Areas Related To Circles

Given: Radius (r₁) of 1st circle = 19 cm

Radius (r₂) or 2nd circle = 9 cm

Now,

Let the radius of 3rd circle be r

Circumference of 1st circle = 2πr₁ = 2π (19) = 38π cm

Circumference of 2nd circle = 2πr₂ = 2π (9) = 18π cm

Circumference of 3rd circle = 2πr

Given:

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

2πr = 38π + 18π = 56π cm

r =

= 28

Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm

Radius (r1) of 1st circle = 8 cm

Radius (r2) of 2nd circle = 6 cm

Let the radius of 3rd circle be r

Area of 1^stcircle = r1² = π (8)² = 64π

Area of 2^nd circle = πr2² = π (6)² = 36 π

Given that,

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

πr² = πr1² + πr2²

πr² = 64 π + 36 π

πr² = 100 π

r² = 100

r =

But we know that the radius cannot be negative. Hence, the radius of the circle having area equal to the sum of the areas of the two circles is 10 cm

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

Given:

Each circle is 10.5 cm wider than the previous circle.

Hence,

Radius (r₂) of 2nd circle = 10.5 + 10.5

21 cm

Radius (r₃) of 3rd circle = 21 + 10.5= 31.5 cm

Radius (r₄) of 4th circle = 31.5 + 10.5= 42 cm

Radius (r₅) of 5th circle = 42 + 10.5= 52.5 cm

Area of gold region = Area of 1^stcircle

= πr1²

= π (10.5)²

= 346.5 cm²

Area of red region = Area of 2nd circle - Area of 1^stcircle

= πr₂² – πr₁²

= π (21)² – π (10.5)²

= 441π – 110.25π

= 330.75π

= 1039.5 cm²

Area of blue region = Area of 3^rdcircle - Area of 2^ndcircle

= πr₃² – πr2²

= π (31.5)² – π (21)²

= 992.25 π - 441 π

= 551.25 π

= 1732.5 cm²

Area of black region = Area of 4^thcircle - Area of 3rd circle

= πr₄² – πr₃²

= π (42)² – π (31.5)²

= 1764 π – 992.25 π

= 771.75 π

= 2425.5 cm²

Area of white region = Area of 5^thcircle - Area of 4^thcircle

= πr₅² – πr₄²

= π (52.5)² – π (42)²

= 2756.25 π - 1764 π

= 992.25 π

= 3118.5 cm²

Therefore, areas of gold, red, blue, black, and white regions are 346.5 cm², 1039.5 cm², 1732.5 cm², 2425.5 cm², and 3118.5 cm² respectively

The diameter of the wheel of the car = 80 cm

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

Circumference of wheel = 2πr

⇒ 2π (40) = 80π cm

Speed of car = 66 km/hour

= 110000 cm/min
As distance= speed × time

Distance traveled by 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)
(In one revolution of the wheel, a wheel covers the distance equal to its circumference)

= Distance travelled in 10 minutes

n × 80 π = 1100000

n =

=

= 4375

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

Let the radius of the circle be r

Circumference of circle = 2πr

Area of circle = πr²

Given that, the circumference of the circle and the area of the circle are equal.

This implies 2πr = πr²

2 = r

Therefore, the radius of the circle is 2 units

Hence, the correct answer is A

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

Area of sector of angle θ = ×πr²

Area of sector OACB = × (6)²

=

= cm²

Hence,

The area of the sector of the circle making 60° at the centre of the circle is cm²

To find: Area of the quadrant of the circle
Given: Circumference of the circle
Let the radius of the circle be r

Circumference = 22 cm

Therefore, 2πr = 22

Quadrant of circle means 1/4 th of the circle.

Area of such quadrant of the circle =

=

=

= cm²
Hence, The area of the quadrant of the circle is cm²

So 15 minutes = 90º

5 min = 5 × 6°
= 30°
So 5 minutes subtend an angle of 30º.

Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.
Area swept by min hand = Area of sector

Let AB be the chord of the circle subtending 90° angle at centre O of the circle.



= 3.14 × 25
= 78.5 cm²

OAB is a right angles triangle,
Area of triangle OAB = 1/2 (base x height)
Area of ΔOAB = x OA x OB

= x 10 x 10

= 50 cm²

Area of minor segment ACB = Area of minor sector OACB -Area of ΔOAB

= 78.5 - 50

= 28.5 cm²
ii) Let AB be the chord of the circle subtending 90° angle at centre O of the circle.

