Question

# What is the amount of air that can be held by a spherical ball of diameter 14 inches?

Hint:

### Volume of sphere is

## The correct answer is: 1436.755

### Explanation:

- We have given spherical ball of diameter 14 inches
- We have to find the amount of air that can be held by a spherical ball of diameter 14 inches.

Step 1 of 1:

The diameter of the spherical ball is 14 inches

The radius will be

So, The volume will be

### Related Questions to study

### A wire when bent in the form of a equilateral triangle encloses an area of 36 √3 sq.cm.Find the area enclosed by the same wire when bent to form a square and a rectangle whose length is 2 cm more than its width.

Area of an equilateral triangle = (√3 / 4) × side

^{2}

Area of a square = Side

^{2}

Area of a rectangle = length × breadth

Step-by-step solution:-

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

^{2}

∴ 36 √3 = (√3 / 4) × side

^{2}…................................................ (From given information)

∴ 36 = 1 / 4 × side

^{2}

∴ 36 × 4 = side

^{2}

∴ 144 = side

^{2}

∴ 12 = side ................................................................................. (Taking square root both the sides)

i.e. Side of the triangle = 12 cm

Length of the wire used to make the triangle = Perimeter of this triangle = 3 × side = 3 × 12 = 36 cm ..................... (Equation i)

We Know that the same wire is used to make the square and the triangle.

∴ Perimeter of the square = perimeter of the triangle

∴ Perimeter of the square = 36 ...................................................................................................................... (From Equation i)

∴ 4 × side = 36 .................................................................................. (Perimeter of a square = 4 × side)

∴ side = 36/4 = 9 cm ..................................................................................................... (Equation ii)

∴ Area of the square = side

^{2}

∴ Area of the square = 9

^{2}........................................................................................................................ (From Equation ii)

∴ Area of the square = 81 cm

^{2}

For the rectangle,

Let the width be x cm

∴ Width = x cm ............................................................................................................................................ (Equation iii)

∴ Length = width + 2 cm = x + 2 ..................................................................................................................... (Equation iv)

We Know that the same wire is used to make the rectangle and the triangle.

∴ Perimeter of the rectangle = perimeter of the triangle

∴ Perimeter of the rectangle = 36 ...................................................................................................................... (From Equation i)

∴ 2 (length + breadth) = 36 .................................................................................. (Perimeter of a rectangle= 2 × length + breadth)

∴ 2 (x + x + 2) = 36 ..................................................................................................... (From Equations iii & iv)

∴ 2 (2x + 2 ) = 36

∴ 4x + 4 = 36

∴ 4x = 36 - 4

∴ 4x = 32

∴ x = 32/4 = 8 cm ................................................................................... (Equation v)

Substituting Equation v in Equations iii & iv, we get-

Width = x = 8

Length = x + 2 = 8 + 2 = 10

∴ Area of the rectangle = length × breadth

∴ Area of the rectangle = 8 × 10

∴ Area of the rectangle = 80 cm

^{2}

Final Answer:-

∴ Areas of the given Square and rectangle are 81 cm

^{2}& 80 cm

^{2}, respectively.

### A wire when bent in the form of a equilateral triangle encloses an area of 36 √3 sq.cm.Find the area enclosed by the same wire when bent to form a square and a rectangle whose length is 2 cm more than its width.

Area of an equilateral triangle = (√3 / 4) × side

^{2}

Area of a square = Side

^{2}

Area of a rectangle = length × breadth

Step-by-step solution:-

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

^{2}

∴ 36 √3 = (√3 / 4) × side

^{2}…................................................ (From given information)

∴ 36 = 1 / 4 × side

^{2}

∴ 36 × 4 = side

^{2}

∴ 144 = side

^{2}

∴ 12 = side ................................................................................. (Taking square root both the sides)

i.e. Side of the triangle = 12 cm

Length of the wire used to make the triangle = Perimeter of this triangle = 3 × side = 3 × 12 = 36 cm ..................... (Equation i)

We Know that the same wire is used to make the square and the triangle.

