Area Of Circle, Sector And Segment

Given:

Difference between the circumference and the radius of circle = 37 cm

Let the radius of the circle be ‘r’.

Circumference of the circle = 2πr

So, Difference between the circumference and the radius of the circle = 2πr – r = 37

2πr – r = 37

2 × × r – r = 37

× r – r = 37

r = 37 ×

r = 7 cm

∴ Circumference of circle = 2 × × 7

= 2 × 22

= 44 cm

Hence the circumference of the circle is 44 cm.

Given:

Circumference of circle = 22 cm

Let the radius of the circle be ‘r’.

∵ Circumference of circle = 2πr

∴ 22 = 2 × π × r

⇒ 22 = 2 × × r

⇒ 22 × × = r or = r

or r =

∵ Area of circle = πr²

∴ Area of its quadrant = πr²

=

=

Hence the area of the quadrant of the circle iscm.

Given:

Let the two circles be C₁ and C₂ with diameters 10 cm and 24 cm respectively.

Area of circle, C = Area of C₁ + Area of C₂ …… (i)

∵ Diameter = 2 × radius

∴ Radius of C₁, r₁ = = 5 cm

and Radius of C₂, r₂ = = 12 cm

∵ Area of circle = πr² …… (ii)

∴ Area of C₁ = πr₁²

=

=

= cm²

Similarly, Area of C₂ = πr₂²

=

= 22/7 × 144

= cm²

∴ Using equation (i), we have

Area of C = +

= cm²

Now, using equation (ii), we have

× r² =

r² =

r² = 169

r =

r = 13 cm

⇒ Diameter = 2 × r

= 2 × 13

= 26 cm

Hence, the diameter of the circle is 26 cm.

Given:

Area of circle = 2 × Circumference of circle …… (i)

Let the radius of the circle be ‘r’.

Then, the area of the circle = πr²

and the circumference of the circle = 2πr

Using (i), we have

πr² = 2 × 2πr

πr² = 4πr

r = 4 cm

∵ Diameter = 2 × radius

∴ Diameter = 2 × 4

= 8 cm

Hence, the diameter of the circle is 8 cm.

Given:

Perimeter of square circumscribes a circle of radius ‘a’.

Side of square = Diameter of circle

Diameter of circle = 2 × radius

= 2a

So, Side of square = 2a

∵ Perimeter of square = 4 × side

∴ Perimeter of square = 4 × 2a

= 8a

Hence, the perimeter of the square is 8a.

Given:

Diameter of circle = 42 cm

⇒ Radius of circle = cm = 21 cm

Angle subtended at the centre = 60°

∵ Length of arc = × 2πr

=

= 22 cm

Hence, the length of the arc is 22 cm.

Given:

Let the two circles with radii 4 cm and 3 cm be C₁ and C₂ respectively.

⇒ r₁ = 4 cm and r₂ = 3 cm

Area of circle, C = Area of C₁ + Area of C₂ …… (i)

∵ Area of circle = πr² …… (ii)

∴ Area of C₁ = πr₁²

=

= × 16 = cm²

Similarly, Area of C₂ = πr₂²

= × 3 × 3

= × 9 = cm²

So, using (i), we have

Area of C = + = cm²

Now, using (ii), we have

πr² =

× r² =

r² = × = 25

r = √25 = 5

r = 5 cm

∵ Diameter = 2 × radius

∴ Diameter = 2 × 5 = 10 cm

Hence, diameter of the circle with area equal to the sum of two circles of radii 4 cm and 3cm is 10 cm.

Given:

Circumference of circle = 8π

∵ Circumference of a circle = 2πr

∴ 8π = 2πr

r = 4

∵ Area of circle = πr²

∴ Area of circle = π × 4 × 4

= 16π

Hence, the area of the circle is 16π.

Given:

Diameter of the semicircular protractor = 14 cm

Radius of the protractor = cm = 7 cm

∵ Perimeter of semicircle = πr + d

∴ Perimeter of semicircular protractor = × 7 + 14 = 22 + 14

= 36 cm

Hence, the perimeter of the semicircular protractor is 36 cm.

Given:

Perimeter of circle = Area of circle …… (i)

∵ Perimeter of circle = 2πr and Area of circle = πr²

∴ Using (i), we have

2πr = πr²

2 =

2 = r or r = 2

Hence, the radius of the circle is 2 cm.

Given:

Radius of one of the circles, C₁ = 19 cm = r₁

Radius of the other circle, C₂ = 9 cm = r₂

Let the other circle be C with radius ‘r’.

Circumference of C = Circumference of C₁ + Circumference of C₂ …………(i)

∵ Circumference of circle = 2πr

∴ Circumference of C₁ = 2πr₁ = 2 × × 19 =

and Circumference of C₂ = 2πr₂ = 2 × × 9 =

Using (i), we have

2πr = + =

2 × × r =

r = × × = 28

r = 28 cm

Hence, the radius of the circle is 28 cm.

Given:

Radius of one of the circles, C₁ = 8 cm = r₁

Radius of the other circle, C₂ = 6 cm = r₂

Let the other circle be C with radius ‘r’.

Area of C = Area of C₁ + Area of C₂ …… (i)

∵ Area of circle = πr²

∴ Area of C₁ = πr₁² = × 8 × 8 =

and Area of C₂ = πr₂² = × 6 × 6 =

Using (i), we have

πr² = + =

× r² =

r² = × = 100

r² = 100

r = √100 = 10 or r = 10

Hence, the radius of the circle is 10 cm.

Given:

Radius of circle = 6 cm

Angle of the sector = 30°

∵ Area of sector = × πr²

= × 3.14 × 6 × 6

= 3 × 3.14 = 9.42 cm²

Hence, the area of the sector is 9.42 cm².

Given:

Radius of circle = 21 cm

Angle subtended by the arc = 60°

∵ Length of arc = × 2πr

= × 2 × × 21 = 22 cm

Hence, the length of the arc is 22 cm.

Given:

Ratio of circumferences of two circles = 2:3

Let the two circles be C₁ and C₂ with radii ‘r₁’ and ‘r₂’.

∵ Circumference of circle = 2πr

∴ Circumference of C₁ = 2πr₁

and Circumference of C₂ = 2πr₂

⇒ =

⇒ =

Squaring both sides, we get

⇒ =

Multiplying both sides by ‘’, we get

⇒ =

∵ Area of circle = πr²

⇒ =

Hence, the ratio between the areas of C₁ and C₂ is 4:9.

Given:

Ratio of areas of two circles = 2:3

Let the two circles be C₁ and C₂ with radii ‘r₁’ and ‘r₂’.

∵ Area of circle = πr²

∴ Area of C₁ = πr₁²

and Area of C₂ = πr₂²

⇒ =

⇒ =

Taking square root on both sides, we get

⇒ =

⇒ =

Multiplying and dividing L.H.S. by ‘π’, we get

⇒ =

Multiplying and dividing L.H.S. by ‘2’, we get

⇒ =

As Circumference of circle = 2πr

⇒ =

Hence, the ratio between the circumferences of C₁ and C₂ is 2:3.

Given:

A square is inscribed in a circle.

Let the radius of circle be ‘r’ and the side of the square be ‘x’.

⇒ The length of the diagonal = 2r

∵ Length of side of square =

∴ Length of side of square = = √2r

Area of square = side × side = x × x = √2r × √2r = 2r²

Area of circle = πr²

Ratio of areas of circle and square = = =

Hence, the ratio of areas of circle and square is π:2.

Given:

Circumference of circle = 8 cm

Central angle = 72°

∵ Circumference of a circle = 2πr

∴ 2πr = 8

2 × × r = 8

r = 8 × ×

r = cm

∵ Area of sector = × πr²

=

= 1.02 cm²

Given:

Angle made by the pendulum = 30°

Length of the arc made by the pendulum = 8.8 cm

Then the length of the pendulum is equal to the radius of the sector made by the pendulum.

Let the length of the pendulum be ‘r’.

∵ Length of arc = × 2πr

∴ We have,

× 2πr = 8.8

× 2 × 3.14 × r = 8.8

r = 8.8 ×

r = 16.8 cm

Hence, the length of the pendulum is 16.8 cm.

Given:

Length of minute hand = 15 cm

Here, the length of the minute hand is equal to the radius of the sector formed by the minute hand.

Angle made by the minute hand in 1 minute = = 6°

Angle made by the minute hand in 20 minutes = 20 × 6 = 120°

Here, the area swept by the minute hand is equal to the area of the corresponding sector made.

∵ Area of sector = × πr²

= × 3.14 × 15 × 15 = 235.5 cm²

Hence, the area swept by it in 20 minutes is 235.5 cm².

Given:

Angle of the sector = 56°

Area of the sector = 17.6 cm²

Let the radius of the circle be ‘r’.

∵ Area of sector = × πr²

∴ 17.6 = × × r²

r² = × × 17.6

r² = 36

r = √36

r = 6 cm

Hence, the radius of the circle is 6 cm.

Given:

Radius of the circle = 10.5 cm

Area of the sector = 69.3 cm²

∵ Area of the sector = × πr²

∴ 69.3 = × × 10.5 × 10.5

θ = 69.3 × 360 × × ×

θ = 72°

Hence, the central angle is 72°.

Given:

Radius of circle = 6.5 cm

Perimeter of sector = 31 cm

Now, Perimeter of sector = 2 × radius + Length of arc

∵ Length of arc = × 2r × 2πr

∴ Perimeter of sector = 2 × r + × 2r × π

= 2r × [1 + × π]

31 = 2 × 6.5 × [1 + × ]

31 = 13 × [1 + × ]

= 1 + ×

- 1 = ×

= ×

θ = × 360 × …………………. (i)

∵ Area of sector = × πr²

∴ using (i), we have

Area = × 360 × × × × 6.5 × 6.5

= 18 × 3.25 = 58.5 cm²

Hence, the area of the sector is 58.5 cm².

Given:

Radius of circle = 17.5 cm

Length of arc = 44 cm

∵ Length of arc = × 2πr

∴ 44 = × 2 × × 17.5

θ = 44 × 360 × × ×

θ = = 144°

Now, Area of sector = × πr²

= × × 17.5 × 17.5 = 385 cm²

Hence, the area of the sector is 385 cm².

Given:

Length of the rectangular cardboard = 14 cm

Breadth of the rectangular cardboard = 7 cm

∵ Area of rectangle = length × breadth

∴ Area of cardboard = 14 × 7 = 98 cm²

Let the two circles with equal radii and maximum area have a radius of ‘r’ cm each.

Then, 2r = 7

r = cm

∵ Area of circle = πr²

∴ Area of two circular cut outs = 2 × πr²

= 2 × × ×

= 11 × 7 = 77 cm²

Thus, the area of remaining cardboard = 98 – 77 = 21 cm²

Hence, the area of remaining cardboard is 21 cm².

Given:

Side of the square = 4 cm

Radius of the quadrants at the corners = 1 cm

Radius of the circle in the centre = 1 cm

∵ 4 quadrants = 1 circle

∴ There are 2 circles of radius 1 cm

Area of square = side × side

= 4 × 4 = 16 cm²

Area of 2 circles = 2 × πr²

= 2 × × 1 × 1 = cm²

∵ Area of shaded region = Area of square – Area of 2 circles

= 16 -

= = cm² = 9.7 cm²

Hence, the area of shaded region is 9.72 cm².

