Rd Sharma 2021 for Class 9 Maths Chapter 18 - Surface Area And Volume Of A Cuboid And Cube

Answer:

Dimensions are given as

Length

Breadth

Height

We have to find lateral surface area and total area

Hence its, lateral surface area,

Total surface area,

The lateral surface area of the cuboids is and total surface area of it is.

Answer:

Edge of the given cube,

We have to find lateral and total surface area

Lateral surface area,

Total surface area,

The lateral surface area of the cube is and its total surface area is.

Answer:

Let the length of the edge of the cube be

We have to find the ratio of total surface area and lateral surface area

Total surface area of the cube,

Lateral surface area of the cube,

The desired ratio,

The ratio of the total surface area and the lateral surface area of a cube is.

Answer:

The dimensions of the cubical block are,

We are asked to find the number of square sheet paper whose side is 40 cm

Let the total surface area of the block be.

So, Mary would require in total of colored paper.

But the paper is available in square sheets of side,

Area of a single square sheet,

The number of square sheets required=

Mary would require square sheets of paper.

Answer:

Dimensions of the room are,

Let,

S The total surface area to whitewash

A₁ The lateral surface area of the room

A₂ The surface area of ceiling

R The rate of whitewashing per

We know that,

We are asked to find the cost of whitewashing

Now, the total surface area to whitewash,

Total cost of whitewashing,

Hence the cost of whitewashing the room and the ceiling is .

Answer:

Let,

Side of each cube

Surface area of each cube

So,

Hence,

Sum of surface areas of three cubes,

The length (say l) of the newly formed cuboids is;

Its breadth (say b) and height (say h) will be the same as that of each cube.

Total surface area of the new cuboids is;

Required Ratio,

The total surface area of the new cuboids to that of the sum of the surface area of the three cubes is

Answer:

We can define the following notations as follows

Side of cube

Volume of cube

Side of cube

Volume of cube

Let;

Number of cubes formed

Surface area of a single small cube

We know,

And;

The number of cubes formed,

Total surface area of all the small cubes formed

The total surface area of all the small cubes is .

Answer:

The hall is cubical.

Let,

Length of the cuboids

Breadth of the cuboids

Height of the cuboids

We have,

We need to find the Height of the hall

It is given that,

The sum of the areas of the floor and the flat roof is equal to the sum of the areas of the four walls

Using abbreviations, we can write the same as,

Height of the wall is .

Answer:

The water tank is cubical.

So let,

Side of the cube

Total surface area covered by tiles

Side of each square tile

Area of each square tile

Number of tiles required

Cost of each tile

We are asked to find the total cost of the tiles 

We have,

.So,

We have,

So;

Now,

The cost of tiles is per dozen.

Hence,

Total cost for the tiles

Hameed would spend for the tiles.

Answer:

Let,

Initial edge of the cube

Initial surface area of the cube

Increased edge of the cube

Increased surface area of the cube

We need to find the percentage increase in the total surface area of the cube

We know that,

And

Now,

Percentage increase in

Percentage increase in the surface area of the cube is .

Answer:

We know,

Length of the iron tank

Width of the iron tank

Depth of the iron tank

Width of the iron sheet

Rate of the iron sheet

We need to find the cost of iron sheet used

Total surface area of the iron tank,

Length of the iron sheet required,

Cost of the required iron sheet,

The total cost of iron sheet used is .

Answer:

The shelter is a box-like structure and hence cubical.

We have,

Length of the cuboid

Breadth of the cuboid

Height of the cuboid

The total surface area of the cuboid,

The area of the base,

The quantity of tarpaulin required,

So the quantity of tarpaulin required is.

Answer:

The external dimensions of the wooden box,

Length

Breadthand height is

Thickness of the wood

We are asked to find the cost of painting

So, the internal dimensions of the box are,

Length

Breadth

Height

The internal surface area of the box,

We are given the rate of painting per square meter is

So the total cost of painting is,

The total cost of painting is .

Answer:

We are given the dimensions of the room as l = 12.5 m, b = 9 m, h = 7 m

The lateral surface area of the room,

Surface area of each door,

Surface area of each window,

There are 2 doors and 4 windows in the room.

Hence, total area to be painted,

Rate of painting the wall at the rate of,

So, total cost of painting,

The total cost of painting is .

Answer:

The paint in the container can paint the area,

Dimensions of a single brick,

Length

Breadth

Height

We need to find the number of bricks that can be painted

Surface area of a brick,

Number of bricks that can be painted

Hencebricks can be painted out of the container.

Answer:

The dimensions of the rectangular box are in the ratio.

So let the dimensions be,

Length

Breadth

Height

We are asked to find the dimensions of the box

The total surface area of the box,

The cost of covering it at the rate of per

The cost of covering it at the rate of per

We know that, the difference between above two costs is.

So,

So the dimensions of the box are;

Hence the dimensions of the box are.