∴ Perimeter of the square = perimeter of the triangle

∴ Perimeter of the square = 36 ...................................................................................................................... (From Equation i)

∴ 4 × side = 36 .................................................................................. (Perimeter of a square = 4 × side)

∴ side = 36/4 = 9 cm ..................................................................................................... (Equation ii)

∴ Area of the square = side

^{2}

∴ Area of the square = 9

^{2}........................................................................................................................ (From Equation ii)

∴ Area of the square = 81 cm

^{2}

For the rectangle,

Let the width be x cm

∴ Width = x cm ............................................................................................................................................ (Equation iii)

∴ Length = width + 2 cm = x + 2 ..................................................................................................................... (Equation iv)

We Know that the same wire is used to make the rectangle and the triangle.

∴ Perimeter of the rectangle = perimeter of the triangle

∴ Perimeter of the rectangle = 36 ...................................................................................................................... (From Equation i)

∴ 2 (length + breadth) = 36 .................................................................................. (Perimeter of a rectangle= 2 × length + breadth)

∴ 2 (x + x + 2) = 36 ..................................................................................................... (From Equations iii & iv)

∴ 2 (2x + 2 ) = 36

∴ 4x + 4 = 36

∴ 4x = 36 - 4

∴ 4x = 32

∴ x = 32/4 = 8 cm ................................................................................... (Equation v)

Substituting Equation v in Equations iii & iv, we get-

Width = x = 8

Length = x + 2 = 8 + 2 = 10

∴ Area of the rectangle = length × breadth

∴ Area of the rectangle = 8 × 10

∴ Area of the rectangle = 80 cm

^{2}

Final Answer:-

∴ Areas of the given Square and rectangle are 81 cm

^{2}& 80 cm

^{2}, respectively.

### Volume of one big sphere is equal to the volume of 8 small spheres. Calculate the ratio of volume of big sphere to volume of small sphere.

- We have given volume of one big sphere is equal to the volume of 8 small spheres
- We have to find the ratio of volume of big sphere to volume of small sphere

We have given volume of one big sphere is equal to volume of 8small spheres.

Let volume of big sphere be denoted as V

_{b}and volume of small sphere be denoted as V

_{s}

So,

Therefore the ratio is 8:1

### Volume of one big sphere is equal to the volume of 8 small spheres. Calculate the ratio of volume of big sphere to volume of small sphere.

- We have given volume of one big sphere is equal to the volume of 8 small spheres
- We have to find the ratio of volume of big sphere to volume of small sphere

We have given volume of one big sphere is equal to volume of 8small spheres.

Let volume of big sphere be denoted as V

_{b}and volume of small sphere be denoted as V

_{s}

So,

Therefore the ratio is 8:1

### In the figure, ABCD is a rectangle and PQRS is a square. Find the area of the shaded portion.

we can see from the diagram that the Area of shaded portion can be calculated by-

Step- 1: Finding the area of rectangle ABCD (with length = 20 cm & breadth = 15 cm) and

Step- 2: Subtracting the area of square PQRS (with side = 5 cm) from it.

∴ Area of shaded portion = Area of rectangles ABCD - Area of □ PQRS

∴ Area of shaded portion = length × breadth - (side)2

∴ Area of shaded portion = (20 × 15) - 52

∴ Area of shaded portion = 300 - 25

∴ Area of shaded portion = 275 cm2

Final Answer:-

∴ Area of the shaded portion is 275 cm2

### In the figure, ABCD is a rectangle and PQRS is a square. Find the area of the shaded portion.

we can see from the diagram that the Area of shaded portion can be calculated by-

Step- 1: Finding the area of rectangle ABCD (with length = 20 cm & breadth = 15 cm) and

Step- 2: Subtracting the area of square PQRS (with side = 5 cm) from it.

∴ Area of shaded portion = Area of rectangles ABCD - Area of □ PQRS

∴ Area of shaded portion = length × breadth - (side)2

∴ Area of shaded portion = (20 × 15) - 52

∴ Area of shaded portion = 300 - 25

∴ Area of shaded portion = 275 cm2

Final Answer:-

∴ Area of the shaded portion is 275 cm2

### A room is 10 m long and 6 m wide. How many tiles of size 20 cm by 10 cm are required to cover its floor?

Area of a rectangle = length × breadth

Step-by-step solution:-

Covering the floor of a room with tiles means putting tiles on the whole surface of the room.

∴ If we know the area of the whole room and the area of each tile, we can find the number of tiles required to cover the floor.

