Area Of Circular Region

Since the cow is fastened to the post, it can move in the area encircling the post with the length of the rope as the radius of that circular path.

Here, r = 2.1m

Maximum area = area of the circle

⇒ πr² =value of π =

= 13.86m²

Given, the perimeter of the circle = 35.2cm

We have to find out the radius and the area of that circle

Perimeter = 2 π r, where r is the radius of the circle

⇒ 35.2 = 2 π r

⇒

⇒

Area = πr² where r is the radius of circle

=

=

Given, Area of circular table cloth = 5544cm²

Since the coloring tape would be placed on the edge of the table cloth so to calculate the length of coloring tape required, we will calculate the perimeter of table cloth.

Now, area of a circle = πr², where r is the radius of the circle

⇒ 5544 = πr²

⇒ (∵ π = )

⇒ 1764 = r²

⇒ r =

Perimeter = 2πr, where r is the radius of the circle

⇒ Perimeter

= 264 cm

Therefore, the length of coloring tape required is 264 cm

Given, the cost of fencing = Rs 924

Now cost of fencing = perimeter of playground × rate of fencing ground per meter

⇒ 924 = perimeter × 21

⇒

= 44 m

But, Perimeter = 2πr where r is the radius of the play ground

⇒

= 7 m

Canvas required will be calculated by calculating the area of the circular field

Area of circular field = πr² where r is the radius of the play ground

⇒

⇒ 154 m²

Given, area of circle = 616 m²

But, Area = πr², where r is the radius of circle

⇒πr² = 616

⇒ r² = 196

⇒ r = 14 m

Perimeter of circle = 2π r, where r is the radius of circle

⇒

⇒ 88m

let R be the radius of circle drawn by Palash and r be the radius of circle drawn by Piyali

Given the ratio of their radius = 4:5

⇒

Area of the two circles of Palash and Piyali would be πR² and πr², respectively.

Ratio of the area =

Perimeter of the rectangular shape figure formed by wire = 2(l + b)

Where, l = length and b = breadth of rectangular figure

Perimeter = 2(48 + 40) = 2 × 88 = 176 m

⇒The length of the wire used for making a circle = 176m

Perimeter of circle = 2πr where r is the radius of circle

⇒2πr = 176

⇒

Area of rectangular shape = l × b = 48 × 40 = 1920 m²

Area of circular shape = πr², where r is the radius of circle

⇒

= 2464 m²

Since the area enclosed by circular shape is more, wire bend as circle covers maximum place.

Given, the field is rectangular in shape with length (l) = 60m and breadth (b) = 42m

In the center lies a circular pool with radius (r) = 14m

Since grass is planted within the field we will calculate area of the rectangular field

Area of rectangular field = l × b = 60 × 42 where l and b are length and breadth of rectangular field, respectively.

= 2520 m²

Area of circular pool = πr² where r is the radius of circular pool

=

= 616 m²

The grass will be planted in the area = Area of rectangular field - Area of circular pool

= 2520 – 616 = 1904 m²

Cost of planting grass per meter is Rs75/ m² = area of field × rate of planting per sq. meter

= 1904 × 75 = Rs 1,42, 800

Given, the perimeter of a circular park = 352 m

Width of a road outside circular park = 7 m

Now,

Perimeter of circle = 2πr where r is the radius of cirlce

⇒

⇒ 8 × 7 = 56 m

⇒ r = 56m

Inner radius of the park (r) = 56 m

Outer radius of the Park including the road (R) = width of circular path + r

R = 7 + 56 = 63 m

R = 63 m

Area of the road = π (R² - r²) where R is the outer radius of the circle and r is the smaller radius of circle

= π(R + r) (R - r) [a² - b² = (a-b)(a-b)]

=

=

= 22 × 119

Area of the road = 2618 m²

Cost of concreting the path = area of path × rate of concreting the path per square meter