Given:

Length of rectangular sheet of paper = 40 cm

Breadth of rectangular sheet of paper = 28 cm

Radius of the semicircular cut out = 14 cm

∵ Area of rectangle = length × breadth

∴ Area of rectangular sheet of paper = 40 × 28

= 1120 cm²

∵ Area of semicircle = πr²

∴ Area of semicircular cut out = × × 14 × 14

= 22 × 14 = 308 cm²

Thus, the area of remaining sheet of paper = Area of rectangular sheet of paper – Area of semicircular cut out

= 1120 – 308 = 812 cm²

Hence, the area of remaining sheet of paper is 812 cm².

Given:

Side of square = 7 cm

Radius of the quadrant = 7 cm

Area of square = side × side

= 7 × 7 = 49 cm²

∵ Area of circle = πr²

∴ Area of a quadrant = πr²

= × × 7 × 7

= = 38.5 cm²

Thus, the area of shaded region = Area of square – Area of quadrant

= 49 – 38.5 = 10.5 cm²

Hence, the area of the shaded region is 10.5 cm².

Given:

Radius of circle = 7 cm

Let the sectors with central angles 80°, 60° and 40° be S₁, S₂, and S₃ respectively.

Then, the area of shaded region = Area of S₁ + Area of S₂ + Area of S₃ …………………….. (i)

∵ Area of sector = × πr²

∴ Area of S₁ = × × 7 × 7

= cm²

Similarly, Area of S₂ = × × 7 × 7

= cm²

and Area of S₃ = × × 7 × 7

= cm²

Thus, using (i), we have

Area of shaded region = + +

=

= = 77 cm²

Hence, the area of shaded region is 77 cm².

Given:

Radius of inner circle = 3.5 cm

Radius of outer circle = 7 cm

∠POQ = 30°

Let the sector made by the arcs PQ and AB be S₁ and S₂ respectively.

Then, Area of shaded region = Area of S₁ – Area of S₂ ………….(i)

∵ Area of sector = × πr²

∴ Area of S₁ = × × 7 × 7

= cm²

Similarly, Area of S₂ = × × 3.5 × 3.5

= cm²

Thus, using (i), we have

Area of shaded region = -

=

= = cm²

Hence, the area of shaded region iscm².

Given:

Side of square = 14 cm

Diameter of each semicircle = 14 cm

Radius of each semicircle = = 7 cm

∵ Both the semicircles have same radius.

∴ We consider one circle of radius 7 cm.

Area of shaded region = Area of square – Area of circle ….……. (i)

Area of square = side × side

= 14 × 14 = 196 cm²

Area of circle = πr²

= × 7 × 7 = 22 × 7 = 154 cm²

Thus, using (i), we have

Area of shaded region = 196 – 154 = 42 cm²

Hence, the area of shaded region is 42 cm².

Give:

Radius of the circle = 42 cm

Central angle of the sector = ∠AOB = 90°

Perimeter of the top of the table = Length of the major arc AB + 2 × radius …………………. (i)

Length of major arc AB = × 2πr

= × 2 × × 42

= × 2 × 22 × 6

= × 264 = 3 × 66 = 198 cm

Thus, using (i), we have

Perimeter of the top of the table = 198 + 2 × 42

= 198 + 84 = 282 cm

Hence, the perimeter of the top of the table is 282 cm.

Given:

Side of square = 7 cm

Radius of each quadrant = 7 cm

Area of square = side × side = 7 × 7 = 49 cm²

∵ Area of quadrant = πr²

∴ Area of 2 quadrants = 2 × × πr²

= × × 7 × 7

= 77 cm²

Area of shaded region = Area of 2 quadrants – Area of square

= 77 – 49 = 28 cm²

Hence, the area of shaded region is 28 cm².

Given:

Radius of Circle = 3.5 cm

OD = 2 cm

∵ Area of Quadrant = πr²

∴ Area of Quadrant OABC = × × 3.5 × 3.5

= 9.625 cm²

∵ Area of Triangle = × Base × Height

∴ Area of COD = × 3.5 × 2

= 3.5 cm²

Area of Shaded Region = Area of Quadrant OABC – Area of COD

= 38.5 – 3.5 = 35 cm²

Hence, the area of shaded region is 35 cm².

Given:

Side of square = 14 cm

Diameter of semi circle = 14 cm

⇒ Radius of semi circle = = 7 cm

∵ There are 2 semi circles of same radius.

∴ We consider it as one circle with radius 7 cm.

So,

Perimeter of 2 semicircles = Perimeter of circle = 2πr

= 2 × × 7

= 2 × 22 = 44 cm

Perimeter of shaded region = Perimeter of 2 semicircles + 2 × Side of Square = 44 + 2 × 14 = 44 + 28 = 72 cm

Hence, the area of the shaded region is 72 cm.

Given:

Radius of the circle = 7 cm

Diameter of the circle = 14 cm

Here, diagonal of square = 14 cm

∵ Side of a square =

⇒ Side = = 7√2 cm

⇒ Area of square = side × side

= 7√2 × 7√2

= 49 × 2 = 98 cm²

Area of circle = πr²

= × 7 × 7 = 22 × 7 = 154 cm²

Thus, the area of the circle outside the square

= Area of circle – Area of square = 154 – 98 = 56 cm²

Hence, the area of the required region is 56 cm².

(i) Given:

Diameter of semicircles APB and CQD = 7 cm

⇒ Radius of semicircles APB and CQD = cm = r₁

Diameter of semicircles ARC and BSD = 14 cm

⇒ Radius of semicircles ARC and BSD = cm = 7 cm = r₂

Perimeter of APB = Perimeter of CQD

Area of APB = Area of CQD ………….. (i)

Perimeter of ARC = Perimeter of BSD

Area of ARC = Area of BSD ………….. (ii)

∵ Perimeter of semicircle = πr …………… (iii)

∴ Perimeter of APB = πr₁

= × = 11 cm

Then, using (i), we have

Perimeter of CQD = 11 cm

Now, using (iii), we have

Perimeter of ARC = πr₂

= × 7 = 22 cm

Then, using (ii), we have

Perimeter of BSD = 22 cm

Perimeter of shaded region

= (Perimeter of ARC + Perimeter of APB) + (Perimeter of BSD + Perimeter of CQD)

= (22 + 11) + (22 + 11) = 33 + 33 = 66 cm

Hence, the perimeter of the shaded region is 66 cm.

(ii) Now,

∵ Area of semicircle = πr² …………. (iv)

∴ Area of APB = πr₁²

= × × × = cm²

Then, using (i), we have

Area of CQD = cm²

Now, using (iv), we have

Area of ARC = πr₂²

= × × 7 × 7 = 11 × 7 = 77 cm²

Then, by using (ii), we have

Area of BSD = 77 cm²

Area of shaded region

= (Area of ARC-Area of APB) + (Area of BSD- Area of CQD)

= (77 - ) + (77 - )

= () + () = + = = 115.5 cm²

Hence, the area of the shaded region is 115.5 cm².

Given:

Diameter of semicircle PSR = 10 cm

⇒ Radius of semicircle PSR = = 5 cm = r₁

Diameter of semicircle RTQ = 3 cm

⇒ Radius of semicircle RTQ = = 1.5 cm = r₂

Diameter of semicircle PAQ = 7 cm

⇒ Radius of semicircle PAQ = = 3.5 cm = r₃

∵ Perimeter of semicircle = πr

∴ Perimeter of semicircle PSR = πr₁

= 3.14 × 5 = 15.7 cm

Similarly, Perimeter of semicircle RTQ = πr₂

= 3.14 × 1.5 = 4.71 cm

and Perimeter of semicircle PAQ = πr₃

= 3.14 × 3.5 = 10.99 cm

Perimeter of shaded region = Perimeter of semicircle PSR

+ Perimeter of semicircle RTQ

+ Perimeter of semicircle PAQ

= 15.7 + 4.71 + 10.99 = 31.4 cm

Hence, the perimeter of the shaded region is 31.4 cm.

Given:

OA = Side of square OABC = 20 cm

∵ Area of square = Side × Side

∴ Area of square OABC = 20 × 20 = 400 cm²

Now,

∵ Length of diagonal of square = √2 × Side of Square

∴ Length of diagonal of square OABC = √2 × 20 = 20√2 cm

⇒ Radius of the quadrant = 20√2 cm

∵ Area of quadrant = πr²

∴ Area of quadrant OPBQ = × 3.14 × 20√2 × 20√2

= × 400 × 2

= 3.14 × 200 = 628 cm²

Area of shaded region = Area of quadrant OPBQ – Area of square OABC = 628 – 400 = 228 cm²

Hence, the area of the shaded region is 228 cm².

Given:

AO = OB

Perimeter of the figure = 40 cm………….. (i)

Let the diameters of semicircles AQO and APB be ‘x₁’ and ‘x₂’ respectively.

Then, using (1), we have

AO = OB

Also, AB = AO + OB = AO + AO = 2AO

⇒ x₂ = 2x₁

So, diameter of APB = 2x₁

and diameter of AQO = x₁

Radius of APB = x₁

and Radius of AQO = ………….. (ii)

Perimeter of shaded region = perimeter of AQO + perimeter APB + diameter of APB ………………… (iii)

∵ Perimeter of semicircle = πr

∴ Perimeter of semicircle AQO = × = cm

Perimeter of semicircle APB = × x₁ = cm

Now, using (iii), we have

40 = + + x₁

40 =

40 × 7 = 40x₁

280 = 40x₁

x₁ = = 7 cm

∴ using (ii), we have

Radius of APB = 7 cm = r₁

And Radius of AQO = cm = 3.5 cm = r₂

Now,

∵ Area of semicircle = πr²

∴ Area of semicircle APB = πr₁²

= × × 7 × 7 = 11 × 7 = 77 cm²

Similarly,

Area of semicircle APB = πr₂²

= × × 3.5 × 3.5 = 19.25 cm²

Thus, Area of shaded region = Area of APB + Area of AQO

= 77 + 19.25 = 96.25 cm²

Hence, the area of the shaded region is 96.25 cm².

Given:

Circumference of circle = 44 cm

Let the radius of the circle be ‘r’ cm

∵ Circumference of circle = 2πr

∴ 44 = 2πr

= × r

r = 22 × = 7 cm

Now, Area of quadrant = × πr²

= × × 7 × 7

= = = 38.5 cm²

Hence, the area of the quadrant is 38.5 cm².

Given:

Side of square = 14 cm

Let the radius of each circle be ‘r’ cm

Then, 2r + 2r = 14 cm

4r = 14 cm

r =


=

Area of square = side × side

= 14 × 14

= 196 cm²

∵ Area of circle = πr²

∴ Area of 4 circles = 4 × πr²

= 4 × × ×

= 22 × 7

= 154 cm²

Area of shaded region = Area of the square – Area of 4 circles

= 196 -154

= 42 cm²

Hence, the area of the shaded region is 42 cm².