Let the number of tiles required be n

Now, from the given information,

length of the room = l1 = 10m

breadth of the room = b

^{1}= 6m

length of each tile = l2 = 20 cm = 0.2 m ............................................... (1m = 100 cm. ∴ 20 cm = 20/100 m = 0.2 m)

breadth of each tile = b

^{2}= 10 cm = 01. m ............................................ (1m = 100 cm. ∴ 10 cm = 10/100 m = 0.1 m)

∴ Area of the room = length × breadth

∴ Area of the room = 10 × 6

∴ Area of the room = 60 m

^{2}............................................................... (Equation i)

Area of each tile = length × breadth

∴ Area of each tile = 0.2 × 0.1

∴ Area of each tile = 0.02 m2 ............................................................... (Equation ii)

Area of the room = Area of each tile × total number of tiles required.

∴ 60 = 0.02 × total number of tiles required ................... (From Equations i & ii)

∴ 60 / 0.02 = total number of tiles required

∴ 3,000 = total number of tiles required

Step-by-step solution:-

Covering the floor of a room with tiles means putting tiles on the whole surface of the room.

∴ If we know the area of the whole room and the area of each tile, we can find the number of tiles required to cover the floor.

Let the number of tiles required be n

Now, from the given information,

length of the room = l

^{1}= 10m

breadth of the room = b

^{1}= 6m

length of each tile = l

^{2}= 20 cm = 0.2 m ............................................... (1m = 100 cm. ∴ 20 cm = 20/100 m = 0.2 m)

breadth of each tile = b2 = 10 cm = 01. m ............................................ (1m = 100 cm. ∴ 10 cm = 10/100 m = 0.1 m)

∴ Area of the room = length × breadth

∴ Area of the room = 10 × 6

∴ Area of the room = 60 m

^{2}............................................................... (Equation i)

Area of each tile = length × breadth

∴ Area of each tile = 0.2 × 0.1

∴ Area of each tile = 0.02 m

^{2}............................................................... (Equation ii)

Area of the room = Area of each tile × total number of tiles required.

∴ 60 = 0.02 × total number of tiles required ................... (From Equations i & ii)

∴ 60 / 0.02 = total number of tiles required

∴ 3,000 = total number of tiles required

Final Answer:-

∴ 3,000 tiles would be required to cover the floor of the given room.

### A room is 10 m long and 6 m wide. How many tiles of size 20 cm by 10 cm are required to cover its floor?

Area of a rectangle = length × breadth

Step-by-step solution:-

Covering the floor of a room with tiles means putting tiles on the whole surface of the room.

∴ If we know the area of the whole room and the area of each tile, we can find the number of tiles required to cover the floor.

Let the number of tiles required be n

Now, from the given information,

length of the room = l1 = 10m

breadth of the room = b

^{1}= 6m

length of each tile = l2 = 20 cm = 0.2 m ............................................... (1m = 100 cm. ∴ 20 cm = 20/100 m = 0.2 m)

breadth of each tile = b

^{2}= 10 cm = 01. m ............................................ (1m = 100 cm. ∴ 10 cm = 10/100 m = 0.1 m)

∴ Area of the room = length × breadth

∴ Area of the room = 10 × 6

∴ Area of the room = 60 m

^{2}............................................................... (Equation i)

Area of each tile = length × breadth

∴ Area of each tile = 0.2 × 0.1

∴ Area of each tile = 0.02 m2 ............................................................... (Equation ii)

Area of the room = Area of each tile × total number of tiles required.

∴ 60 = 0.02 × total number of tiles required ................... (From Equations i & ii)

∴ 60 / 0.02 = total number of tiles required

∴ 3,000 = total number of tiles required

Step-by-step solution:-

Covering the floor of a room with tiles means putting tiles on the whole surface of the room.

∴ If we know the area of the whole room and the area of each tile, we can find the number of tiles required to cover the floor.

Let the number of tiles required be n

Now, from the given information,

length of the room = l

^{1}= 10m

breadth of the room = b

^{1}= 6m

length of each tile = l

^{2}= 20 cm = 0.2 m ............................................... (1m = 100 cm. ∴ 20 cm = 20/100 m = 0.2 m)

breadth of each tile = b2 = 10 cm = 01. m ............................................ (1m = 100 cm. ∴ 10 cm = 10/100 m = 0.1 m)

∴ Area of the room = length × breadth

∴ Area of the room = 10 × 6

∴ Area of the room = 60 m

^{2}............................................................... (Equation i)

Area of each tile = length × breadth

∴ Area of each tile = 0.2 × 0.1

∴ Area of each tile = 0.02 m

^{2}............................................................... (Equation ii)

Area of the room = Area of each tile × total number of tiles required.