= 2618 × 20

= Rs 52360

cost of fencing semicircular land = Rs 2664

Rate is fencing = Rs 18.50/m

Let r be the radius of semicircular field

Then, perimeter of this field = (πr + 2r) m where r is the radius of the circle

Cost of fencing semicircular land = perimeter of the semicircular land × rate of ploughing per sq meter

2664 = (πr + 2r) × 18.50

⇒ 2664 = r( π + 2) × 18.50 (taking r common from the bracket)

⇒ (putting π as)

⇒

⇒

⇒ r = 28 m

Perimeter of the semicircular land = r ( π + 2) where r is the radius

=

= 144 m

Area of semicircular field = where r is the radius

=

= 1232 m²

Cost of ploughing the field = area of field × rate of ploughing per square meter

= 1232 × 32 = Rs 39,424

Given, speed of Rajat = 9 m /sec

Where distance is the perimeter of circular field

But perimeter = 2π r where r is the radius of circle

⇒ Time taken =

According to the given condition

Where r is radius of circular field and 2r = diameter

⇒

⇒ 2π r – 270 = 2r (taking LCM and cancelling 9 from both side)

⇒ 2π r – 2r = 270

⇒ r (π -1) = 135

⇒

Area of the circular field = πr² where r is the radius

= m²

= 12474 m²

Let R be the radius of outer circumference and r be the radius of inner circumference

Outer circumference of circular field = 2π R

And inner circumference of circular field = 2π r

Given,

Length of Outer circular field – length of inner circular field = 132m

circumference of circular field - circumference of circular field = 132 m

⇒ 2π (R- r) = 132m

⇒

⇒R – r = 21 m

This is the radius of the circular path

Given, area of path = 14190 m²

⇒ π (R- r)² = 14190

⇒

⇒ (R-r) = √ 4515

⇒ Radius of circular path = 67.19 m

NOTE: Area of square = (side)²

Area of circle = πr², where ‘r’ is radius of the circle.

i) Given, ABCD is a square

The length of the radius of circle = 7 cm

Diameter of the circle = 14cm

Let the side of square ABCD = x cm

By Pythagoras theorem (being a square there is angle of 90° between two adjacent sides)

AB² + BC² = AC²

⇒ x² + x² = 14²

⇒ 2 x² = 196

⇒ x² = 98

⇒ Area of square = x² = 98cm²

Area of circle = πr² where r is the radius of the circle

=

⇒ Area of circle = 154 cm²

Area of shaded region = area of circle – area of square

= 154 – 98 cm²

= 56 cm²

ii) Given, Radius of each circle = 3.5cm

And A,B, C, D are center of each circle

Hence, ABCD forms a square with each side of 7cm length

Area of square = (side)²

= 7² = 49 cm²

Area of each circle = π r², where r is the radius of circle

=

Area of four circle = 4× area of each circle

= 4 × 38.5 = 154 cm²

Area of shaded region = area of four circles – area of square

= 154 – 49 cm²

= 105 cm²

Given, radius of circle = 3.5cm

Now, Area of each quarter sector of the circle

Quarter sector means there is angle of 90° at the center of circle

Length of sector = (∵length of sector =)

Where r is the radius of circle

= 269.5 cm

Perimeter of quarter circle = length of sector + 2 × length of radius

= 269.5 + 7

= 276.5 cm

Area of sector = (∵ area of sector =)

Where r is the radius of the circle

= 471.62 cm²

Given, each side of square = 12cm

⇒ Area of Square ABCD = side²

⇒ (12)²

⇒ 144cm²

Let Radius of each quadrant be r

⇒ r = 6 cm (given)

Now, the sum of the area of the four quadrants at the four corners of the square

⇒ 4 × area of each quadrant ( area of circle = πr²where r is the radius)

⇒ area of quadrant =

⇒

⇒

⇒ 113.14cm²

Now, area of shaded portion

= Area of square, ABCD - sum of the areas of four quadrants at the four corners of the square

⇒ 144 – 113 .14 cm²

= 30 .86 cm²

Given: Area of circle = 154 cm²

We know that

Area of a circle = πr²

⇒ πr² = 154

⇒ r² = 7× 7

⇒ r = 7 cm

Also, Diameter of the circle = 2× radius

⇒ CE = 2× 7 = 14 cm

This acts as the diagonal of the inscribed square.