Given:

Length of rectangle = 8 cm

Breadth of rectangle = 6 cm

Area of rectangle = length × breadth

= 8 × 6 = 48 cm²

Consider ABC,

By Pythagoras theorem,

AC² = AB² + BC²

= 8² + 6² = 64 + 36 = 100

AC = √100 = 10 cm

⇒ Diameter of circle = 10 cm

Thus, radius of circle = = 5 cm

Let the radius of circle be r = 5 cm

Then, Area of circle = πr²

= × 5 × 5 = = = 78.57 cm²

Area of shaded region = Area of circle – Area of rectangle

= 78.57 - 48

= 30.57 cm²

Hence, the area of shaded region is 30.57 cm².

Given:

Perimeter of square = Circumference of circle ………………. (i)

Area of Square = 484m²

Let the side of square be ‘x’ cm.

∵ Area of Square = side × side

∴ 484 = x × x

x² = 484

x = √484 = 22cm

∵ Perimeter of square = 4 × side

= 4 × 22 = 88 cm

∴ Using (i), we have

Circumference of circle = 88 cm

Also, Circumference of Circle = 2πr

2πr = 88

2 × × r = 88

r = 88 × ×

r = 2 × 7 = 14 cm

Area of Circle = πr² = × 14 × 14

= 22 × 2 × 14 = 616 cm²

Hence, the area of Circle is 616 cm².

Given: Radius of circle = r

Diagonal of Square = 2r

∵ Side of Square =

∴ Side = = √2r

Area of Square = Side × Side

= √2r × √2r

= 2r²

Hence, the area of square is ‘2r²’ square units.

Given:

Rate of fencing a circular field = Rs. 25/m

Cost of fencing a circular field = Rs. 5500

Rate of ploughing the field = 50p/m² = Rs. 0.5/m²

Let the radius of circular field be ‘r’ and the length of the field fenced be ‘x’ m.

Then, 25 × x = 5500

x = = 220 m

∵ Circumference of circular field = 2πr

∴ 220 = 2πr

220 = 2 × × r

r =

r = 35 m

Area of the circular field = πr²

= × 35 × 35

= 22 × 5 × 35

= 3850m²

Now, cost of ploughing the field = Rate of ploughing the field × Area of the field = 0.5 × 3850

= Rs. 1925

Hence, the cost of Ploughing the field is Rs. 1925.

Given:

Length of the rectangular park = 120 m

Breadth of the rectangular park = 90 m

Area of the park excluding the circular lawn = 2950m²

Area of the rectangular park = length × breadth

= 120 × 90

= 10800m²

Area of circular lawn = Area of rectangular park – Area of park excluding the lawn

= 10800 – 2950

= 7850m²

∵ Area of circle = πr²

∴ 7850 = 3.14 × r²

r² = = 2500

r = √2500 = 50 m

Hence, the radius of the circular lawn is 50m.

Given:

OP = 21 m = r₁

OR = 14 m = r₂

Let the quadrants made by outer and inner circles be Q₁ and Q_2, with radius r₁ and r₂ respectively.

Then, Area of flower bed = Area of Q₁ – Area of Q₂

∵ Area of Quadrant = πr²

∴ Area of Q₁ = πr₁²

= × × 21 × 21

= m²

Similarly, Area of Q₂ = πr₂²

= × × 14 × 14

= m²

Thus, Area of flower bed = -

= = 192.5 m²

Hence, the area of the flower bed is 192.5 m².

Given:

AC = 54 cm

BC = 10 cm

⇒ AB = AC-BC = 54-10 = 44 cm

Radius of bigger circle = = = 27 cm = r₁

Radius of Smaller circle = = = 22 cm = r₂

∵ Area of Circle = πr²

∴ Area of Bigger Circle = πr₁²

= × 27 × 27

= cm²

Similarly, Area of Smaller Circle = πr₂²

= × 22 × 22

= cm²

Area of shaded region = Area of Bigger Circle – Area of Smaller Circle = - = = 770 cm²

Hence, Area of Shaded Region is 770 cm².

Given:

AB ∥ CD

∠BCD = 90°

AB = BC = 3.5 cm = EC

DE = 2 cm

DC = DE + EC = 2 + 3.5 = 5.5 cm

Area of Trapezium = × Sum of Parallel Sides × h

= × (AB + DC) × BC

= × (3.5 + 5.5) × 3.5

= × 9 × 3.5

= 15.75 cm²

Area of Quadrant BFEC = × πr² = × × 3.5 × 3.5

= 9.625 cm²

Thus, Area of remaining part of metal sheet

= Area of Trapezium – Area of Quadrant BFEC

= 15.75 – 9.625 = 6.125 cm²

Hence, the area of the remaining part of metal sheet is 6.125 cm².

Given:

Radius of Circle = 35 cm

∠AOB = 90°

∵ Area of Sector = × πr²

= × × 35 × 35

= cm²

∵ ∆ AOB is right-angled triangle.

∴ Area of ∆ AOB = × OA × OB

= × 35 × 35

= cm²

Now, Area of Minor Segment ACB

= Area of Sector – Area of ∆AOB

= - = = 350 cm²

Area of Circle = πr²

= × 35 × 35

= 22 × 5 × 35

= 3850 cm²

Thus, Area of Major Segment = Area of Circle – Area of Minor Segment = 3850 – 350 = 3500 cm²

Hence, the area of the major segment is 3500 cm².

In order to solve such type of questions we basically need to find the radius of the give circle and simply use it to find the area of the given circle.

Given the circumference or perimeter of the circle = 39.6 cm.

And we know, Perimeter or circumference of circle = 2πr

Where, r = Radius of the circle

Therefore, 2πr = 39.6

(put value of π = 22/7)

On rearranging we get,

⇒ r = 6.3 cm

So, the radius of the circle = 6.3 cm

And we also know, Area of the circle = πr²

Where, r = radius of the circle

⇒ Area of the circle = π(6.3)²

(putting value of r)

= 22×5.67

= 124.74 cm²

The area of the circle = 124.74 cm².

In order to solve such type of questions we basically need to find the radius of the give circle and simply use it to find the are circumference or perimeter of the given circle.

Given the area of the circle = 98.56 cm²

And we also know, Area of the circle = πr²

Therefore, πr² = 98.56

(put value of π = 22/7)

On rearranging we get,

⟹ r = √31.36

⇒ r = 5.6 cm

So, the radius of the circle = 5.6 cm

And we know, Perimeter of circle = 2πr

(put value of r)

⇒ Circumference or Perimeter of circle = 2π(5.6)

= 35.2 cm

The circumference or perimeter of the circle is 35.2 cm

Given, the circumference of a circle exceeds its diameter by 45 cm.

⇒ Circumference of circle = Diameter of circle + 45

Let ‘d’ = diameter of the circle

⇒ Circumference = d + 45 → eqn1

And we know, Circumference of a circle = 2πr → eqn2

Where r = radius of circle

Also, we know that the radius of the circle is half of its diameter.

Put value of circumference in equation 1 from equation 2

⇒ 2πr = d + 45 → eqn4

Put value of r in equation 4 from equation 3

⇒ πd = d + 45

⇒ πd – d = 45

⇒ (π – 1)d = 45 (taking d common from L.H.S)

On rearranging, we get

⇒ d = 21 cm

Therefore, the diameter of the circle is 21 cm.

Thus, the radius of the circle

⇒ r = 10.5 cm

Now put the value of r in equation 2, we get

= 66 cm

The circumference of the circle is 66 cm.

In this question the wire is first bent in the shape of square and then same wire is bent to form a circle. The point to be noticed is that the same wire is used both the times which implies that the perimeter of square and that of circle will be equal.

Let the square be of side ‘a’ cm and radius of the circle be ‘r’

Given the area enclosed by the square = 484 cm²

Also, we know that Area of square = Side × Side

Area of the square = a²

⇒ a² = 484

⟹ a = √484

⇒ a = 22 cm

Therefore, side of square, ‘a’ is 22 cm.

Also, circumference of the circle = Perimeter of square → eqn1

Perimeter of square = 4 × side

Perimeter of square = 4×22

⇒ Perimeter of square = 88 cm → eqn2

Also, we know, Circumference of circle = 2πr → eqn3

Put values in equation 1 from equation 2 & 3, we get

2πr = 88

On rearranging,

⇒ r = 14 cm

So, the radius ‘r’ of the circle is 14 cm.

Area of circle = πr²

Where r = radius of the circle

= π(14²)

= 4312/7

= 616 cm²

Area of the circle is 616 cm².

In this question the wire is first bent in the shape of equilateral triangle and then same wire is bent to form a circle. The point to be noticed is that the same wire is used both the times which implies that the perimeter of equilateral triangle and that of circle will be equal.

Let the equilateral triangle be of side ‘a’ cm and radius of the circle be ‘r’.

Given: Area enclosed by equilateral triangle = 123√3 cm²

Also, we know that Area of equilateral triangle

Where ‘a’ = side of equilateral triangle

⟹ a = √484

⇒ a = 22 cm

Therefore, side of equilateral triangle, ‘a’ is 22 cm.

Also, circumference of the circle = Perimeter of equilateral triangle → eqn1

Perimeter of equilateral triangle = 3 × side

= 3 × 22

= 66 cm → eqn2

Also, we know Circumference of circle = 2πr → eqn3

Put values in equation 1 from equation 2 & 3, we get

2πr = 66

(put π = 22/7)

On rearranging,

⇒ r = 10.5 cm

So, the radius ‘r’ of the circle is 10.5 cm.

Area of circle = πr²

Where r = radius of the circle

⇒ Area of circle = π(10.5²)

= 346.5 cm²

Area of the circle is 346.5 cm².

In this question the length of chain used as boundary of the semicircular park is the perimeter of the semicircular park. By using this we will first calculate the radius of the semicircular park and then area of semicircle consequently.

Length of chain = 108 m

Length of chain = Perimeter or circumference of semicircle

Therefore, Circumference or Perimeter of semicircle = 108 m

Also, Circumference or Perimeter of semicircle = πr

Where r = radius of semicircle

⇒ πr = 108

(put π = 22/7)

On rearranging,

⇒ r = 34.46 m

Therefore, radius of semicircle is 34.36 m

As, Area of semicircle

Put value of ‘r’ in equation 1, we get

Area of semicircle

(put π = 22/7)

On rearranging,

= 1855.63 m²

The area of the semicircular park is 1855.63 m².

Given Sum of the radius of the circles = 7 cm

the difference of their circumference = 8 cm

Let the radius one circle be ‘r₁’ cm and other be ‘r₂’ cm and circumference be ‘C₁’ and ‘C₂’ respectively.

Also, circumference of circle = 2πr

Where r = radius of the circle

C₁ = 2πr₁ and C₂ = 2πr₂

r₁ + r₂ = 7 → eqn1

C₁ – C₂ = 8 → eqn2

(Note: Her it is considered that r₁>r₂)

We can rewrite equation 2 as,

2πr₁ – 2πr₂ = 8

⇒ 2π(r₁ – r₂) = 8

(taking 2π common from L.H.S)

Put the value of r₁ from equation 3 in equation 1

(taking 11 as LCM on R.H.S)

Put value of r₂ in equation 3

(from equation 3)

(taking 22 as LCM on R.H.S)

(by putting value of r₁)

= 182/7

= 26 cm

(by putting value of r₂)

= 126/7

= 18 cm

The circumference of circles are 26 cm and 18 cm.