∴ 60 = 0.02 × total number of tiles required ................... (From Equations i & ii)

∴ 60 / 0.02 = total number of tiles required

∴ 3,000 = total number of tiles required

Final Answer:-

∴ 3,000 tiles would be required to cover the floor of the given room.

### A Solid is in the form of a right circular cone mounted on a hemisphere. The radius of hemisphere is 3.5 cm and height of the cone is 4 cm . The solid is placed in a cylindrical vessel , full of water , in such a way that the whole solid is submerged in water. If the radius of cylindrical vessel is 5 cm and its height is 10.5 cm . Find the volume of water left in the cylindrical vessel

- We have given a Solid is in the form of a right circular cone mounted on a hemisphere. The radius of hemisphere is 3.5 cm and height of the cone is 4 cm . The solid is placed in a cylindrical vessel , full of water , in such a way that the whole solid is submerged in water. If the radius of cylindrical vessel is 5 cm and its height is 10.5 cm
- We have to find volume of water left in the cylindrical vessel.

We have given radius of the hemisphere is 3.5cm

Now the solids is in the form of a right circular cone mounted on a hemisphere, then radius of the base of the cone will be equal to radius of the hemisphere.

Radius of the base of the cone is

Height of the cone is

So,

Volume of the solid = volume of the cone + volume of hemisphere.

So,

= 141.109

Now the radius of the base of the cylindrical vessel is 5cm

Height of the cylindrical vessel is 10.5cm

So, Volume of the water in the cylindrical vessel will be

= 825cm^{3}

Now, when the solid is completely submerged in the cylindrical vessel full of water,

The

Volume of the water left in the vessel = volume of the water in the vessel – volume of solid

= (825 - 141.16)cm^{3}

= 683.84cm^{3}

### A Solid is in the form of a right circular cone mounted on a hemisphere. The radius of hemisphere is 3.5 cm and height of the cone is 4 cm . The solid is placed in a cylindrical vessel , full of water , in such a way that the whole solid is submerged in water. If the radius of cylindrical vessel is 5 cm and its height is 10.5 cm . Find the volume of water left in the cylindrical vessel

- We have given a Solid is in the form of a right circular cone mounted on a hemisphere. The radius of hemisphere is 3.5 cm and height of the cone is 4 cm . The solid is placed in a cylindrical vessel , full of water , in such a way that the whole solid is submerged in water. If the radius of cylindrical vessel is 5 cm and its height is 10.5 cm
- We have to find volume of water left in the cylindrical vessel.

We have given radius of the hemisphere is 3.5cm

Now the solids is in the form of a right circular cone mounted on a hemisphere, then radius of the base of the cone will be equal to radius of the hemisphere.

Radius of the base of the cone is

Height of the cone is

So,

Volume of the solid = volume of the cone + volume of hemisphere.

So,

= 141.109

Now the radius of the base of the cylindrical vessel is 5cm

Height of the cylindrical vessel is 10.5cm

So, Volume of the water in the cylindrical vessel will be

= 825cm^{3}

Now, when the solid is completely submerged in the cylindrical vessel full of water,

The

Volume of the water left in the vessel = volume of the water in the vessel – volume of solid

= (825 - 141.16)cm^{3}

= 683.84cm^{3}

### A path of uniform width 4m runs around the outside a rectangular field 24m by 18 m.Find the area of the path ?

Area of a rectangle = Length × breadth

Step-by-step solution:-

From the adjacent diagram,

Let the outer rectangle represent the path around the field.

∴ Length of the outer rectangle = length of the field + 2 (width of the path)

∴ Length of the outer rectangle = 24 + 2 (4)

∴ Length of the outer rectangle = 24 + 8

∴ Length of the outer rectangle = 32 m ........................................................... (Equation i)

Also, breadth of the outer rectangle = breadth of the field + 2 (width of the path)

∴ breadth of the outer rectangle = 18 + 2 (4)

∴ breadth of the outer rectangle = 18 + 8

∴ breadth of the outer rectangle = 26 m .................................................. (Equation ii)

Area of the outer rectangle = length × breadth

∴ Area of the outer rectangle = 32 × 26 ......................................................... (From Equations i & ii)

∴ Area of the outer rectangle = 832 m

^{2}............................................................ (Equation iii)

Area of the field = length × breadth

∴ Area of the room = 24 × 18 ......................................................................... (From given information)

∴ Area of the room = 432 m

^{2}.......................................................................... (Equation iv)

Now, Area of the path = Area of outer rectangle - Area of the inner field

∴ Area of the path = 832 - 432 ............................................................ (From Equations iii & iv)

∴ Area of the path = 400 m

^{2}

Final Answer:-

∴ Area of the path is Rs. 400 m

^{2}.