Answer:

We have,

Cost of matting the floor

Rate of matting per square meter

Length of the floor

Let,

Area of the floor

Width of the room

So,

Now, we have,

The cost of preparing the walls

The rate of preparing the walls

Let,

Lateral surface area of the room

Height of the room

So,

Hence, height of the room is .

Answer:

The length and breadth of the hall are in the ratio 4 : 3.

Hence l = 4x, b = 3x, h = 5.5 m

Rate of decorating the wall, R = 6.6 per square meter

Total cost of decoration C = Rs.5082

We have to find the length and breadth of the room

Surface area of the walls,

Cost of decoration = A × R

Hence,

Length

Breadth

The length and breadth of the hall are and respectively.

Answer:

External dimensions of the bookshelf are,

Length,

Breadth,

Height,

External surface area of the bookshelf excluding the front face,

Area of the front face,

Area to be polished,

Rate of polishing

Total cost of polishing,

Now, above diagram will make it clear that for each row of bookshelf,

Length

Breadth

Height

Hence, area to be painted in one row,

Area to be painted in three rows,

Rate of painting

Total cost of painting,

Total Expense

Therefore, the total expenses are.

Answer:

Dimensions of the water tank, l = 6 m, b = 5 m, h = 4.5 m

We need to find the capacity of the tank

Capacity of the tank,

The tank can hold of water.

Answer:

We have,

Length of the vessel

Width of the vessel

Capacity of the vessel

Let: Minimum required height of the vessel

So,

Thus, to hold of liquid, the vessel must be minimum high.

Answer:

We have, the dimensions of the cubical pit are,

Length

Breadth

Depth

Rate of digging

Volume of the pit,

The cost of digging,

The cost of digging the pit is.

Answer:

We know that, areas of three adjacent faces of the cuboid are respectively.

Where,

Length of the cuboid

Breadth of the cuboid

Height of the cuboid

Let,

Volume of the cuboid

We have, areas of three adjacent faces of the cuboid are respectively,

So their product,

The volume of the cuboid is.

Answer:

Let,

Length of the room

Breadth of the room

Height of the room

Volume of the room

We have, b = 2h, l = 2b and volume of room is 512 dm³

We have to find the dimensions

We know that

We have,

Therefore,

Hence, the dimensions of the cuboid are,

Length, Breadth, Height
Therefore, length is equal to 16 dm, breadth is 8 m  and Height is 4 dm.

Answer:

Let,

Sides of the three small cubes

Volumes of the three small cubes

Side of the new cube formed

Volume of the new cube formed

Surface area of the new cube formed

Diagonal of the new cube formed

We have,

We need to find the volume, surface area and diagonal of the new cube

Now,

We know,

So, the volume, surface area and the diagonal of the new cube will be respectively.

Answer:

We have volume of each cube

Let,

Side of each cube

The two cubes are joined together and we are asked to find the surface area of new cuboid

We know that,

When the two cubes are joined end to end,

Dimensions of the resulting cuboid are,

Length

Breadth

Height

Hence, its surface area

The surface area of the resulting cuboid will be .

Answer:

Let,

Side of the cube

Volume of the cube

Sides of the three smaller cubes

Volumes of the three smaller cubes

We have,

,

We know that,

Edge of the third smaller cube is .

Answer:

Dimensions of the cinema hall are,

Length

Breath

Height

Each person requires of air (say, v)

We are asked to find the number of persons who can sit in the cinema hall

Let,

Volume of the hall, then

The number of people that can sit in the hall,

Maximum people can sit in the hall.

Answer:

We are given that 1 cubic cm of marble weighs 0.25 kg

Let,

Volume of the block

Length of the block

We have,

Width of the block

Thickness of the block

Weight of the block

We need to find the length of the block

We have, of marble occupies of volume.

So, of marble will occupy the volume,

The length of the block is.

Answer:

External dimensions of the box are,

Length

Breadth

Height

Thickness of the wood

We need to find the volume used

So, internal dimensions of the box are,

Length

Breadth

Height

Capacity of the box,

Volume of the wood,

Maximum of liquid can be placed in the box.

Volume of the wood used in the box is .

Answer:

External dimensions of the closed wooden box,

Length

Breath

Height

Thickness of the wood

We need to find number of bricks that can be put inside the box of dimension

Internal dimensions of the box,

Length

Breadth

Height

Capacity of the box,

Volume of each brick,

Number of bricks that can be put in the box,

The box can contain maximum bricks.

Answer:

“When an object is completely immersed in a liquid, the volume of the liquid displaced is equal to the volume of the object”

Using this principle, we shall now solve this problem.

We have,

Edge of the immersed cube

Length of the rectangular container

Breadth of the rectangular container

Let,

The rise in water level

Volume of the immersed cube

As per the above mentioned principle,

The rise in water level is .

Answer:

We have, dimensions of the plot that is dug,

Length

Breadth

Depth

Length of the field

Breadth of the field

We need to find the level of field raised

Here, the volume of earth taken out,

So the rise in the level of the field

The level of the field is raised by .