So, the diagonal of the square BCDE= 14 cm

Let the side of the square be x cm

We know that each angle of a square is 90°.

Using Pythagoras theorem,

CD² +DE² = CE²

⇒x²+ x² = 14²

⇒ 2x² = 196

⇒x² =98

Side of the square =7√2 cm

We know that Area of a Square = side× side

⇒ Area of BCDE =98 cm²

Perimeter of a square = 4× side

⇒ Perimeter of BCDE = 4× 7√2 = 28√2 cm

(i) Given: Radius of the circle = 12 cm and angle subtended by the arc = 90°

We know that the length of the arc

⇒ Length of arc AB

Length of AB:

∵ ∠ O = 90°

Using Pythagoras theorem,

OA² +OB² = AB²

⇒ 12² + 12² = AB²

⇒ AB² = 288

⇒ AB=12√2 cm

⇒ AB =16.92 cm

Perimeter of the circular shaded sector = Length of arc AB + length of AB

⇒ Perimeter of the circular shaded sector=18.857 + 16.92 =35.777 cm

Now, Area of the segment AB = area of sector ABO – area of triangle ABO

We know that the area of the minor sector

⇒ Area of ABO

⇒ Area of ABO = 113.14 cm²

In ∆ABO,

∠ O = 90°, AO = BO = 12 cm {radius of the circle}

⇒ Area of ∆ABO = 72 cm²

∴Area of the segment AB = area of sector ABO – area of triangle ABO

⇒Area of the segment AB = 113.14 – 72

⇒Area of the segment AB = 41.14cm²

(ii) Given: Radius of the circle = 42 cm and angle subtended by the arc = 60°

We know that the length of the arc

⇒ Length of arc

Length of AC:

In ∆ABC,

∠ B = 60°, AB = BC = 42 cm {radius of the circle}

⇒∠ ABC = ∠ ACB {angles opposite to equal sides are equal}

By the angle sum property of the triangle,

∠ BAC + ∠ ACB + ∠ B = 180°

⇒ 2∠ BAC = 180° - 60°

⇒∠ BAC = 60°

Hence, ∆ ABC is an equilateral triangle.

∴ AC = 42 cm

Perimeter of the circular shaded sector = Length of arc AC + length of AC

⇒ Perimeter of the circular shaded sector = 42 + 44 = 86 cm

Now, Area of the segment AC = area of sector ACB – area of triangle ACB

We know that the area of the minor sector

⇒ Area of ACB =

⇒ Area of ACB = 924 cm²

We know that

Area of a equilateral triangle where a is the side of it.

⇒ Area of ∆ABC = 763.83cm²

∴Area of the segment AC = area of sector ABC – area of triangle ABC

⇒ Area of the segment AC = 924 – 763.83

⇒ Area of the segment AC = 160.17 cm²

The bangle is like two concentric circles.

Given: Area of the bangle = 269.5 cm²

Outer diameter of bangle = 28 cm

⇒ Outer radius = 14 cm

Area of the bangle = Area of the larger circle – area of the smaller circle

∵The area of a circle = πr²

⇒ Area of the bangle = πR² - πr²

⇒ 196 - r² = 85.75

⇒ r² = 110.25

⇒ r = 10.5

Inner radius = 10.5 cm

⇒ Inner diameter = 2× 10.5 = 21 cm

Given: ∆ABC is an equilateral triangle with side 10 cm, Arcs centered at B and C have radius 5 cm.

We need to find the area of the colored portion.

Area of colored portion = Area of ∆ABC – Area of sectors centered at B and C.

∵ ∆ABC is an equilateral triangle

We know that

Area of a equilateral trianglewhere a is the side of it.