### A path of uniform width 4m runs around the outside a rectangular field 24m by 18 m.Find the area of the path ?

Area of a rectangle = Length × breadth

Step-by-step solution:-

From the adjacent diagram,

Let the outer rectangle represent the path around the field.

∴ Length of the outer rectangle = length of the field + 2 (width of the path)

∴ Length of the outer rectangle = 24 + 2 (4)

∴ Length of the outer rectangle = 24 + 8

∴ Length of the outer rectangle = 32 m ........................................................... (Equation i)

Also, breadth of the outer rectangle = breadth of the field + 2 (width of the path)

∴ breadth of the outer rectangle = 18 + 2 (4)

∴ breadth of the outer rectangle = 18 + 8

∴ breadth of the outer rectangle = 26 m .................................................. (Equation ii)

Area of the outer rectangle = length × breadth

∴ Area of the outer rectangle = 32 × 26 ......................................................... (From Equations i & ii)

∴ Area of the outer rectangle = 832 m

^{2}............................................................ (Equation iii)

Area of the field = length × breadth

∴ Area of the room = 24 × 18 ......................................................................... (From given information)

∴ Area of the room = 432 m

^{2}.......................................................................... (Equation iv)

Now, Area of the path = Area of outer rectangle - Area of the inner field

∴ Area of the path = 832 - 432 ............................................................ (From Equations iii & iv)

∴ Area of the path = 400 m

^{2}

Final Answer:-

∴ Area of the path is Rs. 400 m

^{2}.

### The perimeter of a rectangle is 28 cm and its length is 8 cm. Find its:

(i) breadth (ii) area

i. We will use the formula for perimeter of a rectangle and substitute the value of perimeter given in the question.

Let the breadth be x cm

Perimeter of the given rectangle = 28

∴ 2 (length + breadth) = 28

∴ 2 × (8 + x) = 28

∴ 2 × 8 + 2 × x = 28 .............. (Opening the bracket and multiplying 2 with the whole term)

∴ 16 + 2x = 28

∴ 2x = 28 - 16

∴ 2x = 12

∴ x =

∴ x = breadth = 6 cm ....... (Equation i)

ii. We will use the formula for area of rectangle and substitute the value of breadth from equation i and length from given information.

Area of the given rectangle = length × breadth

∴ Area of the given rectangle = 8 × 6 ............ (From Equation i & given information)

∴ Area of the given rectangle = 48 cm2

Final Answer:-

∴ For a rectangle with perimeter 28 cm and length 8 cm, its breadth is 6 cm and area is 48 cm2

### The perimeter of a rectangle is 28 cm and its length is 8 cm. Find its:

(i) breadth (ii) area

i. We will use the formula for perimeter of a rectangle and substitute the value of perimeter given in the question.

Let the breadth be x cm

Perimeter of the given rectangle = 28

∴ 2 (length + breadth) = 28

∴ 2 × (8 + x) = 28

∴ 2 × 8 + 2 × x = 28 .............. (Opening the bracket and multiplying 2 with the whole term)

∴ 16 + 2x = 28

∴ 2x = 28 - 16

∴ 2x = 12

∴ x =

∴ x = breadth = 6 cm ....... (Equation i)

ii. We will use the formula for area of rectangle and substitute the value of breadth from equation i and length from given information.

Area of the given rectangle = length × breadth

∴ Area of the given rectangle = 8 × 6 ............ (From Equation i & given information)

∴ Area of the given rectangle = 48 cm2

Final Answer:-

∴ For a rectangle with perimeter 28 cm and length 8 cm, its breadth is 6 cm and area is 48 cm2

### How many square tiles of the side 20 cm will be needed to pave a footpath which is 2 m wide and surrounds a rectangular plot 40 m long and 22 m wide.

