Areas Related To Circles

Given,

Radius of circle = 4.2cm

Circumference of circle = 2πr

= 26.4cm

Area of circle = 2πr²

=55.44cm²

Given,

Area of circle = 301.84cm²

= πr² = 301.84

= 96.24

= r² = = 9.81cm

Circumference of the circle = 2πr

= 61.6 cm

Circumference of the circle = 44cm

2πr = 44cm

Area of circle = πr² =

Area of circle = 154 cm²

Let diameter of circle = X cm

So, acc. to given condition

Circumference = x+16.8 cm

⇒2πr = x + 16.8
Diameter = 2r

Circumference= x + 16.8 (x = 2r)

Length of string = radius of area which horse can graze

r = 28m

so,

Area where the horse can graze = πr²

= 2464 m²

Area of square = 121 cm²

a² = 121

a = = 11cm

Perimeter of square = length of wire

4a = 4 ×11 = 44cm

Perimeter of circle = 2πr

2πr = 44

= 7cm

Area of circle = πr²

Area of circle = = 154cm²

Given,

Length of field = 40m

Breadth of field = 36m

Length of rope (radius) =14m

So,

Area horse can graze =

Area horse can graze = = 154 m²

Area of rectangle = length × breadth

Area of rectangle = 40 × 28

Area of rectangle = 1120 cm²

Diameter of semi circular portion = 28cm

Radius of semi circular portion = = 14cm

So,

Area of semi circular portion =

= 308 cm²

Area of remaining portion = 1120 – 308 = 812cm²

Ratio of circumferences of two circles with radius r₁ and r₂ respectively

Ratio of area = = 4:9

Side of square = 10cm

Radius of inscribed circle =

Radius of inscribed circle = = 5cm

Area of inscribed circle = πr² =

= 78.5 cm²

Radius of circumscribed circle =

Area of circumscribed circle = πr²

Let radius of first circle = r₁cm

Let radius of second circle = r₂cm

So,

r₁ + r₂ = 140cm …. (i)

2πr₁ – 2πr₂ = 88cm

r₁ - r₂ = 14cm …. (ii)

By adding equation 1 & 2

r₁ = 77cm

From equation 1

77 + r₁ = 140 cm

r₂ = 140 – 77 = 63cm

r₂ = 63cm

So,

Diameter of first circle = 2×r₁ = 2×77 = 154cm

Diameter of second circle = 2×r₂ = 2×63 = 126cm

Area of inscribed circle = 154cm²

= πr² = 154cm²

Radius of inscribed circle = 7cm

(a = side of triangle)

a =

Perimeter of equilateral triangle = 3a

(given)

= 72.66 = 72.7cm²

Total cost of fencing = Rs 2640

Per meter rate of fencing = Rs 12

So,

Circumference of field =

Radius of field = 35m

Area of field = = 3850m²

Cost of plugging 1 m² field = 0.50 Rs

Total cost of plugging the field = 3850×0.50 = Rs 1925.00

_{When a square inscribed in a circle then,}

_{Diameter of circle = diagonal of square}

_{Let side of the square be = a cm}

_{Diagonal of square be =}

Area of square = a² cm²

Diameter of circle =

؞ radius of circle =

Area of circle =

Ratio of area of circle and square =

= π : 2

Total area of rectangular park = 120×100 = 12000 m²

Area of park excluding circular lawn = 8700m²

So,

Area of circular lawn = 1200 – 8700 = 3300m²

= πr² = 3300

r = 32.40 m

Radius of first circle = 8cm

Area of first circle = πr²

Radius of second circle = 6cm

Area of second circle =

Total area =

r² = 100

r = 10cm

Radius of the first circle = 19cm

Circumference of first circle = 2πr

= 2π × 19cm

Radius of second circle = 2πr

=2π × 9cm

Total circumference = 2π × 19 + 2π × 9

= 2π (19+9)

2πr = 176

Area of circle =

Total distance covered = 1km = 100000cm

Distance covered by circular wheel in 1 revolution = circumference of circle

Circumference of circle = 2πr

Total no. of revolution = 450

= 2πr × 450 = 100000

Area enclosed between two concentric circle = 770 cm²

Radius of outer circle = 21cm

Let radius of inner circle = r cm

Area enclosed = area of outer circle – area of inner circle

Area enclosed = 770

π 21² – π r² = 770

π (441 – r²) = 770

r = 14 cm

Given,

Angle = 30°

Radius of circle = 4cm

180° = π radius

Arc length = radius × angle subtended by arc at center

Arc length =

Radius of circle = 5cm

Formula:

Arc length = r×q

r = radius of circle

q = angle subtended by arc at the center

Arc length = 20π cm

Angle subtend at center = 144°

Arc length = radius × angle

Arc length = 15cm

Angle subtend = 45°

Radius of circle = a

Length of arc =

So,

Angle subtended at the center = 45°

Given,

Radius of sector = 4cm

Angle of sector = 30°

Area of sector =

Radius of sector = 8cm

Angle = 135°

Area of sector =

Area of sector =

Area of sector = 5π cm²

Radius = 5cm

5π =

_{Length of the chord = 4cm}

_{Radius of circle = 4cm}

(This chord and radius makes an equilateral triangle)

So,

Q = 60° (in equilateral triangle)

Area of sector =

Given,

Radius of circle = 35cm

Angle subtend by arc = 72°

Length of arc = r×q

Since,

180° = π radius

1° =

Length of the arc =

Area of sector =

= 770 cm²

Given,

Perimeter of sector of circle = 272m

Radius of sector = 5.7m

Perimeter of sector =

(Equation first)

Area of sector = (Second equation)

Put value of from equation first to second,

Given,

Perimeter of sector = 27.2m

Radius of sector = 5.6m

(Equation first)

Area of sector = (Equation second)

Put value of from equation first to equation second

Given,

Radius of sector = 21cm

Angle of sector = 120°

Length of arc =

Area of sector =

Length of minute hand =

Angle subtend by minute hand in 1 minute = = 6°

Angle subtend by minute hand in 5 minute (7-7.05) = 5×6 = 30°

So,

Area described by minute hand in 5 minute =

= 5.5cm²

Given,

Length of minute hand = 10cm

Angle subtend by minute hand in 25 minute (8-8.25) = 25×6 = 150°

So,

Area described by minute hand between (8-8.25) =

Given,

Angle of sector = 56°

Area of sector = 4.4cm²

From formula,

Given,

Radius of circle = 6cm

Length of chord = 10cm

Angle subtend by chord = 110°

I. Circumference of circle = 2πr

= 2×3.14×6 = 37.68cm

II. Are of circle = πr²

= 3.14×6× = 113.1cm²

III. Length of arc = radius × angle subtend

IV. Area of sector =

Angle subtend at centre of circle = θ

Angle OAB = 90°

(At point of contract, tangent is perpendicular to radius)

OAB is right angle triangle

Perimeter of shaded region = AB+ BC+(CA arc)

Area of shaded region = (area of triangle AOB) – (area of sector)

Given,

Radius of the sector = r cm

Angle subtend = θ

Area of sector = A cm²

Perimeter of sector = 50cm

Area of sector =

Perimeter of sector =

(i)

First equation

Second equation

Put value of from equation first to equation second

Area = 25r-r²

The length of minute hand = 14cm

Time = 5 minute

Angle subtend by minute hand at center in 60 minute = 360

In one minute =

In five minute =5×6 = 30°

Area swept in 5 minute =

= 51.30cm²

Given,

Radius of circle = 21cm

Angle subtend by arc = 60°

Length of the arc =

Area of sector formed by arc =

Given: AB is a chord of a circle with centre O and radius 4cm. AB is of length 4cm and divides the circle into two segments.

To find: the area of the minor segment.

Solution:

Radius of circle = 4cm

Length of chord = 4cm

(Hence it makes an equilateral triangle at centre, in which all angle must be = 60°)

Area of sector =

=

Area of equilateral ∆OAB =

=

Area of minor segment = area of sector – area of ∆OAB

_Given,

_{Radius of circle = 4cm}

_{Length of chord = 4cm}

(Hence it makes an equilateral triangle at centre, in which all angle must be = 60°)