Area of major sector OACB =

=3/4 x 3.14 x 10 x 10

=235.5cm²

Radius (r) of circle = 21 cm

The angle subtended by the given arc = 60°

Length of an arc of a sector of angle θ =

(i) Length of arc ACB =

= × 2 × 22 × 3

= 22 cm

(ii) Area of sector OACB =


=

= × × 21 × 21

= 231 cm²

(iii) In ΔOAB,

⇒ ∠OAB = ∠OBA (Angles opposite to equal sides are equal)

⇒ ∠OAB + ∠AOB + ∠OBA = 180°

⇒ 2∠OAB + 60° = 180°

⇒ ∠OAB = 60°

Hence,

ΔOAB is an equilateral triangle

Area of equilateral triangle = (Side)²

⇒ Area of ΔOAB = (Side)²

= × (21)²

= cm²

Area of segment ACB = Area of sector OACB - Area of ΔOAB

= (231 - ) cm²

Radius (r) of circle = 15 cm

Area of sector OPRQ = (60/360)×Πr²

= 1/6 × 3.14 × (15)²

= 117.75 cm²

In ΔOPQ,

∠OPQ = ∠OQP (As OP = OQ)

∠OPQ + ∠OQP + ∠POQ = 180°

2 ∠OPQ = 120°

∠OPQ = 60°

ΔOPQ is an equilateral triangle.

Area of ΔOPQ = ×(Side)²

= × (15)²

=

= 56.25 ×

= 97.3125 cm²

Area of segment PRQ = Area of sector OPRQ - Area of ΔOPQ

= 117.75 - 97.3125

= 20.4375 cm²

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

= Π(15)² – 20.4375

= 3.14 ×225 – 20.4375

= 706.5 – 20.4375

= 686.0625 cm²

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

SV = VT

In ΔOVS,
As,

= Cos 60^o

=

OV = 6 cm

= Sin 60^o

=

SV = 6√3 cm

ST = 2 × SV

= 2 × 6√3

= 12√3 cm

Area of ΔOST = × 12√3 × 6

= 36√3

= 36 × 1.73

= 62.28 cm²

Area of sector OSUT = × π × (12)²

= × 3.14 × 144

= 150.72 cm²

Area of segment SUT = Area of sector OSUT - Area of ΔOST

= 150.72 - 62.28

= 88.44 cm²

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

= × r²

= × 3.14 × 5 × 5

= 19.625 m²

The area that can be grazed by the horse when the length of rope is 10 m long = × (10)²

= × 3.14 × 100

= 78.5 m²

(ii) Increase in grazing area = Area grazed by a horse now - Area grazed previously
(78.5 - 19.625)

= 58.875 m²

(i) To Calculate total length of wire required to make brooch, we need to find Circumference of Brooch and the length of 5 diameters of Brooch.

Diameter of Circle = 35 mm



Radius of Circle = 35/2 mm

Circumference of Circle = 2 π r



Total Length of wire required = Circumference of Circle + 5 x Diameter of Circle

Total Length of wire required = 110 + 5 x 35

Total Length of wire required = 285 mm

(ii)

A complete Circle subtends an angle of 360º

10 sectors = 360º

1 sector will subtend = 36º

There are 8 ribs in the given umbrella. The area between two consecutive ribs is subtending = 45^o at the centre of the assumed flat circle

Now we know that area of a sector of a circle subtending an angle x is given by

It can be seen from the figure that each blade of wiper would sweep an area of a sector of 115° in a circle with 25 cm radius

Area of sector =



Area swept by 2 blades :


= 1254.96 cm²

=1255 cm² (approx)

It can be seen from the figure that the lighthouse spreads light across a sector of 80° in a circle of 16.5 km radius

Area of sector OACB = × r²

= × 3.14 × 16.5 × 16.5

= 189.97 km²

It can be concluded that these designs are segments of the circle.

Let us take segment APB.

Chord AB is a side of the regular hexagon.

And,

Each chord will substitute = 60^o at the centre of the circle

In ΔOAB,

∠OAB = ∠OBA (As OA = OB)

∠AOB = 60°

∠OAB + ∠OBA + ∠AOB = 180°

2∠OAB = 180° - 60°
= 120°

∠OAB = 60°

Hence,

ΔOAB is an equilateral triangle.

Area of ΔOAB = (Side)²

= × (28)²

= 196

= 333.2 cm²

Area of sector OAPB = × r²

= × × 28 × 28

= cm²

Now,

Area of segment APB = Area of sector OAPB - Area of ΔOAB

= ( - 333.2) cm²

Area of design = 6 × ( - 333.2)

= 2464 – 1999.2

= 464.8 cm²

Cost of making 1 cm² designs = Rs 0.35

Cost of making 464.76 cm² designs = 464.8 × 0.35

= Rs 162.68

Hence, the cost of making such designs would be Rs 162.68.

