle = 2πr

given, Area = Circumference

2πr = πr²

2 = r

So, option(d) is the correct answer.

Area of circle = 616cm²

πr² = 616

r² = 196

r = 14 cm

Given θ = 60^o

area of the minor sector =

=

=

= cm²

So, option(c) is the correct answer.

Radius of the minor sector= 15 cm

Area of minor sector = 150

Area of a minor sector = r l

150 = × 15 × l

l = 20 cm

So, option(b) is the correct answer.

r be the radius of the circle

Area of the circle = πr²

10% increase in radius = 1.1r

Area of circle with new radius = π(1.1r)²

= 1.21πr²

increase in area = 1.21πr² - πr²

= 0.21 πr²

∴ 21% increase in the area of circle

So, option(c) is the correct answer.

let r₁, r₂ are the radii of two circles

given,

πr₁²: πr₂² = 1 : 4

r₁²: r₂² = 1 : 4

r₁ : r₂ =1 : 2

r₂ = 2 r₁

circumference of 1^st circle = 2πr₁

circumference of 2^nd circle = 2πr₂ = 2π(2r₁) = 4πr₁

ratio of circumference = 2πr₁ : 4πr₁ = 1 : 2

So, option(b) is the correct answer.

Given radius of semi-circle = 8 cm

For a largest triangle in semi-circle

base will be its diameter of semi-circle

height will be radius of semi-circle

so, its base = 2 × radius = 2 × 8 = 16 cm

height = 8 cm

Area of the triangle = × base × height

= × 16 × 8

= 64 cm²

So, option(c) is the correct answer.

Given circumference = 44 cm

2πr = 44

πr = 22

r = 7 cm

Given that square inscribed in a circle.

so, diagonal of square will be equal to diameter of circle.

diagonal = 2 × radius = 2 × 7 = 14 cm

diagonal = √2 × side of square

14 = √2 × side

side = 7√2 cm

The angle moved by minute hand = 90^o

length of minute hand (r) = 14 cm

Area of covered by minute hand =

=

=

= 153.93 cm²

= 154 cm²

So, option(a) is the correct answer.

The angle moved by minute hand = 120^o

length of minute hand (r) = 15 cm

Area of covered by minute hand =

=

=

= 235.619 cm²

= 235.5 cm²

So, option(a) is the correct answer.