⇒ Area of ∆ABC = 43.3cm²

For the sectors,

Radius = 5 cm and angle subtended = 60°

{∵∆ABC is an equilateral triangle and each angle of it is 60°}

We know that the area of the minor sector

⇒ Area of sector centered at

⇒ Area of sector centered at B =13.095 cm²

Total area of both the sectors centered at B and C = 13.095 + 13.095 = 26.19 cm²

⇒ Area of colored portion = 43.3 – 26.19 = 17.11 cm²

Given: Length of side of equilateral triangle = 21 cm

Area of the colored portion = Area of the circum circle - Area of ∆BCD

Height of equilateral triangle

The centroid of equilateral triangle is at A and lies on height BE.

⇒ BA = 7√3 cm

So, the radius of the circum circle of this triangle = 7√3 cm.

∵ Area of a circle = πr²

Area of the circum circle

⇒ Area of the circum circle = 462 cm²

We know that

Area of a equilateral triangle where a is the side of it.

⇒ Area of ∆BCD = 190.96cm²

Now, Area of the colored portion = 462 – 190.96

⇒ Area of the colored portion = 271.04 cm²

Given: Area of the circumscribing circular region = 462 cm²

Circle is centered at A.

∵ Area of a circle = πr²

⇒ r = 7√3

So, the radius of the circum circle of this triangle = 7√3 cm.

The centroid of equilateral triangle is at A and lies on height BE.

⇒ BE = 10.5√3 cm

Height of equilateral triangle = 10.5√3 cm

Height of equilateral triangle

⇒ Side of the triangle = 21 cm

Given: Area of inscribing circle = 38.5 cm²

Let BCD is a triangle of perimeter 32 cm. AB, AD and AC are the internal bisectors of respective angles of the triangle. The three internal bisectors meet at A. Perpendiculars drawn from A on sides BC, CD and DB are AG, AE and AF respectively.

AG = AE = AF

Legth of inner radius of triangle = AG

Let AG be r units.

∵ Area of a circle = πr²

⇒ r = 3.5 cm

Now, Area of ∆BCD = area of ∆BCA + area of ∆DCA +area of ∆BDA

⇒ Area ∆BCD = 56cm²

Let BCD is a triangle of sides BC, BD and DC equal to 20 cm, 15 cm, 25 cm. AB, AD and AC are the internal bisectors of respective angles of the triangle. The three internal bisectors meet at A. Perpendiculars drawn from A on sides BC, CD and DB are AG, AE and AF respectively.

AG = AE = AF

Legth of inner radius of triangle = AG

Let AG be r units.

The given triangle forms a right angled triangle with ∠ B = 90°

Using Pythagoras theorem,

BE² +DE² = BD²

⇒ BE² + 12.5² = 15²

⇒ BE² = 68.75

⇒ BE = 8.3 cm

The centroid of triangle is at A and lies on height BE.

⇒ AE = 2.767 cm

r = 2.767 cm

∵ Area of a circle = πr²

⇒ Area of incircle = 24 cm²

∵The given triangle forms a right angled triangle

∴ BC will act as the diameter of the circumcircle centered at O.

Diameter = 25 cm

⇒ Radius = 12.5 cm

∵ Area of a circle = πr²

⇒ Area of the circumcircle

⇒ Area of circumcircle = 491 cm²

Given: Length of side of equilateral triangle = 4√3 cm

Height of equilateral triangle

The centroid of equilateral triangle is at A and lies on height BK.

⇒ BA = 4 cm

So, the radius of the circum circle of this triangle = 4 cm.

Now, diameter of the circle = 2× radius = 8 cm

⇒ IJ = 8 cm

From the figure, IJ = EH = side of the square

⇒ Side of the square = 8 cm

We know that the diagonal of a square =√2× side

⇒ FH = 8√2 cm

It is given that the wire was cut into two equal parts which means perimeter of square and circle will be equal.

Perimeter of a square = 4× side

Circumference of a circle = 2π r

Let s be the side of the square and r be the radius of the circle.

According to the question,

4s = 2π r

⇒ 2s = π r

..(1)

Also, it is given that the area of circle exceeds that of the square by 33 sq. cm.