Consider the ring as shown in the figure below,

The inner radius of ring is ‘r’ and the outer radius is ‘R’.

Area of inner Circle = πr² and Area of outer Circle = πR²

Where r = 12 cm and R = 23 cm

Area of ring = Area of outer circle – Area of inner circle

Area o ring = πR² – πr² (put values of r & R)

⇒ Area of ring = π(23²) – π(12²)

⇒ Area of ring = π(23² – 12²) (taking π common from R.H.S)

⇒ Area of ring = π(529 – 144)

= 1210 cm²

Area of ring is 1210 cm².

Given radius of circular park = R = 17 m

Width of the circular path outside the park = d = 8 m

Therefore, the radius of the outer circle = R’ = R + d

Outer radius = R’ = 17 + 8

R’ = 25 m

Area of inner circle = πR² and,

Area of outer circle = πR’²

Area of path = Area of outer circle – Area of inner circle

= πR’² – πR² (put values of R’ & R)

= π(25²) – π(17²)

= π(25² – 17²) (taking π common from R.H.S)

= π(625 – 289)

(put π = 22/7)

= 7392/7

= 1056 m²

The area of the path is 1056 m².

Consider the race track as shown below,

The inner and outer radius of track is ‘r’ cm and ‘R’ cm respectively.

Let inner and outer circumference be ‘C₁’ and C₂’ respectively.

C₁ = 352 m and C₂ = 396 m.

We know,

Circumference of circle = 2πr

Where r = radius of the circle

C₁ = 2πr and C₂ = 2πR

⇒ 2πr = 352 and 2πR = 396

On rearranging,

⇒ r = 56 m and R = 63 m

So, the width of the race track = R – r,

⇒ Width of the race track = 63 – 56

⇒ Width of the race track = 7 m

Area of race track = area of outer circle – area of inner circle

⇒ Area of track = πR² – πr² (put values of r and R)

⇒ Area of track = π(63²) – π(56²)

⇒ Area of track = π(63² – 56²) (taking π common from R.H.S)

⇒ Area of track = π(3969 – 3136)

⇒ Area of track = π×833

= 22×119

= 2618 m²

The width of tack is 7 m and area of track is 2618 m².

Consider the circle shown above,

Given radius of the circle = R = 21 cm → eqn1

And angle of the sector = θ = 150^o→ eqn2

Length of arc of a sector

Where ‘R’ = radius of sector (or circle)

θ = angle subtended by the arc on the centre of the circle

Put the values of R and θ from equation 1 and 2 in equation 3

= 138600/2520

= 55 cm

Area of a sector

Where ‘R’ = radius of sector (or circle)

θ = angle subtended by the arc on the centre of the circle

Put the values of R and θ from equation 1 and 2 in equation 3

= 1455300/2520

= 577.5 cm²

The length of arc is 55 cm and area of sector is 577.5 cm².

Consider the circle shown above,

We know , Area of sector

Where R = radius of the circle and θ = central angle

Given R = 10.5 cm and Area of sector = 69.3 cm²

Let the angle subtended at centre = θ

Put the values of R and area of sector in equation 1

⇒θ = 72^°

The central angle of the sector is 72^°.

Consider the Circle shown above,

We know, Length of arc of sector

Where R = radius of circle and θ = central angle of the sector

Given, Length of arc = ℓ = 16.5 cm and θ = 54^o. Let the radius be x cm

Put the values of ℓ and θ in equation 1

On rearranging

⇒ x = 17.5 cm

Also, we know circumference of the circle = 2πR

⇒ Circumference of the circle = 2πx (put value of x in this equation)

⇒ Circumference of the circle = 2π(17.5)

⇒ Circumference of the circle = 110 cm

Also, we know Area of the circle = πR²

⇒ Area of the circle = πx²

⇒ Are of the circle = π(17.5²)

⇒ Area of the circle = 962.5 cm²

The radius of circle is 17.5 cm, circumference is 110 cm and area is 962.5 cm²

Consider the above figure,

From here we can conclude that the portion or the segment below the chord AB is the minor segment and the segment above AB is major segment.

Also we know,

Area of minor segment = Area of sector – Area of ∆AOB → eqn1

Now, Area of sector

Where R = radius of the circle and θ = central angle of the sector

Given, R = 7 cm and θ = 90^°

Putting these values in the equation 2, we get

⇒ Area of sector = 38.5 cm²→ eqn3

Area of △AOB = 1/2 × base × height

As triangle is isosceles therefore height and base both are 7 cm.

⟹ Area of △AOB = 1/2×7×7

= 24.5 cm²→ eqn4

Putting values of equation 2 and 4 in equation 1 we get

Area of minor segment = 38.5 – 24.5

⇒ Area of minor segment = 14 cm²

Area of major segment = πR² – Area of minor segment → eqn5

Put the value of R, and Area of minor segment in equation 5

= π(7²) – 14

= 49π - 14

= (22×7) - 14

= 154 - 14

= 140 cm²

Area of minor segment is 14 cm² and of major segment is 140 cm².

Consider the figure shown above.

In this, the triangle AOB is an equilateral triangle as all the sides are equal; therefore, it is obvious that the central angle of the sector is 60 degrees. Now by simply applying the formula of length of an arc, we can easily calculate the length of arc of the sector AOB.

Given Radius of circle = R = 12 cm,

Length of chord AB = 12 cm

∴ Central angle = θ = 60° (∵ ∆AOB is an equilateral triangle)

Where R = radius of the circle and θ = central angle of the sector

Put the values of R and θ in equation 1

= 2×3.14×2

= 12.56 cm

Now, Length of major arc = 2πR – Length of minor arc

⇒ Length of major arc = 2π(12) – 12.56 (put π = 3.14)

⇒ Length of major arc = (2×3.14×12) – 12.56

⇒ Length of major arc = 75.36 – 12.56

⇒ Length of major arc = 62.8 cm

Now, Area of minor segment = Area of sector – Area of triangle → eqn1

= 75.36 cm²→ eqn2

⇒ Area of triangle = 1.73×36

⇒ Area of triangle = 62.28 cm²→ eqn3

Put the values of equation 2 and 3 in equation 1,

∴ Area of minor segment = 75.36 – 62.28

= 13.08 cm²

Length of major arc is 62.8 cm and of minor arc is 12.56 cm and area of minor segment is 13.08 cm².

Consider the figure shown above.

In this, the triangle AOB is an isosceles triangle. So here we will construct a perpendicular bisector from O on AB and as this triangle is isosceles therefore this perpendicular will also act as median and angle bisector.

Therefore,

Draw a perpendicular bisector from O which meets AB at D and bisects AB, as ABO is an isosceles triangle therefore OD acts as a median.

So, consider right angle triangle AOD right angled at D

Let ∠AOD = θ ⇒ Perpendicular = AD and Hypotenuse = AO = R

Given Radius of circle = R = 5√2 cm

Length of chord AB = 10 cm, AD = 5 cm

⇒ θ = 45°

∴ ∠AOD = 45°

Area of minor segment = Area of sector – Area of right angle triangle

→ eqn1

Where R = radius of the circle and θ = central angle of the sector

∴ Area of sector = 39.25 cm²

Area of right angle triangle = 1/2 × base × height

As this is isosceles right-angle triangle

∴ height = base = 5√2 cm

Area of right angle triangle = 1/2 ×5√2×5√2 = = 25 cm²

Put the value of area of sector and area of right angle triangle in equation 1,

⇒ Area of minor segment = 39.25 -25

= 14.25 cm²

Area of major segment = πR² – area of minor segment

⇒ Area of major segment = 157 – 14.25 = 142.75 cm²

Area of major segment is 142.75 cm² and of minor segment is 14.25 cm ².

Given R = 42 cm and central angle of sector = 120°

Area of minor segment = Area of sector – Area of triangle → eqn1

Where R = radius of the circle and θ = central angle of the sector

∴ Area of sector = 1848 cm²

Area of right angle triangle = 1/2×base×height×sin θ

Where θ = central angle of the sector

Area of triangle = 1/2×42×42×√3/2

Area of triangle = (42×42×√3)/4

(put √3 = 1.73)

= 762.93 cm²

Put the values of area of triangle and area of sector in equation 1

⇒ Area of minor segment = 1848 – 762.93

= 1085.07 cm²

Area of major segment = πR² – Area of minor segment

Put the value of area of minor segment and R in above equation

= π(42²) – 1085.07

⇒Area of major segment = 22/7×42×42-1085.07

(put π = 22/7)

⇒ Area of major segment = 5544 – 1085.07

∴ Area of major segment = 4458.93 cm²

Area of major segment is 4458.93 cm² and of minor segment is 1085.07 cm².

Area of minor segment = Area of sector – Area of triangle → eqn1

Where R = radius of the circle and θ = central angle of the sector

∴ Area of sector = 471 cm²

Where a = side of the triangle

Area of triangle = √3/4×30×30

Area of triangle = √3/4×900

Area of triangle = (900×√(3 ))/4

(put √3 = 1.732)

Area of triangle = (1.732×900)/4

∴ Area of triangle = 389.7 cm²

Put the values of area of triangle and area of sector in equation 1

Area of minor segment = 471 – 389.7

⇒ Area of minor segment = 81.3 cm²

Area of major segment = πR² – Area of minor segment

Put the value of area of minor segment and R in above equation

⇒ Area of major segment = π×(30²) – 81.3 (put π = 3.14)

⇒ Area of major segment = 3.14×30×30 – 81.3

⇒ Area of major segment = 2826 – 81.3

= 2744.7 cm²

Area of major segment is 2744.7cm² and of minor segment is 81.3 cm².

Given radius of circle = R = 10.5 cm

Let the area of major sector be ‘A₁’ and that of minor sector be ‘A₂’

We know, Area of circle = Area of major sector + Area of minor sector

⇒ Area of circle = A₁ + A₂

We also know, Area of circle = πR²

Where R = radius of circle, put value of area of circle in equation 2.

(taking 5 as L.C.M on R.H.S)

= 288.75 cm²

The area of major sector is 288.75 cm².

In an hour the minute hand completes one rotation therefore in 24 hours the minute hand will complete 24 rotations similarly the hour hand completes one rotation in 12 hours therefore in 24 hours it will complete 2 rotations. Now we have to just calculate the perimeter of the circle traced by minute hand and hour hand and multiply it with the number of rotations of minute hand and hour hand in 2 days respectively.

Length of short/hour hand = r = 4 cm

Length of long/minute hand = R = 6 cm

∴ The perimeter of circle traced by short hand = p = 2πr → eqn1

∴ The perimeter of circle traced by Long hand = P = 2πR → eqn2

Now put the value of ‘r’ and ‘R’ in the equation 1 and 2 respectively.