Area of a rectangle = length × breadth

Step-by-step solution:-

Paving the floor of a footpath with tiles means putting tiles on the whole surface of the footpath.

∴ If we know the area of the footpath and the area of each tile, we can find the number of tiles required to cover the footpath.

Let the number of tiles required be n

Now, from the given information,

length of the plot = l1 = 40 m

breadth of the plot = b1 = 22 m

Side of each tile = 20 cm = 0.2 m (1m = 100 cm)

∴ Area of the plot = length × breadth

∴ Area of the plot = 40 × 22

∴ Area of the plot = 880 m2 ............................................................... (Equation i)

In the adjacent diagram-

For the outer rectangle-

Length = Length of plot + 2 (width of footpath)

Length = 40 + 2 (2)

Length = 40 + 4

Length = 44 m .................................................................................. (Equation ii)

Breadth = Breadth of plot + 2 (width of footpath)

Breadth = 22 + 2 (2)

Breadth = 22 + 4

Breadth = 26 m .................................................................................. (Equation ii)

Area of Outer rectangle = length × breadth

∴ Area of Outer rectangle = 44 × 26 ..................................................... (From Equations ii & iii)

∴ Area of Outer rectangle = 1,144 m

^{2}.................................................. (Equation iv)

Area of footpath = Area of outer rectangle - Area of inner plot

∴ Area of footpath = 1,144 - 880 ......................................................... (From Equations i & iv)

∴ Area of footpath = 264 m

^{2}............................................................... (Equation v)

Area of each tile = Side

^{2}

∴ Area of each tile = (0.2)

^{2}

∴ Area of each tile = 0.04 m

^{2}............................................................... (Equation vi)

Area of the footpath = Area of each tile × total number of tiles required.

∴ 264 = 0.04 × total number of tiles required ................... (From Equations v & vi)

∴ 264 / 0.04 = total number of tiles required

∴ 6,600 = total number of tiles required

Final Answer:-

∴ 6,600 tiles would be required to cover the footpath.

### How many square tiles of the side 20 cm will be needed to pave a footpath which is 2 m wide and surrounds a rectangular plot 40 m long and 22 m wide.

Area of a rectangle = length × breadth

Step-by-step solution:-

Paving the floor of a footpath with tiles means putting tiles on the whole surface of the footpath.

∴ If we know the area of the footpath and the area of each tile, we can find the number of tiles required to cover the footpath.

Let the number of tiles required be n

Now, from the given information,

length of the plot = l1 = 40 m

breadth of the plot = b1 = 22 m

Side of each tile = 20 cm = 0.2 m (1m = 100 cm)

∴ Area of the plot = length × breadth

∴ Area of the plot = 40 × 22

∴ Area of the plot = 880 m2 ............................................................... (Equation i)

In the adjacent diagram-

For the outer rectangle-

Length = Length of plot + 2 (width of footpath)

Length = 40 + 2 (2)

Length = 40 + 4

Length = 44 m .................................................................................. (Equation ii)

Breadth = Breadth of plot + 2 (width of footpath)

Breadth = 22 + 2 (2)

Breadth = 22 + 4

Breadth = 26 m .................................................................................. (Equation ii)

Area of Outer rectangle = length × breadth

∴ Area of Outer rectangle = 44 × 26 ..................................................... (From Equations ii & iii)

∴ Area of Outer rectangle = 1,144 m

^{2}.................................................. (Equation iv)

Area of footpath = Area of outer rectangle - Area of inner plot

∴ Area of footpath = 1,144 - 880 ......................................................... (From Equations i & iv)

∴ Area of footpath = 264 m

^{2}............................................................... (Equation v)

Area of each tile = Side

^{2}

∴ Area of each tile = (0.2)

^{2}

∴ Area of each tile = 0.04 m

^{2}............................................................... (Equation vi)

Area of the footpath = Area of each tile × total number of tiles required.

∴ 264 = 0.04 × total number of tiles required ................... (From Equations v & vi)

∴ 264 / 0.04 = total number of tiles required

∴ 6,600 = total number of tiles required

Final Answer:-

∴ 6,600 tiles would be required to cover the footpath.

### A circle has a diameter of 120 metres. Your average walking speed is 4 kilometres per hour. How many minutes will it take you to walk around the boundary of the circle 3 times? Round your answer to the nearest integer.