Answer:

We have,

Length of the field (L) = 18 m

Width of the field (B) = 15 m

Length of the pit (l) = 7.5 m

Breadth of the pit (b) = 6 m

Depth of the pit (h) = 0.8 m

We have to find the level of field raised

Volume of the earth dug out

The area on which the earth has to be spread,

The rise in the level of the field

The level of the field has been raised to .

Answer:

Let,

Number of days the water will last for

One villager requires of water per day.

Population of the village is 4000

Measurement of the tank is

We need to find for how many days the water of the tank will last

The requirement of water for all 4000 villagers for days,

But we are given;

The water of this tank will last for.

Answer:

We have,

Number of boxes

In the above structure we need to find the total volume

Edge of each cube

Volume of each cube

Hence, total volume of the structure,

The volume of the structure built by the child is.

Answer:

We have,

Volume of the godown

Volume of each crate

We need to find the maximum number of crates in the godown that can be placed

Hence, the number of crates that can be stored,

But, we can not place this amount of crates in the godown, as this is not an integer.

So, we can place maximumcrates in the godown.

Answer:

We have,

Length of the wall

Height of the wall

Thickness of the wall

Dimension of the brick is

We need to find the number of bricks

Here,

Volume of the wall,

Dimensions of the brick are,

So, number of bricks in the wall,

As this is not an integer, we should take least integer greater than.

So, we need bricks to build the wall.

Answer:

We have,

Volume of the cuboid

Surface area of the cuboid

Dimensions of the cuboid

We need to prove,

We know that,

And

Hence,

Answer:

Let,

Length of the cuboid

Breadth of the cuboid

Height of the cuboid

Volume of the cuboid

Areas of three adjacent faces of the cuboid

We know that, areas of three adjacent faces of the cuboid are lb, bh, and hl respectively

Hence,

Hence,

Answer:

River is flowing at the speed of,

Width of the river

Depth of the river

We need to find the water flowed in one minute

So, in one minute, the river covers the distance (say, l) of .

Quantity of water that will fall into the sea in one minute,

So, in one minute of water will fall into the sea.

Answer:

We are given;

Velocity of the water

So, in 30 min, it will go the distance (say, l) 5,00,000 dm.

Width of the canal

Depth of the canal

In 30 min, quantity of water flown,

If 8 cm of standing water is desired, then the area that will be irrigated,

In 30 min, it will irrigate the area of .

Answer:

Volume of the gold sheet,

Area it covers,

Let,

Thickness of the sheet

We know that,

Thickness of the sheet is .

Answer:

We have,

External dimensions of the iron box are,

Length

Breadth

Height

Thickness of ironand of iron weighs

We are asked to find the volume of the metal used in the box and weight of the empty box

Internal dimensions of the box,

Length

Breadth

Height

Let,

External volume of the box

Internal volume of the box

Volume of the iron

So,

We have, of iron weighs,

So, weight of of iron,

In that open box, there areof iron, and weight of the empty box is .

Answer:

“When an object is completely immersed in a liquid, the volume of the liquid displaced is equal to the volume of the object”

Using this principle, we shall now solve this problem.

We have,

Length of the container

Breadth of the container

Height to which the water raised

Volume of the water displaced,

= volume of the water raised + volume of the water over flown

We need to calculate the volume and edge of the cube

Let,

Volume of the cube submerged

Edge of the cube submerged

According to the principle mentioned above,

Volume of the cube is and edge of the cube is .

Answer:

It is given that

Length of the tank (l) = 80 m

Breadth of the tank (b) = 25 m

= 2500 cm

Area of the cross section of the pipe is 25 cm²

The water flows at the rate of 16 km/hr.

We are asked to find the level of tank raised in 45 minutes

In 45 minutes, the water through the pipe will go,

Area of the cross-section of the pipe is 25 cm².

So, the quantity of the water poured in 45 minutes,

Let,

Height to which the water is raised

So,

In 45 minutes, the water level rises by .

Answer:

Dimensions of the reservoir are,

Length

Breadth

Depth

Side of the cross-section of the pipe

The flow of water through the water is;

We are asked to find the time in which the reservoir can be emptied

Here, volume of the water in the reservoir,

Since the side of the cross-section of the pipe

So, the area of the cross-section of the pipe,

Velocity of the water,

Let,

Time required emptying the reservoir

So,

Using that pipe, the water is emptied in.

Answer:

The longest rod that can be fitted in the cubical vessel is its diagonal.

Side of the cube

So, the diagonal of the cube,

So, the length of the longest rod that can be fitted in the cubical box is.

Hence, the correct choice is (c).

Answer:

Let, Side of each cube

So, the dimensions of the resulting cuboid are,

Length

Breadth

Height

Total surface area of the cuboid,

Sum of the surface areas of the three cubes,

Required ratio,

Thus, the required ratio is.

Hence the correct choice is (a).

Answer:

Let,

Side of the cube

Length of the diagonal

We have to find the surface area of the cube

Surface area of the cube,

Thus, surface area of the cube is.

Hence, the correct choice is (b).