Area of sector =

Area of ∆OAB =

Area of minor segment = area of sector – area of ∆OAB

Length of chord PQ = 12cm

Angle subtend at the center = 120°

Let radius of circle = r cm

Area of sector =

Length of triangle POQ = r cos 60

Length of base PQ = 2×RQ

Put value of r in respective place,

Area of minor segment = area of sector – area of ∆POQ

Radius of the circle = 14cm

Angle subtend at center = 90°

By Pythagoras theorem = AB² = OA² + OB²

= 14² +14²

Area of sector OAB =

=

=

Area of triangle AOB =

So area of minor segment – OACB =area of sector – area of triangle

= 154 – 98 = 56cm²

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

= 44×14 – 56 = 560cm²

Length of chord = 10cm

Radius of circle =

(This triangle POQ satisfy Pythagoras theorem)

= PQ² = PO² + OQ²

= 100 = 50 + 50

So,

Angle AOQ = 90°

Area of sector =

Area of triangle POQ =

Area of minor segment =

Radius of circle = 14cm

Angle = 60°

Area of sector =

= 102.57cm²

Area of triangle OAB =

So,

Area of minor segment = 102.57 – 84.77 = 17.80cm²

Given,

AB = 60m

BC = 28m

Area of rectangular portion = 28m×60m = 1680m²

Diameter of semicircle = length of side BC

Radius =

Area of semicircle =

Total area of plot = 1680+308 = 1988m²

Given:

AB = 36m

BC = 24.5m

Area of rectangular portion = 36×24.5 = 882m²

Radius of semicircular portion =

Area of both semicircular portion =

= 471.625

Area of play ground = 882+471.625 = 1353.62

Given,

Circumference of outer circle = 525m

Let radius of outer circle = R₂m

Let radius of inner circle = R₁m

So,

R₂-R₁ = 14 (equation 1)

= 2πR₂ = 525

Put value of R₁ in equation first

= 83.52 – R₁ = 14

= - R₁ = 14 – 83.52

= R₁ = 69.52m

Area of path =

Cost of leveling the path = 6733.76×.50 = Rs 3388

Length of rectangle = 20m²

Breadth of rectangle = 15m²

Area of rectangle = 20×15 = 300m²

Radius of quadrant = 3.5m²

Area of quadrant =

Area of quadrant =

Area of 4 quadrant = = 2×19.25 = 38.50m²

Area of remaining part = (area of rectangle-area of 4 quadrant)

Area of remaining part = 300 – 38.50 = 261.5m²

Given,

Radius of each circle = 5cm

So,

Side of square = 10cm

Area of square = (10)² = 100cm²

Area of each quadrant of circle with radius 5cm =

Area of 4 quadrants =

Area of remaining portion = 100-25π = 21.5cm²

Side of square = 50m

Area of square = (5)² = 2500m²

Radius of quadrant circle = 25m

Area of one quadrant =

Area of 4 quadrants =

So,

Area which left un-grazed = 2500-1964.28 = 535.72m²

Given,

Circumference of park = 352m

Width of road = 7m

Let radius of park = r

2πr = 352

Area of circle =

Radius of circle included path of width, 7m = 56+7 = 63m

Area of circle included path =

So,

Area of path = 12474 – 9856 = 2618m²

Radius of each circle = a meter

If we join the centre of each circle it makes a square of side = 2a

Area of square = (2a)² = 4a²m²

Area of each quadrant of circle =

Area of 4 quadrants =

So,

Area between circles = 4a² – πa²

Side of water tank = 40m

Side of semi circular grassy plots =

Area of one grassy plot =

Area of grassy plots = 4×200π = 800π

Area of grassy plots = 800 × 3.14 = 2512 cm²

Cost of surfing 1m² plot = 1.25 Rs

Cost of surfing 2512m² = 2512 × 1.25 = 3140 Rs

Length of rectangular park = 100m

Breadth of rectangular park = 50m

Radius of flower bed along length of park =

Area of flower bed along length of park =

Radius of flower bed along width =

Area of flower bed along width =

Total area of flower beds = 7850+1962.50 = 4812.50m²

So,

Cost of leveling semicircular flower beds = 9812.50×.60 = Rs 5887.50

Area of inner circle with radius r = πr²

Radius of outer circle = r+h

Area of outer circle = π(r+h)²

Area of circular path with width = h

= π(r+h)² – πr²

By using (a+b)² = a² + b² +2ab

= π(r² + h² + 2rh) – πr²

=πr² + πh² + 2πrh – πr²

= πh(2r+h)… Proved

Given,

Inside perimeter of track = 400m

Length of straight portion = 90m

Width of path = 14m

Total length of straight path = 90+90 = 180m

Remaining length = 400 – 180 = 220m

This length includes two semi circles or a complete circle.