⇒ p = 2π(4) & P = 2π(6) (put π = 3.14)

⇒ p = 2×3.14×4 & P = 2×3.14×6

∴ p = 25.12 cm & P = 37.68 cm

Therefore, distance covered by short hand in one rotation = 25.12 cm

Distance covered by long hand in one rotation = 37.68 cm

Number of rotation of short hand in one day = 2

Number of rotation of long hand in one day = 24

Therefore number of rotation of small hand in two days = 4

Number of rotation of long hand in two days = 48

Total distance covered by long hand in 2 days = P × no. of rotations in 2 days

⇒ Total distance covered by long hand in 2 days = 37.68×48

⇒ Total distance covered by long hand in 2 days = 1808.64 cm → eqn3

Total distance covered by short hand in 2 days = p × no. of rotations in 2 days

⇒ Total distance covered by short hand in 2 days = 25.12×24

⇒ Total distance covered by short hand in 2 days = 100.48 cm → eqn4

Now total distance covered by tip of both hands in 2 days = eqn3 + eqn4

⇒ Total distance covered by both hands in 2 days = 1808.64 + 100.48

⇒ Total distance covered by both hands in 2 days = 1909.12 cm

The distance covered by both hands tip in 2 days is 1909.12 cm

Quadrant is a sector in which the central angle is 90 degrees, and this is the key to solve this question. As we know the central angle of the sector so we can easily calculate the area of quadrant by first calculating the radius of the circle as the circumference of the circle is given and then applying the formula of area of sector.

So, we know Circumference of a circle = 2πR → eqn1

Where R = radius of the circle

Given Circumference of the circle = 88 cm, θ = 90°

Put the given values in equation 1

⟹ 88 = (44×R)/7

⟹ 88 = 44R/7

⟹ (88×7)/44 = R

⟹ 616/44 = R

⇒ R = 14 cm

Put the values of R and θ in the above equation

= 154 cm².

The area of quadrant is 154 cm².

Here the increase in the length of the rope simply means that there is increase in the radius of the circle within which cow can graze. Now to find the additional area available for grazing can be easily be found by simply subtracting the initial area available for grazing from the new area available.

Initial radius = r = 16 cm

Increased radius = R = 23 cm

Additional ground available = Area of new ground – Initial area → eqn1

Initial area of ground = π(r²)

⇒ Initial area of ground = π(16²)

⇒ Initial area of ground = 256π → eqn2

Area of new ground = πR²

⇒ Area of new ground = π(23²)

⇒ Area of new ground = 529π → eqn3

Now put the values of equation 2 and 3 in equation 1

⇒ Additional area of ground available = 529π – 256π

⇒ Additional area available = (529 – 256)π (Taking π common)

⇒ Additional ground available = 273π

(22×273)/7

= 6006/7

= 858 cm²

The additional ground available is 858 cm².

Here the horse is tethered to one corner implies or means that the area available for grazing is a quadrant of radius 21 m. Now we need to find the area of this quadrant to find out the area available for grazing and then subtract it from the total area of the rectangular field to obtain the area left ungrazed.

Given length of rectangular field = ℓ = 70 m

Breadth of rectangular field = b = 52 m

∴ Area of the field = ℓ × b

⇒ Area of the field = 70×52

⇒ Area of the field = 3640 m²

We know in a rectangle all the angles are 90 degrees.

∴ Area available for grazing = area of quadrant

Where R = radius of circle & θ = central angle

Given R = 21 m and θ = 90°

Put the given values in the above equation,

(put π = 22/7)

= (22×63)/4

= 1386/4

⇒ Area available for grazing = 346.5 m²

Area left ungrazed = Area of field – Area available for grazing

⇒ Area left ungrazed = 3640 – 346.5

⇒ Area left ungrazed = 3293.5 m²

The area available for grazing is 346.5 m² and area left ungrazed is 3293.5 m².

Here the horse is tethered to one corner implies or means that the area available for grazing is a sector of radius 21 m with central angle as 60 degrees as the field is in shape of equilateral triangle . Now we need to find the area of this sector to find out the area available for grazing and then subtract it from the total area of the triangular field to obtain the area left ungrazed.

Given the side of field = a = 12 m

∴ Area of field = Area of equilateral triangle

⇒ Area of field = 62.352 m²

We know in an equilateral triangle all the angles are 60 degrees.

∴ Area available for grazing = Area of the sector

Where R = radius of circle and θ = central angle of sector

Given R = 7 m and θ = 60°

Put the given values in the above equation,

⇒ Area available for grazing = 25.666 m²

Area that cannot be grazed = Area of field – Area available for grazing

⇒ Area that cannot be grazed = 62.352 – 25.666

⇒ Area that cannot be grazed = 36.686 m²

The area that cannot be grazed is 36.656 m².

Here the 4 cows are tethered to each corner implies or means that the area available for grazing is a quadrant of radius 25 m with central angle as 60 degrees as the field is in shape of square . Now we need to find the area of this sector to find out the area available for grazing for all the cows and then subtract it from the total area of the square field to obtain the area left ungrazed.

The reason why we have taken the radius as 25 m is , basically we have considered that each cow is tethered to a rope which is equal to half of the side of the square as we had to maximize the area each cow gets to graze without sharing thus the maximum radius within which a cow can graze maximum unshared area is simply the half of the side of square.

Given the side of field which is in shape of square = a = 50 m

∴ Area of the field = Area of Square

⇒ Area of field = a²

⇒ Area of field = (50²)

⇒ Area of field = 2500 m²

We know in an square all the angles are 90 degrees.

∴ Area available for grazing for one cow = area of sector/quadrant

Where R = radius of circle & θ = central angle of sector

Given R = 25 m & θ = 90°

Put the given values in the above equation,

⇒ Area available for grazing for one cow = 490.625 m²

⇒ Area available for 4 cows = 4 × Area available for one cow

⇒ Area available for 4 cows = 4 × 490.625

⇒ Area available for 4 cows = 1962.5 m²

Area left ungrazed = Area of field – Area available for grazing for 4 cows

⇒ Area that cannot be grazed = 2500 – 1962.5

⇒ Area that cannot be grazed = 537.5 m²

The area left ungrazed is 537.5 m².

Here in the given figure ‘O’ is the centre of circle on which three vertices of rhombus lie, this implies that OP, OR are both radius of the circle. Also we know that in rhombus all the 4 sides are equal in length. Thus OP, OR, PQ, RQ, they all are radii of circle. Also OQ is equal to radius of circle. Now rhombus being a parallelogram therefore diagonal OQ will divide the rhombus into two equal halves this means that the area of triangle OQR will be equal to half of the area of rhombus. Also we can see that triangle OQR is an equilateral triangle and hence we can easily calculate its area in terms of radius of circle and equate it to half of the area of rhombus and calculate the radius of given circle.

Let the radius of the circle = x cm

Now join OQ

Consider ∆OQR,

OQ = OR = RQ = x cm

⇒ ∆OQR is an equilateral triangle

Where a = side of equilateral triangle

Also we know OQ is a diagonal of rhombus OPQR and as in a parallelogram diagonal divides it into two equal area or halves , similarly OQ is also dividing the rhombus into two equal areas therefore,

⇒ Area of ∆OQR = Area of ∆OPQ → eqn2

Area of OPQR = Area of ∆OQR + Area of ∆OPQ

Area of OPQR = 2 × Area of ∆OQR (from eqn2) → eqn3

Put the values of area of OPQR and equation 1 in equation 3

As every quadratic equation has two roots, similarly x² = 64 also have two roots i.e. x = 8 and x = -8. As we know that ‘x’ represents radius of circle therefore it cannot be a negative value, hence we discard the negative root.

Therefore radius of the circle = x = 8 cm.

The radius of circle is 8 cm.

(i)

Consider the above figure, Join PR,

Now PR = Diameter of the inscribed circle

Also, PR = BC = 10 cm.

So, PR = 10 cm

⇒ r = 5 cm

∴ Area of inscribed circle = πr² (put value of r in this equation)

⇒ Area of inscribed circle = π(5²)

⇒ Area of inscribed circle = 78.57 cm²

The area of inscribed circle is 78.57 cm².

(ii)

Consider the above figure, O is the centre of circle and ABCD is a square inscribed. Now OB and OD are radii of circle.

Consider ∆DBC right angled at c (as C is a vertex of square)

∴ Apply Pythagoras theorem in triangle DBC

Hypotenuse² = Perpendicular² + Base²

In triangle DBC, hypotenuse = DB,

perpendicular = BC and

base = DC

Put the values of BC and DC i.e. 10 cm

⟹ BD² = 200

⟹ BD = √200

⟹ BD = 10√2 cm

Now radius of circle = half of BD

⟹ r = (10√2)/2

⟹ r = 5√2 cm

Hence Area of circumscribing circle = πr²

⟹ Area of circumscribing circle = 3.14×5√2×5√2

(put π = 3.14 and r = 5√2 cm)

⇒ Area of circumscribing circle = 3.14 × 50

⇒ Area of circumscribing circle = 157 cm²

Area of circumscribing circle is 157 cm².

Consider the figure shown below where O is centre of circle, join BC which passes through O, let the side of square be ‘a’ and radius of circle be ‘r’.

Now we know OB and OC are radius of circle

So, OB = OC = r

Consider ∆BDC right angled at D

And we know BC = OC + OB

BC = 2r and BD = DC = a (put these values in eqn1)

⇒ (2r)² = a² + a²

⇒ 4r² = 2a²

Area of inscribed square = side × side

Areaa of inscribed square = a × a

Area of inscribed square = a²→ eqn3

Area of circumscribing circle = πR² where R = radius of circle

⇒ Area of circumscribing circle = πr²→ eqn4

Put the values from equation 3 & 4 in above equation

(from eqn 2)

So, Ratio is π : 2

The ratio is π:2

Consider the figure shown above, AF, BE and CD are perpendicular bisector.

Now we know that the point at which all three perpendiculars meet is called incentre, so O is the incentre, thus O divides all three perpendiculars in a ratio 2:1.

Let AB = BC = CA = a cm

Therefore let AF = h cm

⟹ ∠AFC = 90° and OF = 1/3 × AF

⟹ OF = h/3 cm (putting value of OF)

⇒ h = 3×OF → eqn1

And we can see from figure that OF = radius of circle

Now let radius of circle be = r cm

∴ Area of circle = πR²

where R = radius of circle

Given area of circle = 154 cm ²

⇒ πr² = 154

⟹ r² = 49

⇒ r = 7 cm

Therefore OF = 7 cm

⇒ h = 3×7 (from eqn 1)

⇒ h = 21 cm

we know area of an equilateral triangle =

where a = side of triangle

Also, Area of triangle = 1/2 ×base×height

Equating both the areas we get,

Put the values of BC and AF

(putting value of h = 21 cm)

(rationalize it)

⟹ a = 14√3 cm

∴ Perimeter of equilateral triangle = 3×side of triangle

⟹ Perimeter of ∆ ABC = 3×14√3 (put √3 = 1.73)

⇒ Perimeter of ∆ABC = 42×1.73

⇒ Perimeter of ∆ABC = 72.66 cm

The perimeter of triangle is 72.66 cm

In one revolution a wheel will cover a distance equal to its circumference, so in order to find the number of revolutions we have to first calculate the circumference of the wheel and then divide it with the total distance covered to find out the total number of revolutions

Given radius of wheel = r = 42 cm

Circumference of wheel = 2πR where R = radius of the wheel

= 2π(42) (putting value of r)

Therefore distance covered in one revolution = 264 cm

Total distance covered = 19.8 km = 1980000 cm

Total number of revolutions = n

Distance covered on 1 revolution ×no. of revolutions = Total distance

264×n = 1980000

⇒ n = 7500

Total number of revolutions is 7500.