Answer:

Let,

Volumes of the two cubes

Edges of the two cubes

We know that,

So,

Ratio of their edges is.

So, the correct choice is (c).

Answer:

Let,

Side of the cube

Volume of the cube

Surface area of the cube

We have,

So,

Thus, volume of the cube is.

Hence the correct choice is (c).

Answer:

Length (l), width (b) and height (h) of the rectangular solid are in the ratio 3 : 2 : 1.

So we can take,

We need to find the total surface area of the box

Volume of the box,

Thus,

Surface area of the box,

Thus total surface area of the box is.

Hence, the correct option is (d).

Answer:

Let, the edges of the cuboid be a cm, b cm and c cm.

And, a < b < c

The areas of the three adjacent faces are in the ratio 2 : 3 : 4.

So,

ab : ca : bc = 2 : 3 : 4, and its volume is 9000 cm³

We have to find the shortest edge of the cuboid

Since;

Similarly,

Volume of the cuboid,

As and

Thus, length of the shortest edge is.

Hence; the correct choice is (c).

Answer:

Let, Initial edge of the cube

So,

In the new cube, let,

Edge of new cube

Volume of the new cube,

Volume of the new cube is.

Hence, the correct choice is (d).

Answer:

Let,

Length of the first cuboid

Breadth of the first cuboid

Height of the first cuboid

And,

Length of the new cuboid

Breadth of the new cuboid

Height of the new cuboid

We know that,

Surface area of the first cuboid,

Surface area of the new cuboid,

The surface area of the new cuboid is.

So, the correct choice is (b).

Answer:

The area of the floor

Height of the room

We have to find the volume of the air in the room

So, capacity of the room to contain air,

Volume of the air contained in the room is.

So the correct choice is (d).

Answer:

Dimensions of the wall are,

Length

Breadth

Height

Volume of the hall,

Cost of building the wall at the rate of Rs. 25/m³,

The cost of building the wall is.

Hence, the correct option is (c).

Answer:

Volume of the clay to be spread,

Area on which the clay is spread

Let,

Rise in the level of the ground

We know that,

Rise in the level of the ground is.

Hence, the correct option is (a).

Answer:

Let,

Length of the first cuboid

Breadth of the first cuboid

Height of the first cuboid

Volume of the cuboid is 12 cm³

Dimensions of the new cuboid are,

Length

Breadth

Height

We are asked to find the volume of the new cuboid

We know that,

Volume of the new cuboid,

Thus volume of the new cuboid is.

Hence, the correct option is (d).

Answer:

A cube has total 12 edges.

Let, edge of the cube

Sum of all the edges of the cube = 12a

Volume of that cube,

Volume of the cube is.

Hence, the correct option is (b).

Answer:

We have the cuboid of dimensions.

We are to find how many cubs with edge 3 cm can be cut from the given cuboid

Let us cut this cuboid into following two cuboids

And

So the number of cubes of side 3 cm, that can be cut from the first cuboid,

We can not cut a single cube of side 3 cm from the second cuboid of dimension

Hence this much volume is useless for us.

So, we can cut maximum cubes of side 3 cm from the cuboid of dimensions.

Hence, the correct option is (c).

Answer:

Length of the terrace,

Breadth of the terrace,

Height of the water level

We have to find the quantity of water

Quantity of water,

The quantity of water is.

The correct option is (d).

Answer:

We have;

Here A₁, A₂ and A₃ are the areas of three adjacent faces of a cuboid.

But the areas of three adjacent faces of a cuboid are lb, bh and hl, where,

Length of the cuboid

Breadth of the cuboid

Height of the cuboid

We have to find the volume of the cuboid

Here,

Thus, volume of the cuboid is.

Hence, the correct choice is (c).

Answer:

We have,

Diagonal of the cube

Volume of the cube

Side of the cube

We know that,

So, the correct choice is (d).

Answer:

Dimensions of the cuboid are.

So, the surface area of the cuboid

Volume of the cuboid

Hence, the correct choice is (c).

Answer:

Let,

Length of the cuboid

Breadth of the cuboid

Height of the cuboid

We have,

, diagonal of the cuboid

We are asked to find the surface area

So, the surface area,

Thus, the surface area is

Hence, the correct choice is (c).

Answer:

Let,

Initial edge of the cube

Initial surface area of the cube

Increased edge of the cube

Increased surface area of the cube

We have to find the percentage increase in the surface area of the cube

Since it’s given that

We have,

Percentage increase in surface area,

Increase in surface area is.

Hence, the correct choice is (d).

Answer:

Let,

Volumes of the three cubes

Sides of the three cubes

We know that,

So,

Similarly,

And;

So the height of the resulting structure,

The height of the structure is.

Hence, the correct choice is (a).

Answer:

Let the edge of the cube be a cm.

Lateral surface area of cube = 256 cm²             (Given)

$\Rightarrow 4a^2=256$

$\Rightarrow a^2=64$

$\Rightarrow a=8 \mathrm{cm}$

∴ Volume of the cube = a³ = (8 cm)³ = 512 cm³

Thus, the volume of the cube is 512 cm³.