Area of sector of angle θ = ×R²

Where θ = angle , r = radius of circle

Here θ = p and radius = R

Area of sector of angle p = (R²)

Area of sector of angle p = ()(2R²)

Hence, (D) is the correct answer

As, ΔPQR is in the semicircle and we know angle in the semicircle is a right angle, therefore PQR is a right-angled triangle.
[ If an angle is inscribed in a semi-circle, that angle measures 90 degrees.]
Here, QR is the hypotenuse of ΔPQR as lines from two ends of diameter always make a right angle when they meet at the circumference of that circle.

Hence by Pythagoras theorem we get,
Pythagoras Theorem : It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

QR² = PQ² + PR²

QR² = 24² + 7²

QR² = 576 + 49

QR = 25 cm

Or, radius = 12.5cm

Area of right angled triangle triangle =

= 12 × 7
= 84 cm²

= 245.3125 sq cm

Hence,

Area of shaded region = Area of semicircle - area of ΔPQR
= 245.3125 – 84

Area of shaded region = 161.3125 sq cm

Area of the shaded region = Area of Bigger sector – Area of Smaller Sector

If r₂ and r₁ are the two radii of Bigger and Smaller circles respectively, then:

Area of bigger sector = 40/360 πr₂²

Area of smaller sector = 40/360 πr₁²

It can be observed from the figure that the radius of each semi-circle is 7 cm

Area of each semi-circle =r²

= * * (7)²

= 77 cm²

Area of square ABCD = (Side)²

= (14)²

= 196 cm²

Area of the shaded region= Area of square ABCD - Area of semi-circle APD - Area of semi-circle BPC

= 196 - 77 - 77

= 196 - 154

= 42 cm²

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

Area of sector = r²

= x x 6 x 6

= cm²

Area of triangle OAB = x (12)²

= 36 cm²

Area of circle = πr²

= × 6 × 6

= cm²

Area of shaded region = Area of ΔOAB + Area of circle - Area of sector

= -

= (36 + ) cm²

Each quadrant is a sector of 90° in a circle of 1 cm radius

Area of each quadrant = * r²

= * * (1)²

= cm²

Area of square = (Side)²

= (4)²

= 16 cm²

Area of circle = πr² = π (1)²

= cm²

Area of the shaded region = Area of square - Area of circle - 4 × Area of quadrant

= 16 - - 4 *

= 16 - -

= 16 -

=

= cm²

Radius of the circle "R" = 32 cm

Draw a median AD of the triangle passing through the centre of the circle.
⇒ BD = AB/2

Since, AD is the median of the triangle

∴ AO = Radius of the circle = 2/3 AD [ By the property of equilateral triangle inscribed in a circle]

⇒ 2/3 AD = 32 cm

⇒ AD = 48 cm
In ΔADB,

By Pythagoras theorem,

AB² = AD² + BD²

⇒ AB² = 48² + (AB/2)²

⇒ AB² = 2304 + AB²/4
⇒ 3/4 (AB²)= 2304

⇒ AB² = 3072

⇒ AB= 32√3 cm
Area of ΔABC = √3/4 × (32√3)² cm²
= 768√3 cm²

Area of circle = π R²
= 22/7 × 32 × 32
= 22528/7 cm²

Area of the design = Area of circle - Area of ΔABC

= (22528/7 - 768√3) cm²

To Find: Area of shaded region

Given: Side of square ABCD = 14 cm

Radius of circles with centers A, B, C and D = 14/2 = 7 cm


Area of shaded region = Area of square - Area of four sectors subtending right angle


Area of each of the 4 sectors is equal to each other and is a sector of 90° in a circle of 7 cm radius. So, Area of four sectors will be equal to Area of one complete circle

Area of 4 sectors = Πr²

Area of square ABCD = (Side)²

Area of square ABCD = (14)²

Area of square ABCD = 196 cm²

Area of shaded portion = Area of square ABCD - 4 × Area of each sector

= 196 – 154

= 42 cm²

Therefore, the area of shaded portion is 42 cm²

(i) Distance around the track along its inner edge = AB + semicircle BEC + CD + semicircle DFA

= 106 + ( × 2πr )+ 106 +( × 2πr) [Perimeter of semicircle is half of circle and perimeter of circle is 2πr)

= 212 +( × 2 × × 30) +( × 2 × × 30)

= 212 + (2 × ×30)

=

= m

(ii) Area of the track = (Area of GHIJ - Area of ABCD) + (Area of semi-circle HKI - Area of semi-circle BEC) + (Area of semi-circle GLJ - Area of semi-circle AFD)
Area of semicircle is half of circle and area of circle is πr²
Also area of rectangle = length x breadth