Area of a circle = πr²

Area of a square = side× side

⇒πr²= s² + 33

From (1),

⇒ r = 7 cm

Circumference of a circle = 2π r

∵ the wire was cut into two equal parts which means perimeter of square and circle will be equal.

Total length of wire = 44 + 44 = 88 cm

Given that area of a circle = X sq unit, circumference = Y unit and diameter = Z unit

We know that

Circumference of a circle = π× diameter

⇒ Y = π Z …(1)

Area of a circle = πr²

…(2)

For a square inscribing the circle,

Diagonal of the square = diameter of the circle

Let the radius of the circle be r

⇒ Diagonal of the circle = 2r

We know that the diagonal of the square = √2 side

⇒√2 side = 2r

⇒Side of the inscribed circle = √2r

Area of the inscribed square = side× side

⇒ Area of the inscribed circle = 2r²

For a square circumscribing the circle,

Side of the square = diameter of the circle

⇒ Side of the circumscribed circle = 2r

Area of the circumscribed square = side× side

⇒Area of the circumscribed circle = 4r²

Ratio of area of two square circumscribe and inscribe by a circle

⇒ Ratio = 2:1

Given that the perimeter and area of a circular field are equal.

We know that circumference of the circle = 2πr

Also, the area of a circle = πr²

According to the question,

2πr = πr²

⇒ r = 2 units

For a square inscribing the circle,

Diagonal of the square = diameter of the circle

⇒ Diagonal of the circle = 2r = 4 units

Let the length of side of equilateral triangle be s units.

Height of equilateral triangle

The centroid of equilateral triangle is at A and lies on height BE.

∵ Area of a circle = πr²

Area of the circum circle

⇒ Area of the circum circle

We know that

Area of a equilateral triangle, where a is the side of it.

So the ratio of the area of the equilateral triangle and the circle

Given: Inner diameter = 20 cm

Outer diameter = 22 cm

⇒ Inner radius = 10 cm and outer radius = 11 cm

Area of such concentric circles = Area of the larger circle – area of the smaller circle

∵The area of a circle = πr²

⇒ Area of the iron plate = πR² - πr²

⇒ Area of the iron plate

⇒ Area of the iron plate

⇒ Area of the iron plate

⇒ Area of the iron plate

⇒ Area of the iron plate = 22× 3 = 66cm²

Let the radius of the circular field be r units.

The area of a circle = πr²

Now, increasing the radius to 10%

New radius =1.1r

Area of new circle = π(1.1r)²

⇒ Area of new circle = 1.21πr²

Increase in area =1.21πr² - πr² =0.21πr²

Per centage increase in the area of circular field

⇒ Percentage increase = 21%

Let the radius of the circular field be r units.

Circumference = 2π r

The area of a circle = πr²

Now, decreasing the perimeter to 50%

New perimeter =3πr

⇒ New radius = 1.5r

Area of new circle = π(1.5r)²

⇒ Area of new circle = 2.25πr²

Increase in area =2.25πr² - πr² =1.25πr²

Per centage increase in the area of circular field

⇒ Percentage increase = 125%

Given that the length of the circular field = r meter

The area of the circular field = πr²

Also, given that the area of the other circle = xπr²

We know that area of the circle = π(radius)²

Then, π ×(radius of other circle)² = xπr²

⇒ (Radius of the other circle)² = x r²

⇒ Radius of the other circle = r √x meter

By the pythagoras theorem,

3² + 4² = 9+16 = 25

And 5² = 25

⇒ The given triangle forms a right angled triangle

∴ BC will act as the diameter of the circumcircle centered at O.

Diameter = 5 cm

⇒ Radius = 2.5 cm

∵ Area of a circle = πr²

⇒ Area of the circumcircle

⇒ Area of circumcircle = 19.64 cm²

Given ratio of diameter of three circles =3:5:7

⇒ Ratio of their radius = 3:5:7

∵ Area of a circle = πr²

⇒ Ratio of their areas = 3× 3 : 5× 5 : 7× 7

⇒ Ratio of their areas = 9 : 25 : 49