The lateral surface area of a cube is 256 cm². The volume of the cube is __512 cm³__.

Answer:

The length of the longest pole that can be put in a room is same as the length of the diagonal of the room.

The dimensions of the given room are 10 m × 10 m × 5 m.

Let the length, breadth and height of the room be l, b and h, respectively.

∴ l = 10 m, b = 10 m and h = 5 m

Now,

Length of the longest pole that can be put in the room

= Length of diagonal of the room

$=\sqrt{l^2+b^2+h^2}$

$=\sqrt{{10}^2+{10}^2+5^2}$

$=\sqrt{100+100+25}$

$=\sqrt{225}$

$=15 \mathrm{m}$

Thus, the length of the longest pole that can be put in the room of given dimensions is 15 m.

The length of the longest pole that can be put in a room of dimensions 10 m × 10 m × 5 m is ___15 m___.

Answer:

Volume of the pit = Length × Breadth × Height = 16 m × 12 m × 4 m = 16 m × 1200 cm × 400 cm               (1 m = 100 cm)

Volume of each plank = 4 m × 50 cm × 20 cm

∴ Number of planks that can be stored in the pit

$$\begin{gathered} =\frac{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{pit}}{\mathrm{Volume} \mathrm{of} \mathrm{each} \mathrm{plank}} \\ =\frac{16 \mathrm{m}\times 1200 \mathrm{cm}\times 400 \mathrm{cm}}{4 \mathrm{m}\times 50 \mathrm{cm}\times 20 \mathrm{cm}} \\ =1920 \end{gathered}$$

Thus, the number of planks that can be stored in the given pit are 1920.

The number of planks of dimensions 4 m × 50 cm × 20 cm that can be stored in a pit which is 16 m long, 12 m wide and 4 m deep is __1920__.

Answer:

Let the edges of the two cubes be x units and y units.

Volume of cube 1 : Volume of cube 2 = 729 : 1331            (Given)

$\therefore \frac{x^3}{y^3}=\frac{729}{1331}$                       [Volume of the cube = (Edge)³]

$\Rightarrow {\left(\frac{x}{y}\right)}^3={\left(\frac{9}{11}\right)}^3$

$\Rightarrow \frac{x}{y}=\frac{9}{11} .....\left(1\right)$

$\therefore \frac{\mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{cube} 1}{\mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{cube} 2}=\frac{6x^2}{6y^2}$

$\Rightarrow \frac{\mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{cube} 1}{\mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{cube} 2}={\left(\frac{x}{y}\right)}^2={\left(\frac{9}{11}\right)}^2=\frac{81}{121}$                 [Using (1)]

⇒ Total surface area of cube 1 : Total surface area of cube 2 = 81 : 121

Thus, the ratio of their total surface areas is 81 : 121.

The ratio of the volumes of two cubes is 729 : 1331. The ratio of their total surface areas is ___81 : 121___.

Answer:

Let the length of the cuboid be l cm.

Breadth of the cuboid, b = 4 cm

Height of the cuboid, h = 4 cm

Now,

Total surface area of the cuboid = 148 cm²         (Given)

$\Rightarrow 2\left(lb+bh+hl\right)=148$

$\Rightarrow l\times 4+4\times 4+4\times l=74$

$\Rightarrow 4l+4l=74-16$

$\Rightarrow 8l=58$

$\Rightarrow l=\frac{58}{8}=7.25 \mathrm{cm}$

Thus, the length of the cuboid is 7.25 cm.

The length of a cuboid having breadth = 4 cm, height = 4 cm and total surface area = 148 cm², is ___7.25 cm___.

Answer:

Let the edges of the bigger and smaller cubes be x cm and y cm, respectively.

Edge of the bigger cube, x = 12 cm

Edge of each smaller cube, y = 4 cm

∴ Number of smaller cubes that can be cut from the bigger cube

$=\frac{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{bigger} \mathrm{cube}}{\mathrm{Volume} \mathrm{of} \mathrm{each} \mathrm{smaller} \mathrm{cube}}$

$=\frac{12 \mathrm{cm}\times 12 \mathrm{cm}\times 12 \mathrm{cm}}{4 \mathrm{cm}\times 4 \mathrm{cm}\times 4 \mathrm{cm}}$                    [Volume of cube = (Edge)³]

= 27

Thus, 27 smaller cubes of edge 4 cm that can be cut from a cube of edge 12 cm.

The number of cubes of edge 4 cm that can be cut from a cube of edge 12 cm, is ___27___.

Answer:

Length of the cuboid, l = 20 cm

Let b and h be the breadth and height of the cuboid, respectively.

Breadth of the cuboid, b = 4x

Height of the cuboid, h = 3x

It is given that, the volume of a cuboid is 3840 cm³.