= (106 × 80 – 106 × 60 )+( × × (40)² - × × (30)²) +( × × (40)² - × × (30)² )

=[ 106 (80 – 60) ]+[ × (40)² - × (30)²]

= [106 (20)] +[ (40)² – (30)²]

= 2120 +[ (40 – 30) (40 + 30)]

= 2120 + (22/7) (10) (70)

= 2120 + 2200

= 4320 m²

Radius (r₁) of larger circle = 7 cm
As OD = r₂ = 7 cm

Radius (r₂) of smaller circle = cm

Area of smaller circle = π (r₂)²

= x x

= cm²

Area of semi-circle ACB = 1/2 Πr₁²

= x x (7)²

= 77 cm²

Area of triangle ABC = x AB x OC
As AB = OB + OA
⇒ AB = 7 + 7
⇒ AB = 14 cm

Area of triangle ABC = x 14 x 7

= 49 cm²

Area of the shaded region

= Area of smaller circle + Area of semi-circle ACB - Area of ΔABC

= + 77 – 49

= 28 +

= 28 + 38.5

= 66.5 cm²

Let the side of the equilateral triangle be a

Area of equilateral triangle = 17320.5 cm²

(a)² = 17320.5

(a)² = 17320.5

a² = 4 × 10000

a = 200 cm

Each sector is of measure 60° as each angle of an equilateral triangle is 60º.

Area of each sector = × π × r²

= × π × (100)²

=

= cm²

Area of shaded region = Area of equilateral triangle - 3 × Area of each sector

= 17320.5 – 3 *

= 17320.5 – 15700

= 1620.5 cm²

Radius of circle is 7 cm.
Diameter of the circle is 14 cm.
As 3 circles are there on one side of the square.
The side of the square will be 14 × 3 = 42 cm

Area of square = (Side)²

= (42)²

= 1764 cm²

Area of each circle = πr²

= × (7)²

= 154 cm²

Area of 9 circles = 9 × 154

= 1386 cm²

Area of the remaining portion of the handkerchief = 1764 - 1386

= 378 cm²

(i) Since OACB is a quadrant, it will subtend 90° angle at O

Area of quadrant OACB = * r²

= * * (3.5)²

= * * ()²

=

= cm²

(ii) Area of ΔOBD = * OB * OD

= * 3.5 * 2

= * * 2

= cm²

Area of the shaded region = Area of quadrant OACB - Area of ΔOBD

= -

=

= cm²

In ΔOAB,

OB² = OA² + AB²

= (20)² + (20)²

= 20

Radius (r) of circle = 20 cm

Area of quadrant OPBQ = × 3.14 × (20)²

= × 3.14 × 800

= 628 cm²

Area of OABC = (Side)²

= (20)²

= 400 cm²

Area of shaded region = Area of quadrant OPBQ - Area of OABC

= (628 - 400)

= 228 cm²

Area of the shaded region = Area of sector OAEB - Area of sector OCFD

= * * (21)² - * * (7)²

= * [(21)² – (7)²]

= * [(21 – 7) (21 + 7)]

=

= cm²

As quadrant is 1/4 th of a circle and area of circle is Πr²
Area of quadrant =

=154 cm²

Area of triangle=

= 98 cm²
In right angle triangle ABC,
BC² = AB² + AC²
BC² = (14)² + (14)²
BC² = 2(14)²
BC= 14 √2 cm
Hence diameter is 14 √2 cm.
Radius is (14 √2)/2 = 7 √2 cm
So, area of the semicircle = 1/2 πr²

= 154 cm²

Area of the shaded region = Area of semi Circle - Area of segment BC = 154 - 56

= 98 cm²

First, let us understand how we can acquire the given Area
From the figure first we need to subtract Area of quadrant from the square and then again subtract another quadrant from the square, by this, we will get the Area of the unshaded region
And finally to find Area of the shaded region we can subtract Area of the unshaded region from Area of square.
Radius of quadrant = length of square
= r
= 8 cm
We can interpret from above that:
Area of unshaded region = 2(Area of the square - Area of the quadrant of the circle)
Area of shaded region = Area of square - 2(Area of square - Area of quadrant of circle)

Area of shaded region = 2 (Area of quadrant of circle) - Area of square

= 2 × (1/4)πr² - 8 × 8

= [(1/2) × π × 8²] - 64

= 32π - 64

= 32(π-2)

= 32 × 1.14

= 36.54 cm²

Therefore, area of shaded region is 36.54 cm²