Given radius of wheel = R = 2.1m

Number of revolutions in one minute = 75

Number of revolutions in 1 hour = 75×60

Number of revolutions in 1 hour = 45000

Distance covered in one revolution = Circumference of wheel

Distance covered in 1 revolution = 2πR (where R = radius of wheel)

Distance covered n 1 revolution = 2π(2.1)

= 13.2 m

So, distance covered in 4500 revolutions = 4500×distance covered in 1

Distance covered in 4500 revolution = 4500× 13.2

Distance covered in 4500 revolutions = 59400 m = 59.4 km

∴ Distance covered in 1 hour = 59.4 km

Hence speed of the locomotive = 59.4 km/hr

The speed of locomotive is 59.4 km/hr

Let the diameter of the wheel be ‘d’ cm

Total distance covered in 250 revolutions = 49.5 km = 495000 m

⇒ Distance covered in one revolution = 198 cm → eqn1

Also, Distance covered in one revolution = circumference of wheel

∴ Distance covered in one revolution = πD where d = diameter of wheel

Equate equation 1 and 2 we get,

⟹ d = 9×7

⟹ d = 63 cm

The diameter of the wheel is 63 cm.

Given diameter of wheel = d = 60 cm

Number of revolutions in one minute = 140

Number of revolutions in one hour = 140×60

Number of revolutions in one hour = 8400

Distance covered in one revolution = circumference of wheel

⇒ Distance covered in one revolution = πd

= 188.57 cm

Distance covered in one hour = Distance in 1 revolution × no. of revolutions

⇒ Total distance covered in one hour = 188.57× 8400

⇒ Total distance covered in one hour = 1583988 cm = 15.839 km

∴ speed with which boy is cycling = 15.839 km/hr

The speed with which boy is cycling is 15.839 km/hr

Given diameter of wheel of bus = d = 140 cm

Speed of bus = 72.6 km/hr

∴ Distance covered by bus in one hour = 72.6 km = 7260000 cm

Distance covered in one minute = 121000 cm → eqn1

Let the number of revolutions made by wheel per minute = x

Distance covered by wheel in one revolution = circumference of wheel = 2πR

Distance covered by wheel in one revolution = 2π(70)

(putting value of R)

= 2×22×10 = 440 cm

∴ Total distance = No. of revolution× Distance covered in1 revolution

On putting the required values we get,

121000 = 440×(x)

⇒ x = 275

Number of revolutions made per minute is 275.

Given diameter of front wheel = d = 80 cm

so, Radius of front wheel = r = d/2 = 80/2 = 40 cm

Diameter of rear wheel = D = 2 m = 200 cm

Distance covered by wheel in 1 revolution = Circumference of wheel

⇒ Distance covered by front wheel = 2πr = 2π(40)

⇒ Distance covered by front wheel = 80π

∴ Distance covered by front wheel in 800 revolutions = 80π×800

⇒ Distance covered by front wheel in 800 revolutions = 6400π → eqn1

Similarly

⇒ Distance covered by rear wheel = 2πR = 2π(100)

⇒ Distance covered by rear wheel = 200π → eqn2

Let the number of revolutions made by rear wheel to cover 6400π cm be “x”

∴ (x)×200π = 6400π (from eqn1 and eqn2)

⟹ x = 64000/200

⇒ x = 320

Number of revolution made by rear wheel to cover the distance covered by front wheel in 800 revolutions is 320.

Here the distance between the center of circles touching each other is equal to the side of the square. Therefore, we can say that the radius of ach circle is equal to the half of the side of the square. Now by simply calculating the area of the 4 quadrants and then subtracting it from the area of the square we can easily calculate the area of the shaded region.

Given side of square = a = 14 cm

Central angle of each sector formed at corner = θ = 90°

So, radius of 4 equal circles = r = a/2 = 14/2

∴ Radius of 4 circles = r = 7 cm

where R = radius of circle

⇒ Area of all 4 quadrants = 49π → eqn2

Also, Area of square = side×side = a×a = a² = 14² (putting value of side of square)

⇒ Area of square = 196 cm²→ eqn3

∴ Area of shaded region = Area of square – Area of all 4 quadrants

⇒ Area of shaded region = 196 – 49π (fromeqn3 and eqn2)

= 196 – (7×22)

= 196 – 154

= 42 cm²

The area of shaded region is 42 cm².

Here, first we join the center of all adjacent circles then the distance between the center of circles touching each other is equal to the side of the square formed by joining the center of adjacent circles. Therefore, we can say that the side of the square equal to the twice of the radius of circle. Now by simply calculating the area of the 4 quadrants and then subtracting it from the area of the square we can easily calculate the area of the shaded region.

Given radius of each circle = r = 5 cm

Central angle of each sector formed at corner = θ = 90°

Side of square ABCD = a = 2×r = 2×5 = 10 cm

where R = radius of circle

(putting value of r and θ)

Area of all 4 quadrants = 4×Area of one quadrant

⇒ Area of all 4 quadrants = 25π → eqn2

Area of square = side×side = a×a = a²

⇒ Area of square = 10² (putting value of side of square)

⇒ Area of square = 100 cm²→ eqn3

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

Area of shaded region = 100 – 25π (from eqn3 and eqn2)

= 100 – (25×3.14) (put π = 3.14)

= 100 – 78.5

= 21.5 cm²

The area of shaded region is 21.5 cm².

Here, first we join the centre of all adjacent circles then the distance between the centre of circles touching each other is equal to the side of the square formed by joining the centre of adjacent circles. Therefore, we can say that the side of the square equal to the twice of the radius of circle. Now by simply calculating the area of the 4 quadrants and then subtracting it from the area of the square we can easily calculate the area of the shaded region.

Given radius of each circle = “a” units

Central angle of each sector formed at corner = θ = 90°

Side of square ABCD = 2×a units

where R = radius of circle

∴ Area all 4 quadrants = 4×Area of one quadrant

= πa² sq. units → eqn2

Area of square = side×side = 2a×2a = 4a²

⇒ Area of square = 4a² sq. units → eqn3

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

⇒ Area of shaded region = 4a² – πa² (from eqn3 and eqn2)

Consider the above figure,

Here, first we join the center of all adjacent circles then the distance between the center of circles touching each other is equal to the side of an equilateral triangle formed by joining the center of adjacent circles. Therefore, we can say that the side of the equilateral triangle is equal to the twice of the radius of circle. Now by simply calculating the area of the 3 sectors and then subtracting it from the area of the equilateral triangle we can easily calculate the area of the enclosed region.

Given radius of each circle = r = 6 cm

Central angle of each sector = θ = 60° (∵ ∆ABC is equilateral)

Side of equilateral ∆ABC = a = 2×r = 2×6

∴ Side of equilateral ∆ABC = a = 12 cm

⇒ Area of one sector = 6π cm²→ eqn1

Area of all the 3 sector = 3×Area of one sector

= 3×6π (from eqn1)

= 18π cm²→ eqn2

Area of enclosed region = Area of equilateral ∆ABC – Area of all 3 sectors

⟹ Area of enclosed region = 36√3-18π (from eqn 3 and eqn 2)

⟹ Area of enclosed region = (36×1.732)-(18×3.14)

(put π = 3.14 &√3 = 1.732)

= 62.352 - 56.52

= 5.832 cm²

The area of enclosed region is 5.832 cm².

Consider the figure shown below

Here, first we join the center of all adjacent circles then the distance between the center of circles touching each other is equal to the side of an equilateral triangle formed by joining the center of adjacent circles. Therefore, we can say that the side of the equilateral triangle is equal to the twice of the radius of circle. Now by simply calculating the area of the 3 sectors and then subtracting it from the area of the equilateral triangle we can easily calculate the area of the enclosed region.

Given radius of each circle = “a” units

Central angle of each sector = θ = 60° (∵ ∆ABC is equilateral)

Side of equilateral ∆ABC = 2×a units

∴ Area of all 3sectors = 3×Area of one sector

Area of enclosed region = Area of equilateral ∆ABC – Area of all 3 sectors

⟹ Area of enclosed region

(taking a² common)

Here in order to find the area of the shaded region we have to calculate the area, or the quadrant shown and subtract it from the area of the trapezium. And in order to find the area of the quadrant we have to calculate the radius of the sector EAB by the area of trapezium.

Given Area of trapezium ABCD = 24.5 cm² → eqn1

AD ∥ BC, AD = 10 cm, BC = 4 cm, ∠DAB = 90°

Putting the values in equation 2, we get,

⇒ AB = 3.5 cm

Therefore radius of the sector EAB = r = 3.5 cm

⇒ Area of the quadrant EAB = 9.625 cm² → eqn3

∴ Area of shaded region = Area of trapezium – Area of quadrant EAB

⇒ Area of shaded region = 24.5 – 9.625 (putting values from eqn1 and eqn3)

⇒ Area of shaded region = 14.875 cm²

Here in order to find the area of the shaded region we have to calculate the area, or the quadrant shown and subtract it from the area of the trapezium. And in order to find the area of the quadrant we have to calculate the radius of the sector EAB by the area of trapezium.

Given AB = 30 m, AD = 55 m, BC = 45 m

θ_A = 90°, θ_B = 90°, θ_C = 120°, θ_D = 60°

Radius of each sector = r = 14 m

(i) total area of 4 sectors

Area of sector at corner A = 49π m²→ eqn2

As we know that central angle at A and B are both 90 degrees and radius is also same i.e. 14 m therefore the area of the sector at B will be exactly same as that of sector at A.

∴ Area of sector at corner B = Area of sector at corner A

⇒ Area of sector at corner B = 49π → eqn3

Similarly,

Area of sector at corner C = 65.33π m²→ eqn4

Similarly,

Area of sector at corner D = 32.67π → eqn5

Total area of 4 sectors = eqn2 + eqn3 + eqn4 + eqn5

⇒ Total area of 4 sectors = 49π + 49π + 65.33π + 32.67π

⇒ Total area of 4 sectors = 196π

∴ Total area of 4 sectors = 616 m²

Total area of 4 sectors is 616 m².

(ii) Area of the remaining portion

Here in order to find the area of the remaining portion of the trapezium we have to subtract the area of the 4 sectors from the area of the trapezium.

On putting the values,

Area of trapezium

= 50×30

Area of trapezium = 1500 m²→ eqn1

Area of remaining portion = Area of trapezium – Area of the 4 sectors

⇒ Area of remaining portion = 1500 – 616 (from eqn1 and part (i))

∴ Area of remaining portion = 884 m²

The area of the remaining portion is 884 m².

Area of shaded region can be calculated by subtracting the area of minor sector at vertex B from the sum of areas of the major sector at O and area of equilateral triangle.