$\therefore l\times b\times h=3840 {\mathrm{cm}}^3$

$\Rightarrow 20\times 4x\times 3x=3840$

$\Rightarrow x^2=16$

$\Rightarrow x=4 \mathrm{cm}$

So, 

Breadth of the cuboid, b = 4x = 4 × 4 cm = 16 cm

Height of the cuboid, h = 3x = 3 × 4 cm = 12 cm

∴ Total surface area of the cuboid 

= 2(lb + bh + hl)

= 2(20 × 16 + 16 × 12 + 12 × 20)

= 2 × 752

= 1504 cm²

Thus, the total surface area of the cuboid is 1504 cm².

The volume of a cuboid is 3840 cm³ and the length of the cuboid is 20 cm. If the ratio of its breadth and its height is 4 : 3, then the total surface area of the cuboid is $\underline{ 1504 {\mathrm{cm}}^2 }$.


 

Answer:

Length of the cuboid, l = 12 cm

Let the breadth and height of the cuboid be b and h, respectively.

Breadth of the cuboid, b = 8x

Height of the cuboid, h = 5x

It is given that, the total surface area of cuboid is 392 cm².

∴ 2(lb + bh + hl) = 392 cm²

$\Rightarrow 2\left(12\times 8x+8x\times 5x+5x\times 12\right)=392$

$\Rightarrow 40x^2+156x-196=0$

$\Rightarrow 10x^2+39x-49=0$

$\Rightarrow 10x^2-10x+49x-49=0$

$\Rightarrow 10x\left(x-1\right)+49\left(x-1\right)=0$

$\Rightarrow \left(x-1\right)\left(10x+49\right)=0$

$\Rightarrow x-1=0 \mathrm{or} 10x+49=0$

$\Rightarrow x=1 \mathrm{or} x=-\frac{49}{10}$

Now, x cannot be negative. So, x = 1 cm.

So,

Breadth of the cuboid, b = 8x = 8 × 1 = 8 cm 

Height of the cuboid, h = 5x = 5 ×​ 1 = 5 cm

∴ Volume of the cuboid = l × b × h = 12 × 8 × 5 = 480 cm³

Thus, the volume of the cuboid is 480 cm³.

The total surface area of a cuboid is 392 cm² and the length of the cuboid is 12 cm. If the ratio of its breadth and height is 8 : 5, then the volume of the cuboid is $\underline{ 480 {\mathrm{cm}}^3 }$.

Answer:

Let the length, breadth and height of the cuboid be l units, b units and h units, respectively.

∴ Volume of the original cuboid = lbh cu. units

If the dimensions of the cuboid are decreased by 10%, then

Length of new cuboid, L = l − 10% of l $=l-\frac{l}{10}=\frac{9l}{10}$

Breadth of new cuboid, B = b − 10% of b $=b-\frac{b}{10}=\frac{9b}{10}$

Height of new cuboid, H = h − 10% of h $=h-\frac{h}{10}=\frac{9h}{10}$

∴ Volume of the new cuboid $=\frac{9l}{10}\times \frac{9b}{10}\times \frac{9h}{10}=\frac{729}{1000}lbh$ cu. units

Now,

Percent decrease in volume of cuboid

$=\frac{\mathrm{Decrease} \mathrm{in} \mathrm{volume} \mathrm{of} \mathrm{cuboid}}{\mathrm{Original} \mathrm{volume} \mathrm{of} \mathrm{cuboid}}\times 100\%$

$=\frac{lbh-{\displaystyle \frac{729}{1000}}lbh}{lbh}\times 100\%$

$=\frac{1000-729}{1000}\times 100\%$

$=\frac{271}{1000}\times 100\%$

$=27.1\%$

Thus, the volume of cuboid decreases by 27.1%.

If the dimensions of a cuboid decrease by 10% each, then its volume decreases by ___27.1%___.

Answer:

Let the length, breadth and height of the cuboid be l cm, b cm and h cm, respectively.

Total surface area of the cuboid = 96 cm²         (Given)

⇒ 2(lb + bh + hl) = 96 cm²       .....(1)

Also,

l² + b² + h² = 48 cm²         (Given)

⇒ 2(l² + b² + h²) = 2 × 48 = 96 cm²      .....(2)

Subtracting (1) from (2), we get

2(l² + b² + h²) − 2(lb + bh + hl) = 96 cm² − 96 cm² = 0

⇒ (l² − 2lb + b²) + (b² − 2bh + h²) + (h² − 2hl + l²) = 0

⇒ (l − b)² + (b − h)² + (h − l)² = 0

⇒ l − b = 0, b − h = 0, h − l = 0

⇒ l = b, b = h, h = l

⇒ l = b = h      .....(3)

From (2) and (3), we get

h² + h² + h² = 48 cm²

⇒ 3h² = 48

⇒ h² = 16

⇒ h = 4 cm

Thus, the height of the cuboid is 4 cm.

A cuboid has a total surface area of 96 cm². The sum of the squares of its length, breadth and height (in cm) is 48. The height of the cuboid is ___4 cm___.

Answer:

Let the edge of cube be a cm.