Explanation :-

Given , the diameter of the circle is 120 m .The no. of round need to cover = 3

Given , the speed = 4 km / hr

circumference of circle

The total distance = no. of rounds × circumference of circle

The total distance

We get

hrs = 0.1885 hrs

As we know 1 hour = 60 mins

As time = 0.1885 × 60 mins = 11.3097 mins (11 is the nearest integer)

### A circle has a diameter of 120 metres. Your average walking speed is 4 kilometres per hour. How many minutes will it take you to walk around the boundary of the circle 3 times? Round your answer to the nearest integer.

Explanation :-

Given , the diameter of the circle is 120 m .The no. of round need to cover = 3

Given , the speed = 4 km / hr

circumference of circle

The total distance = no. of rounds × circumference of circle

The total distance

We get

hrs = 0.1885 hrs

As we know 1 hour = 60 mins

As time = 0.1885 × 60 mins = 11.3097 mins (11 is the nearest integer)

### Find the total area in the given figure.

In the adjacent figure, Let rectangles ABCD & □ EFGC be the rectangle and square, respectively.

∴ from the given diagram, length = 50, breadth = 20, side = 30

We have to find the area of entire figure which includes the rectangle & square mentioned above.

∴ Area of the given figure = Area of rectangles ABCD + Area of rectangles EFGC

∴ Area of the given figure = (length × breadth) + (side)2

∴ Area of the given figure = (50×20) + (30)2

∴ Area of the given figure = 1,000 + 900

∴ Area of the given figure = 1,900 square units.

Final Answer:-

∴ Total area in the given figure is 190 square units.

### Find the total area in the given figure.

In the adjacent figure, Let rectangles ABCD & □ EFGC be the rectangle and square, respectively.

∴ from the given diagram, length = 50, breadth = 20, side = 30

We have to find the area of entire figure which includes the rectangle & square mentioned above.

∴ Area of the given figure = Area of rectangles ABCD + Area of rectangles EFGC

∴ Area of the given figure = (length × breadth) + (side)2

∴ Area of the given figure = (50×20) + (30)2

∴ Area of the given figure = 1,000 + 900

∴ Area of the given figure = 1,900 square units.

Final Answer:-

∴ Total area in the given figure is 190 square units.

### A rectangular parking lot is 90 yards long and 35 yards wide. It costs about $.45 to pave each square foot of the parking lot with asphalt. About how much will it cost to pave the parking lot?

Explanation :-

Given, the length and breadth rectangular parking lot are 90 yards and 35 yards

We know 1 yard = 3 foot , convert l and b to foot .

Then we get area of parking lot = 28350 sq. foot

Given , the cost per sq. foot is $ 0.45

Total cost of paving parking lot = area of parking lot in sq. foot ×cost of paving parking lot per foot

Total cost of paving parking lot = 28350 sq. foot × $0.45/ sq. foot = $12757.5

∴The cost of paving the rectangular parking lot is $12757.5

### A rectangular parking lot is 90 yards long and 35 yards wide. It costs about $.45 to pave each square foot of the parking lot with asphalt. About how much will it cost to pave the parking lot?

Explanation :-

Given, the length and breadth rectangular parking lot are 90 yards and 35 yards

We know 1 yard = 3 foot , convert l and b to foot .

Then we get area of parking lot = 28350 sq. foot

Given , the cost per sq. foot is $ 0.45

Total cost of paving parking lot = area of parking lot in sq. foot ×cost of paving parking lot per foot

Total cost of paving parking lot = 28350 sq. foot × $0.45/ sq. foot = $12757.5

∴The cost of paving the rectangular parking lot is $12757.5

### A person is standing exactly at the centre of a circle. The distance of the person from the boundary of the circle is 3 ft. Find the area of the circle.

Explanation :-

Given, the radius of circle is 3 ft

Then we get area of circle =

= 28.274 sq. ft

∴The area of the circle is 28.274 sq. feet.

### A person is standing exactly at the centre of a circle. The distance of the person from the boundary of the circle is 3 ft. Find the area of the circle.

Explanation :-

Given, the radius of circle is 3 ft

Then we get area of circle =

= 28.274 sq. ft

∴The area of the circle is 28.274 sq. feet.

### The length and breadth of a rectangular plot are in the ratio 3: 1 and its perimeter is 128 m. Find the area of the plot ?