So,

2πr = 220m

Then,

Area of path = (area of rectangles ABCD + rectangle EFGH + two semicircles)

= 14×90+14×90+π [(25+14)² – 35²]

[(a² – b²) = (a+b)(a-b)]

Area of path = 6216m²

Length of outer track = 90+90+2πr

r = 35+14 = 49

= 180 +308 = 488m²

Area of semicircle with diameter = 10cm

Area of triangle AED =

Area of square ABCD = 10×10 = 100cm²

Area of figure excluded triangle = 100-24 = 76 cm²

Total area of figure = 39.28+76 = 115.3 cm²

Area of semicircle ACB =

= area of circle with diameter OD = πr²)

Remaining shaded portion in lower semi circle = 77 – 38.5 = 38.5cm²

Total shaded portion area = 77 + 38.5 = 115.5 cm²

Given,

Area of quadrant OACB =

Area of shaded region = area of quadrant OACB – area of quadrant ODEF

Given,

Side of square = 8cm

Radius of quadrant circle = 1.4 cm

Radius of inner-circle = 4.2

Area of square = (side)² = 8² = 64cm

Area of one quadrant of circle =

Area of one quadrant of circle =

So,

Area of 2 quadrant = 2×1.54 = 3.08cm²

Area of inner circle = πr² = 3.14×4.2×4.2 = 55.44 cm²

Area of shaded portion = area of square – (area of quadrants + area of inner circle)

= 64 – (3.08 + 55.44)

= 64 – 58.52 = 5.48 cm²

Given,

AC = 24 cm

BC = 10 cm

By Pythagoras theorem

AB² = AC² + BC²

= 24²+10² = 576 + 100 = 676

AB = = 26cm

Radius of semi-circle with diameter AB = = 13cm

Area of semi-circle = = 265.33cm²

Area of triangle ABC = = 120cm²

So,

Area of shaded region = area of semi-circle – area of triangle

= 265.33 – 120 = 145.33 cm²

Given,

Side of square = 7cm

Area of square = (side)² = 7² = 49cm²

Area of quadrant OAPC =

Area of shaded region = (area of square – area of quadrant)

= 49 – 38.5 = 10.5 cm²

Given,

Diameter of circular pond = 17.5m

Radius of circular pond = = 8.75m

Radius of outer circle = (radius of inner circle + width of circular path)

= 8.75+2 = 10.25m

Area of circular path = (area of outer circle – area of inner circle)

= π(R² – r²)

= π(R+r)(R-r)

= (10.75+8.75)(10.75-8.75)

= × 19.50 × 2 = 3061.50 m²

Given,

Area of regular hexagon = cm²

From formula

So,

Area of circum circle of regular hexagon = π(side)²

= 3.14×4×4cm² = 50.24cm²

Given,

Perimeter of semi-circle = 72m

Width of path around it = 3.5m

Perimeter of semi-circle = πr+2r

= r +2r = 72

= 22r+14r = 72×7

r = = 14cm

Radius including the width of path(R) = r+3.5 = 14+3.5 = 17.5m

So, area of path =

= 173.25m²

Given,

Radius = 7cm

Side of equilateral triangle = 14cm

Area of circle =

Area of circle =

Area of equilateral triangle =

Area of equilateral triangle =

=

We know that an equilateral triangle always subtend an angle of 60 at centre area of sector =

=

= = 25.666cm²

This area is common in both the figure so,

Area of shaded region = (area of circle + area of equilateral triangle - 2×area of sector)

= (154+84.77-2×25.67)

= (238.77-51.33) = 187.44cm²

Given,

Side of square = 14cm

Radius of each circle = = 7cm

Area of square = (side)² = 14² = 196cm²

Area of 4 quadrants of circle =

Area of shaded region = area of square – area of 4 quadrants

= 196 – 154 = 42cm²

Perimeter of shaded region = ×2πr

So, total perimeter of 4 circles = 4×11 = 44cm