Given Radius of circle at O = r = 6 cm

Side of equilateral triangle = a = 12 cm

Central angle at O = 360 -60 = 300°

Central angle at B = 60°

Area of the equilateral triangle =

where a = side of equilateral triangle

Area of the equilateral triangle = (putting the value of a)

Area of the equilateral triangle = (144×√3)/4

Area of the equilateral triangle = 36√3 cm² → eqn1

Area of sector = θ/360×πR² where r = radius of the sector

Area of minor sector at B = 60/360×π×(6²) (given)

∴ Area of minor sector at B = 6π cm²→ eqn2

Similarly,

∴ Area of major sector at O = 30π cm²→ eqn3

Area of shaded region = eqn1 + eqn3 – eqn2

On putting values

⇒ Area of shaded region = 36√3 + 30π-6π

Area of shaded region = 36√3 + 24π

(put π = 3.14 and √3 = 1.73

∴ Area of shaded region = (36×1.73) + (24×3.14)

⇒ Area of shaded region = 62.28 + 75.36

∴ Area of shaded region = 137.64 cm²

Area of the shaded region is 137.64 cm².

Here in order to find the area of the shaded region we have to subtract the area of the semicircle and the triangle from the area of the rectangle.

Given AB = 80 cm, BC = 70 cm, DE = 42 cm, ∠AED = 90°

Here we see that the triangle AED is right angle triangle, therefore, we can apply Pythagoras theorem i.e.

H² = P² + B² (pythagoras theorem)

AD² = DE² + AE²

⇒ 70² = 42² + AE² (putting the given values)

⇒ 4900 = 1764 + AE²

⇒ 4900 – 1764 = AE²

⇒ 3136 = AE²

AE = √3136

∴ AE = 56 cm

Area of ∆AED = 1/2×AE×DE

(Area of triangle = 1/2×base×height)

On putting values we get,

Area of ∆AED = 1/2×56×42

⇒ Area of ∆AED = 28×42

∴ Area of ∆AED = 1176 cm²→ eqn1

R = 35 cm

⇒ Area of semicircle = 11×175

∴ Area of semicircle = 1925 cm²→ eqn2

Area of rectangle = ℓ×b (ℓ = length of rectangle, b = breadth of rectangle)

⇒ Area of rectangle = 80×70 = 5600 cm²→ eqn3

Area of shaded region = Area of rectangle – Area of semicircle – Area of ∆

⇒ Area of shaded region = 5600 -1925 – 1176 (fromeqn1, eqn2 and eqn3)

∴ Area of shaded region = 2499 cm²

Area of the shaded region is 2499 cm².

Here in order to find the area of the shaded region (region excluding the triangle) we have to subtract the area of the triangle from the area of the rectangle and then add the area of the semicircle.

Given AB = 20 cm, DE = 12 cm, AE = 9 cm and ∠AED = 90°

Here we see that the triangle AED is right angle triangle, therefore, we can apply Pythagoras theorem i.e.

AD² = DE² + AE²

AD² = 12² + 9² (putting given values)

⇒ AD² = 144 + 81

⇒ AD² = 225

⇒ AD = √225

∴ AD = 15 cm

Area of ∆AED = 1/2×AE×DE

(Area of triangle = 1/2×base×height)

On putting values we get,

⇒ Area of ∆AED = 9×6

∴ Area of ∆AED = 54 cm²→ eqn1

Here radius of semicircle = BC/2 = 15/2

⇒ R = 7.5 cm

⇒ Area of semicircle = 1.07×56.25

∴ Area of semicircle = 88.3125 cm²→ eqn2

Area of rectangle = ℓ×b

(ℓ = length of rectangle, b = breadth of rectangle)

⇒ Area of rectangle = 20×15

(putting the values of ℓ & b)

∴ Area of rectangle = 300 cm²→ eqn3

Area of shaded region = Area of rectangle + Area of semicircle – Area of ∆

⇒ Area of shaded region = 300 + 88.3125 – 53 (from eqn1, eqn2, eqn3)

∴ Area of shaded region = 334.3125 cm²

Area of shaded region is 334.3125 cm².

Here in order to find the area of the shaded region (region excluding the area of segment AC and quadrant OCD) can be calculated by subtracting the area of triangle and quadrant OBD from the area of the circle.

Given AC = 24 cm, AB = 7 cm and ∠BOD = 90°

Here we see that the triangle ACB is right angle triangle, therefore, we can apply Pythagoras theorem i.e.

BC² = AC² + AB²

⇒ BC² = 24² + 7² (putting the given values)

⇒ BC² = 576 + 49

⇒ BC² = 625

BC = √625

∴ BC = 25 cm

Area of ∆ACB = 1/2×AB×AC (Area of triangle = 1/2×base×height)

On putting values we get,

⇒ Area of ∆AED = 7×12

∴ Area of ∆AED = 84 cm²→ eqn1

Area of circle = πR² (R = radius of circle)

⇒ R = 12.5 cm

∴ Area of circle = π× 12.5²

⇒ Area of circle = 156.25×3.14 (put π = 3.14)

∴ Are of circle = 490.625 cm²→ eqn2

⇒ Area of quadrant OBD = 122.65625 cm²→ eqn3

Area of shaded region = Area of circle – Area of quadrant – Area of ∆

⇒ Area of shaded region = 490.625 – 84 – 122.65625 (from eqn1, 2 and 3)

⇒ Area of shaded region = 283.96875 cm²

Area of shaded region is 283.96875 cm².

Here we will draw median from all the vertices of the equilateral triangle and the point at which they intersect will be the incircle of the triangle and that will be the centre of the circle. Thenwith the help of which we will find out the height of the triangle and subsequently the radius of the circle and ultimately the area of the shaded region (region of equilateral triangle excluding the area of circle inscribed).

As AD = BF = CE = h

Consider ∆ADB, ∠ADB = 90°, BD = 6 cm

12² = AD² + 6² (putting the given values)

144 = AD² + 36

144 – 36 = AD²

AD² = 108

We also know that a point O will divide each median in a ratio of 2:1

Area of the circle = πr²

∴ Area of the circle = 12π cm²→ eqn1

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

⇒ Area of the shaded region = (36×1.73) – (12×3.14)

⇒ Area of the shaded region = 62.28 -37.68

∴ Area of the shaded region = 24.6 cm²

Here we will first find the sides of equilateral triangle ant then subtract the area of the triangle from the area of the circle.

Given radius of circle = r = 42 cm

∴ Area of the circle = πR², where R = radius of the circle

⇒ Area of the circle = π(42²)

⇒ Area of the circle = 22×252

∴ Area of the circle = 5544 cm²→ eqn1

Consider the figure shown,

In ∆ABD, ∠ADB = 90°

Let the sides of the equilateral triangle = a cm

And as we know AD is a median therefore it will bisect the side BC into two equal parts i.e.

BD = DC → eqn3

Also, BC = BD + DC

⇒ BC = BD + BD (from eqn3)

⇒ a = 2BD (BC = a)

Now, we also know that the point ‘O’ which is the intersection of all the three medians i.e. centroid of the triangle. Also we know that the centroid divides the median in the ratio 2:1.

Also, we know AO = radius = r = 42 cm

⇒ AD = 63 cm

Putting the value in equation 4,

Area of covered by design = Area of circle – Area of triangle ABC

⇒ Area covered by design = 5544 – 2288.79

∴ Area covered by design = 3255.21 cm²

Area covered by design is 3255.21 cm².

We know perimeter of a sector = Length of its arc + 2R → eqn1

Where R = radius of the sector.

Perimeter = 25 cm

θ = 90°

⇒ R = 7 cm → eqn2

∴ Area of the quadrant = 38.5 cm²

Area of the quadrant is 38.5 cm ².

Given the radius of the circle = 42 cm

Central angle of the sector = θ = 90°

Area of the minor segment = Area of sector – area of the right angle triangle

⇒ Area of the sector = 25×3.14

∴ Area of the sector = 78.5 cm²→ eqn1

∴ Area of triangle = 50 cm²→ eqn2

Area of the minor segment = 78.5 – 50 (from eqn1, eqn2)

∴ Area of the minor segment = 28.5 cm²

Area of the minor segment is 28.5 cm².

Here we will first find out the area of the road running around the circular garden and then multiplying it with rate per square meter to calculate the cost of leveling.

Here we see in the figure there are two concentric circles so,

Area of road = Area of outer circle- Area of circular garden

Area of circle = πR² (where R = radius of circle) → eqn1

Let the radius of inner circle = r = 100 m

Also, radius of outer circle = R = 110 m (R = r + 10)

Area of outer circle = π(110)²→ eqn2 (putting R in eqn1)

Area of inner circle = π(100)²→ eqn2 (putting r in eqn1)

∴ Area of road = π(110)² – π(100)² (from eqn2 and 3)

⇒ Area of road = π(12100 – 10000)

⇒ Area of road = 2100 π (put π = 3.14)

⇒ Area of road = 2100×3.14

∴ Area of road = 6594 m²

Cost of leveling = Rate of leveling × Area of road

⇒ Cost of leveling = 20×6594

∴ Cost of leveling = Rs.131880

Area of road is 6594 m² and cost of leveling is Rs.131880.

Each angle of triangle = θ = 60°

Area of triangle not included in circles = Area of triangle – Area of all sectors

Area of all 3 sectors area equal as all the three circles are having same radius which is equal to the half of the side of the equilateral triangle.

Let the side of equilateral triangle be = a cm

⇒ a = 7×2

⇒ a = 14 cm

So radius of the circles = 7 cm

∴ Area of all 3 sectors = 77 cm²→ eqn1

Area of triangle not included = (49×1.73) - 77

⇒ Area of triangle not included = 84.77 – 77

∴ Area of triangle not included = 7.77 cm²

Area of triangle not included in circles is 7.77 cm².

Area of whole figure = ar || ABCD + ar || FGHI + ar DCIF + ar ∆DEF + area semicircle

CD = 8 cm, BP = HQ = 4 cm, DE = EF = 5 cm, CI = 8 cm

ar ∥ ABCD = ar ∥ FGHI = base × height

ar ∥ ABCD = ar ∥ FGHI = BP×DC

ar ∥ ABCD = ar ∥ FGHI = 4×8

ar ∥ ABCD = ar ∥ FGHI = 32 cm²→ eqn1 and eqn2

ar DCIF = area of square = side×side

ar DCIF = DC×CI

ar DCIF = 8×8

ar DCIF = 64 cm²→ eqn3

Consider ∆DEF, EF⊥DF and ∆DEF is isosceles

So, FL = LD

FL = LD = 4 cm

In ∆DEL, ∠DLE = 90°

5² = EL² + 4² (putting the values)

25 = EL² + 16

25 – 16 = EL²

⇒ EL² = 9

∴ EL = 3 cm

⇒ Area of ∆DEF = 4×3

∴ Area of ∆DEF = 12 cm²→ eqn4

R = 4 cm

⇒ Area of semicircle = 3.14×8

∴ Area of semicircle = 25.12 cm²→ eqn5

Area of whole figure = eqn1 + eqn2 + eqn3 + eqn4 + eqn5

⇒ Area of whole figure = 32 + 32 + 64 + 12 + 25.12

∴ Area of whole figure = 165.12 cm²

Area of the whole figure is 165.12 cm ².