Surface area of the cube = 384 cm²       (Given)

⇒ 6a² = 384 cm²

⇒ a² = 64

⇒ a = $\sqrt{64}$ = 8 cm

∴ Volume of the cube = a³ = (8 cm)³ = 512 cm³ 

Thus, the volume of the cube is 512 cm³.

The volume of a cube whose surface area is 384 cm², is ___512 cm³___.

Answer:

Let the edge of new cube be a cm.

The sides of the three cubes are 8 cm, 6 cm and 1 cm.

It is given that, the three cubes are melted to form the new cube.

∴ Volume of the new cube = Sum of volumes of the three cubes

⇒ a³ = (8 cm)³ + (6 cm)³ + (1 cm)³                     [Volume of cube = (Side)³]

⇒ a³ = 512 + 216 + 1 = 729 cm³

⇒ a³ = (9 cm)³

⇒ a = 9 cm

∴ Surface area of the new cube = 6a² = 6 × (9 cm)² = 6 × 81 cm² = 486 cm²

Thus, the surface area of the new cube is 486 cm².

Three cubes of sides 8 cm, 6 cm and 1 cm are melted to form a new cube. The surface area of the cube so formed is ___486 cm²____.

Answer:

Let each side of the cube be a cm.

Each face of a cube is a square.

∴ Perimeter of one face of cube = 4a

⇒ 4a = 40 cm

⇒ a = 10 cm

∴ Volume of the cube = a³ = (10 cm)³ = 1000 cm³

Thus, the volume of the given cube is 1000 cm³.

The perimeter of one face of a cube is 40 cm. The volume of the cube is ___1000 cm³____.

Answer:

Let the edge of the cube be a cm.

Volume of the cube = 2744 cm³     (Given)

∴ a³ = 2744 cm³

⇒ a³ = (14 cm)³               (2744 = 2 × 2 × 2 × 7 × 7 × 7)

⇒ a = 14 cm

∴ Surface area of the cube = 6a² = 6 × (14 cm)² = 6 × 196 cm² = 1176 cm²

Thus, the surface area of the cube is 1176 cm².

The volume of a cube is 2744 cm³. Its surface area is ____1176 cm²____.

Answer:

Let the length, breadth and height of the cuboid be l cm, b cm and h cm, respectively.

Therefore, the areas of the adjacent faces of the cuboid are lb, bh and hl, respectively.

Now,

lb = 6 cm²      .....(1)

bh = 8 cm²     .....(2)

hl = 27 cm²    .....(3)

Multiplying (1), (2) and (3), we get

lb × bh × hl = 6 cm² × 8 cm² × 27 cm²

⇒ l²b²h² = 1296 cm⁶

⇒ (lbh)² = (36 cm³)²

⇒ lbh = 36 cm³

Thus, the volume of the cuboid is 36 cm³.

If the areas of three adjacent faces of a cuboid are 6 cm², 8 cm² and 27 cm², then its volume is ___36 cm³___.

Answer:

Let the height of hall be h m.

Length of the hall, l = 15 m

Width of the hall, b = 12 m

Now,

Area of the floor + Area of the flat roof = Sum of the areas of the four walls

⇒ lb + lb = 2h(l + b)

⇒ 2lb = 2h(l + b)

⇒ 15 m × 12 m = h(15 m + 12 m)

⇒ h = $\frac{15\times 12}{27}$ = $\frac{20}{3}$ m

∴ Capacity of the hall = lbh = $15 \mathrm{m}\times 12 \mathrm{m}\times \frac{20}{3} \mathrm{m}$ = 1200 m³

Thus, the capacity of the hall is 1200 m³.

The length of a hall is 15 m and width is 12 m. The sum of the areas of the floor and flat root is equal to the sum of the areas of the four walls. The capacity of the hall is ___1200 m³___.

Answer:

The area of the cardboard needed to make the box of given size is same as the total surface area of the box.

Length of the box, l = 25 cm

Breadth of the box, b = 15 cm

Height of the box, h = 8 cm

∴ Area of the cardboard needed to make the box

= Total surface area of the box

= 2(lb + bh + hl)

= 2(25 cm × 15 cm + 15 cm × 8 cm + 8 cm × 25 cm)

= 2(375 cm² + 120 cm² + 200 cm²)

= 2 × 695 cm²

= 1390 cm²

Thus, the area of the cardboard needed to make the box of given size is 1390 cm².

The area of the cardboard needed to make a box of size 25 cm × 15 cm × 8 cm is ___1390 cm²___.

Answer:

Let the side of the cube be a cm.

We know

Length of the diagonal of cube = $\sqrt{3}\times \mathrm{Side}$

$\therefore \sqrt{3}a=6\sqrt{3} \mathrm{cm}$             (Given)

⇒ a = 6 cm

∴ Volume of the cube = a³ = (6 cm)³ = 216 cm³

Thus, the volume of the cube is 216 cm³.

If the length of the diagonal of a cube is $6\sqrt{3}$ cm, then its volume is ___216 cm³___.

Answer:

Let the length of each edge of the original cube be a units.