Perimeter of a rectangle = 2 (length + breadth)

Area of a rectangle = length × breadth

Step-by-step solution:-

For the given rectangle-

Length : Breadth = 3 : 1

Let x be the common factor.

∴ Length = 3x …............................................ (Equation i)

& Breadth = x …............................................ (Equation ii)

Perimeter of a rectangle = 2 (length + breadth)

∴ Perimeter of a rectangle = 2 (3x + x) ............ (From Equations i & ii)

∴ Perimeter of a rectangle = 2 × 4x

∴ Perimeter of a rectangle = 8x

∴ 128 = 8x ..................... (From given information)

∴ 128 / 8 = x

∴ 16 = x

∴ Length = 3x = 3 × 16 = 48 m ............................ (Equation iii)

& Breadth = x = 16 m ....................................... (Equation iv)

Area of the given rectangle = length × breadth

∴ Area of the given rectangle = 48 ×16

∴ Area of the given rectangle = 768 m

^{2}

Final Answer:-

∴ Area of the give rectangle is 768 m

^{2}.

### The length and breadth of a rectangular plot are in the ratio 3: 1 and its perimeter is 128 m. Find the area of the plot ?

Perimeter of a rectangle = 2 (length + breadth)

Area of a rectangle = length × breadth

Step-by-step solution:-

For the given rectangle-

Length : Breadth = 3 : 1

Let x be the common factor.

∴ Length = 3x …............................................ (Equation i)

& Breadth = x …............................................ (Equation ii)

Perimeter of a rectangle = 2 (length + breadth)

∴ Perimeter of a rectangle = 2 (3x + x) ............ (From Equations i & ii)

∴ Perimeter of a rectangle = 2 × 4x

∴ Perimeter of a rectangle = 8x

∴ 128 = 8x ..................... (From given information)

∴ 128 / 8 = x

∴ 16 = x

∴ Length = 3x = 3 × 16 = 48 m ............................ (Equation iii)

& Breadth = x = 16 m ....................................... (Equation iv)

Area of the given rectangle = length × breadth

∴ Area of the given rectangle = 48 ×16

∴ Area of the given rectangle = 768 m

^{2}

Final Answer:-

∴ Area of the give rectangle is 768 m

^{2}.

### Find the cost of fencing a square park with 16 m side if the cost of fencing is $16/m.

Explanation :-

Given the side length of square park = 16 m

Then perimeter (or) boundary length to be fenced = 4×16 = 64 m.

Given the cost of fencing per metre = $16 / m

Total cost of fencing = total perimeter in m × The cost of fencing per metre

Total cost of fencing = 64 m × $16 / m = $ 1024

∴$ 1024 is the cost of fencing the square park .

### Find the cost of fencing a square park with 16 m side if the cost of fencing is $16/m.

Explanation :-

Given the side length of square park = 16 m

Then perimeter (or) boundary length to be fenced = 4×16 = 64 m.

Given the cost of fencing per metre = $16 / m

Total cost of fencing = total perimeter in m × The cost of fencing per metre

Total cost of fencing = 64 m × $16 / m = $ 1024

∴$ 1024 is the cost of fencing the square park .

### If the area of a triangle is 77 m2 and its height is m, find the length of its base

**Explanation :-**

Given, The height of the triangle is m (i.e h = m)

Let the base length of the triangle be b

Given, area of triangle = 77 sq.m

Then we get area of triangle = ½ × b × h

∴Therefore, The base of the triangle is 112.93 m.

Given, The height of the triangle is m (i.e h = m)

Let the base length of the triangle be b

Given, area of triangle = 77 sq.m

Then we get area of triangle = ½ × b × h

∴Therefore, The base of the triangle is 112.93 m.

**If the area of a triangle is 77 m2 and its height is m, find the length of its base**

**Maths-General**

**ANS :- The base of the triangle is 112.93 m.**

**Explanation :-**

Given, The height of the triangle is m (i.e h = m)

Let the base length of the triangle be b

Given, area of triangle = 77 sq.m

Then we get area of triangle = ½ × b × h

∴Therefore, The base of the triangle is 112.93 m.Given, The height of the triangle is m (i.e h = m)

Let the base length of the triangle be b

Given, area of triangle = 77 sq.m

Then we get area of triangle = ½ × b × h

∴Therefore, The base of the triangle is 112.93 m.

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