θ₁ = 90° θ₂ = 120°, θ₃ = 150°

Radius of circle = r = 6 cm

Area of circle = πR²

⇒ Area of circle = π×6²

⇒ Area of circle = 36π → eqn2

Also, Area of sector (θ₃) = 15π cm²

Area of sector (θ₂) = 12π cm²→ eqn3

Area of sector (θ₁) = 9π cm²→ eqn4

Ratio of three sectors ∷ 9π:12π:15π

Ratio of three sectors ∷ 3:4:5

Total area of design = Area of all the minor segments

Here we will find out the area of one segment and then multiply it with 6 to get the total area of design. And as the figure inscribed in the circle is a regular hexagon this implies that it will be having all edges of same length. Therefore we can say that the angle subtended by each chord which are actually the edges of regular hexagon are equal(theorem).

Let angle subtended by chord AB on centre O be θ

∴ Angle subtended = θ = 60°

Radius of circle = 35 cm

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

Area of minor segment OAB = 641.0833333 – 530.425

∴ Area of minor segment OAB = 110.6583333 cm²→ eqn2

Total area of design = 6 × Area of minor segment OAB

⇒ Total area of design = 6×110.6583333 (from eqn2)

∴ Total area of design = 663.95 cm²

Total area of design is 663.95 cm².

Here we will subtract the area of right angle triangle PQR and semicircle from the area of entire circle.

Given PQ = 24 cm, PR = 7 cm

Consider ∆PQR, ∠QPR = 90°

RQ² = 24² + 7²

⇒ RQ² = 576 + 49

⇒ RQ² = 625

Therefore Radius of the circle = half of RQ

Let radius be ‘r’

∴ r = 12.5 cm

Area of ∆PQR = 7×12

∴ Area of ∆PQR = 84 cm²→ eqn1

∴ Area of semicircle = 245.3125 cm²→ eqn2

Area of circle = πr²

Area of circle = π(12.5²)

⇒ Area of circle = 3.14×156.25 (putting π = 3.14)

∴ Area of circle = 490.625 cm²→ eqn3

Area of shaded region = eqn3 – eqn2 – eqn1

⇒ Area of shaded region = 490.625 – 245.3125 – 84

∴ Area of shaded region = 161.3125 cm²

Area of shaded region is 161.3125 cm².

Given AB = 6 cm, BC = 10 cm

Consider ∆ABC, ∠BAC = 90°

⇒ 10² = 6² + AC² (putting given values)

⇒ 100 = 36 + AC²

⇒ 100 – 36 = AC²

⇒ AC² = 64

∴ AC = 8 cm

Join OB, OA, OC, OE, OF, OD

Here OE = OF = OD = radius of circle = r cm

∠OEC = ∠ODB = ∠OFB = 90° (angle at the point of contact of radius & tangent)

Area ∆ABC = Area of ∆OAC + Area of ∆OCB + Area of ∆OAB → eqn1

∴ Area of ∆OAC = 4r → eqn2

∴ Area of ∆OCB = 5r → eqn3

∴ Area of ∆OAB = 3r → eqn4

⇒ Area of ∆ABC = 3×8

∴ Area of ∆ABC = 24 cm²→ eqn5

Putting all the values in equation we get;

⇒ 24 = 4r + 5r + 3r

⇒ 24 = 12r

∴ r = 2 cm

Area of circle = πr²

Put the value of r, we get,

⇒ Area of circle = π×2²

⇒ Area of circle = 3.14×4 (putting π = 3.14)

∴ Area of circle = 12.56 cm²→ eqn6

Area of shaded region = Area of triangle – Area of circle

⇒ Area of shaded region = 24 – 12.56 (from eqn5 and eqn6)

∴ Area of shaded region = 11.44 cm²

Area of shaded region is 11.44 cm².

Here we will first find out the area of semicircle whose diameter is BC and then subtract the area of right angle triangle ABC from it and then we will subtract this result from the area of semicircles whose diameters are AB and AC.

Consider ∆ABC, ∠BAC = 90°

⇒ BC² = 4² + 3²

⇒ BC² = 16 + 9

⇒ BC² = 25

∴ BC = 5 cm

Area of semicircle whose diameter is AC

∴ Radius = 2 cm

∴ Area of semicircle = 2π cm² –eqn2

Area of semicircle whose diameter is AB

∴ Radius = 1.5 cm

∴ Area of semicircle = 1.125π cm²→ eqn3

Area of semicircle whose diameter is BC

∴ Radius = 2.5 cm

∴ Area of semicircle = 3.125π cm²→ eqn4

⇒ Area of triangle PQR = 3×2

∴ Area of triangle PQR = 6 cm²→ eqn5

Now subtract equation 5 from equation 4,

⇒ Area of semicircle excluding ∆ABC = eqn4 – eqn5

⇒ Area of semicircle excluding ∆ABC = 3.125π – 6

∴ Area of semicircle excluding ∆ABC = 3.8214 cm²→ eqn6

Area of shaded region = eqn3 + eqn2 – eqn6

⇒ Area of shaded region = 2π + 1.125π – 3.8214

⇒ Area of shaded region = 3.125π – 3.8214

⇒ Area of shaded region = 9.8214 – 3.8214

∴ Area of shaded region = 6 cm²

Area of shaded region 6 cm².

Here we will subtract the area of semicircle whose diameter is QS from the area of the semicircle whose diameter PS and add the area of semicircle whose diameter is PQ so as to find out the area of the shaded region.

Given PS = 12 cm

Radius of the circle = 6 cm

PQ = QR = RS

So let PQ = QR = RS = k cm

Also, PQ + QR + RS = PS

⇒ k + k + k = 12

⇒ 3k = 12

∴ k = 4 cm

So, PQ = QR = RS = 4 cm

Area and perimeter of semicircle whose diameter is PS

∴ Radius = 6 cm

∴ Area of semicircle = 18π cm²→ eqn2

Perimeter of semicircle = πr

⇒ Perimeter of semicircle = π×6

∴ Perimeter of semicircle = 6π cm → eqn3

Area of semicircle whose diameter is QS

⇒ Radius = 4 cm

∴ Area of semicircle = 8π cm²→ eqn4

Perimeter of semicircle = πr

⇒ Perimeter of semicircle = π×4

∴ Perimeter of semicircle = 4π cm → eqn5

Area of semicircle whose diameter is PQ

∴ Radius = 2 cm

∴ Area of semicircle = 2π cm²→ eqn6

Perimeter of semicircle = πr

⇒ Perimeter of semicircle = π×2

∴ Perimeter of semicircle = 2π cm → eqn7

Area of the shaded region = eqn2 – eqn4 + eqn6

Area of shaded region = 18π – 8π + 2π

⇒ Area of shaded region = 12π

⇒ Area of shaded region = 12×3.14 (putting π = 3.14)

∴ Area of shaded region = 37.68 cm²

Perimeter of shaded region = eqn3 – eqn5 + eqn7

⇒ Perimeter of shaded region = 6π -4π + 2π

⇒ Perimeter of shaded region = 4π

⇒ Perimeter of shaded region = 4×3.14 (put π = 3.14)

∴ Perimeter of shaded region = 12.56 cm

Perimeter of the shaded region is 12.56 cm and Area of shaded region is 37.68 cm².

Consider the figure as a combination of two semicircles on the ends of the rectangle

Let the length of rectangle be ‘L’ m and breadth be ‘B’ cm

Given L = 90 m, W = 14 m

Perimeter of running track = 400 m

Perimeter of inside of running track = 2L + Arc of two semicircles → eqn1

Arc length of a semicircle = πr where r = radius

⇒ 400 = (2×90) + (2×πr) (putting values in eqn1)

⇒ 400 = 180 + 2πr

⇒ 400 – 180 = 2πr

⇒ 220 = 2πr

∴ r = 35 m

Area of inner of running track = Area of rectangle + 2×area of semicircles → eqn2

Area of rectangle = L×B

Here B = 2r

B = 70 m

⇒ Area of inner rectangle = 90×70

∴ Area of inner rectangle = 6300 m²→ eqn3

⇒ Area of inner semicircles = 1225π

⇒ Area of inner semicircle = 175×22

∴ Area of inner semicircle = 3850 m²→ eqn4

Area of inner of running track = 6300 + 3850 (from3&4)

∴ Area of inner of running track = 10150 m²

Now radius of semicircles of outer of the running track = R = r + W

⇒ R = 35 + 14

∴ R = 49 m

⇒ Area of outer semicircles = 2401π

⇒ Area of outer semicircles = 343×22

∴ Area of outer semicircle = 7546 m²→ eqn5

Breadth of outer running track = B’ = 2R

⇒ B’ = 2×49

∴ B’ = 98 m

Area of outer rectangle = L×B’

⇒ Area of outer rectangle = 90×98

∴ Area of outer rectangle = 8820 m²→ eqn6

Area of entire ground = 8820 + 7546 (from 5&6)

∴ Area of entire ground = 16366 m²

Area of running track = Area of entire ground – Area of inner ground

⇒ Area of running track = 16366 – 10150

∴ Area of running track = 6216 m²

Perimeter of outer boundary = 2L + Arc of outer semicircles

Arc length of an outer semicircle = πR, where R = outer radius

⇒ Perimeter of outer boundary = 180 + (2×22×7)

⇒ Perimeter of outer boundary = 180 + 308

∴ Perimeter of outer boundary = 488 m

Area of running track is 6216 m² and perimeter of outer boundary is 488 m.

Let the radius if circle be r

Given, Area of circle = 38.5 cm²

Area of circle = πr²

⇒ πr² = 38.5

Since π = 22/7

⇒ 22/7 × r²= 38.5

⇒ r² = 38.5 × (7/22)

⇒ r² = 12.25

⇒ r = √12.25 = 3.5 cm

∴ Radius of circle = 3.5 cm

Circumference of circle = 2πr

= 2 × 22/7 × 3.5 cm

= 22 cm

∴ Circumference of the circle is 22 cm

Let the radius if circle be r

Given, Area of circle = 38.5 cm²

Area of circle = πr²

πr² = 38.5

Since = 22/7

∴ πr² = 38.5

⇒ 22/7 × r² = 38.5

⇒ r² = 12.25

⇒ r = √12.25 = 3.50cm

∴ Radius of circle = 3.5 cm

Circumference of circle = 2πr

= 2 × 22/7 × 3.5

= 22 cm

∴ Circumference of the circle is 22 cm

Let the radius if circle be r

Given, Area of circle = 49π cm²

Area of circle = πr²

πr² = 49π

Since = 22/7

∴ πr² = 49π

⇒ r² = 49

⇒ r = √49 = 7 cm

∴ Radius of circle = 7 cm

Circumference of circle = 2πr

= 2 × π × 7 cm

= 14π cm

∴ Circumference of the circle is 14π cm

Let the radius if circle be r

Circumference of circle = 2πr

Difference between the circumference and radius of a circle = 37 cm

⇒ 2πr – r = 37 cm

⇒ 2 × 22/7 × r – r = 37 cm

⇒ 44/7 × r – r = 37 cm

⇒ (44/7 – 1) × r = 37 cm

⇒ 37/7 × r = 37 cm

⇒ r = 37 × 7/37

⇒ r = 7 cm

Area of circle = πr²

= 22/7 × 7 × 7 cm²

= 22/7 × 49 cm² = 22 × 7 cm² = 154 cm²

∴ Area of the circle is 154 cm²