∴ Volume of the original cube = a³

If the length of each edge of a cube increases by 20%, then

Length of each edge of new cube = a + 20% of a = $a+\frac{20}{100}a=a+\frac{1}{5}a=\frac{6a}{5}$

∴ Volume of the new cube $={\left(\frac{6a}{5}\right)}^3=\frac{216a^3}{125}$

Percent increase in volume of cube

$=\frac{\mathrm{Increase} \mathrm{in} \mathrm{volume} \mathrm{of} \mathrm{cube}}{\mathrm{Original} \mathrm{volume} \mathrm{of} \mathrm{cube}}\times 100\%$

$=\frac{{\displaystyle \frac{216}{125}}{a}^3-a^3}{a^3}\times 100\%$

$=\frac{216-125}{125}\times 100\%$

$=\frac{91}{125}\times 100\%$

$=72.8\%$

Thus, the volume of the cube increases by 72.8%.

If the length of each edge of a cube increases by 20%, then the volume of the cube increases by ___72.8%___.

Answer:

When a square hole of side 2 feet is punched through between a pair of opposite faces of a cube of side 6 feet, then the empty space (or hole) so formed in the cube is in the form of a cuboid. The length, breadth and height of the hole are 6 feet, 2 feet and 2 feet, respectively.

Now,

Volume of the cube = (Side)³ = (6)³ = 216 cubic feet

Volume of the hole = Length × Breadth × Height = 6 × 2 × 2 = 24 cubic feet

∴ Volume of the remaining solid

= Volume of the solid cube − Volume of the hole

= 216 − 24

= 192 cubic feet

Thus, the volume of the remaining solid is 192 cubic feet.

From a solid cube of side 6 feet, a square hole of side 2 feet is punched through between a pair of opposite faces. The volume of the remaining solid in cubic feet is ___192___.

Answer:

Let the edge of three small metallic cubes be 3x, 4x and 5x, respectively.

Suppose the edge of the big cube is a cm.

It is given that the three small metallic cubes are melted to form a big cube.

∴ Volume of big cube = Sum of the volumes of three small cubes

⇒ a³ = (3x)³ + (4x)³ + (5x)³

⇒ a³ = 27x³ + 64x³ + 125x³ = 216x³

⇒ a³ = (6x)³

⇒ a = 6x                .....(1)

It is given that the length of diagonal to the big cube is 18 cm.

$\therefore \sqrt{3}a=18 \mathrm{cm}$                          (Length of the diagonal of the cube = $\sqrt{3}$ × Side)

$\Rightarrow 6\sqrt{3}x=18$                              [Using (1)]

$\Rightarrow x=\frac{18}{6\sqrt{3}}=\sqrt{3} \mathrm{cm}$ 

Now,

Edge of the smallest cube = 3x = $3\sqrt{3}$ cm

Surface area of the cube = 6 × (Side)²

∴ Total surface area of the smallest cube = $6\times {\left(3\sqrt{3}\right)}^2$ = 6 × 27 = 162 cm²            

Thus, the the total surface area of the smallest cube is 162 cm².

Three small metallic cubes whose edges are in the ratio 3 : 4 : 5 are melted to form a big cube. If the diagonal of the cube so formed is 18 cm, then the total surface area of the smallest cube is $\underline{ 162 {\mathrm{cm}}^2 }$.

Answer:

We have,

Side of each cube (a) = 6 cm

We need to find the volume of resulting cuboid

Hence, dimensions of the resulting cuboid are,

Length (l) = 2a

Breadth (b) = a

= 6 cm

Height (h) = a

= 6 cm

Hence, volume of the resulting cuboid,

Hence, volume of the resulting cuboid is.

Answer:

The edges of the three cubes are in the ratio 3 : 4 : 5.

So, let the edges be 3x cm, 4x cm, 5x cm.

The diagonal of new cube is

We need to find the edges of three cubes

Here, volume of the resulting cube,

Let,

Edge of the resulting cube

So, diagonal of the cube, so

Hence,

Now;

The edges of the three cubes are,

The edges of the three cubes are .

Answer:

Let,

Side of the cube

Perimeter of each face is 32 cm.

Lateral surface area,

So the lateral surface area of the cube is .

Answer:

Let,

Edge of the cube

Surface area of the cube = 6a²

So,

Side of the cube is .

Answer:

We have,

Total surface area of the cuboid

Lateral surface area of the cuboid

Let,

Area of the base

We know that,

Area of the base is.

Answer:

Side of each cube (a) = 4 cm

We need to find the surface area of the resulting cuboid

Dimensions of the resulting cuboid,

Length (l) = 3a

Breadth (b) = a

Height (h) = a

Surface area of the cuboid,

Surface area of the cuboid is .

Answer:

Let, l → Length of the cuboid

Breadth of the cuboid (b) = 10 cm

Height of the cuboid (h) = 20 cm

Surface area of the cuboid (A) = 1300 cm²

We have to find the length of the cuboid

We know that,

Length of the cuboid is .