Rectangular Parallelopiped Or Cuboid

4 cuboidal shapes found in the environment are books, pencil boxes, mobile phones, and windows.

4 cubical shapes found in the environment are sweets, ice cubes, sugar cubes, and dice.

The above cuboidal figure has 6 surfaces, 12 edges, and 8 vertices.

6 surfaces are ABCD, EFGH, CBGH, ADFE, ABHE, and DCGF.

12 edges are AB, BH, EH, AE, DC, CG, GF, FD, DA, CB, FE, and GH.

8 vertices are A, B, C, D, E, F, G, and H.

Given, length of the room = 5 m

breadth of the room = 4 m

and height of the room = 3 m

Therefore, length of the longest rod = length of the body diagonal

=

=

= √50 = 5√2 m

Let the surface of the cube be ‘a’ m.

We know that all surfaces of a cube are equal.

Therefore, the surface area of the cube = a²

Given surface area of the cube = 64 sq. m

∴ a² = 64

⇒ a = √64

⇒ a = 8 m

We know that the volume of a cube = a³

∴ Volume = 8³ = 8×8×8 = 512 m³

Given , the breadth of the canal = 2m

And the depth of the canal = 8 dcm = m

As we know that 1m = 1000 dcm

Volume of the canal = 240 cubic m

We know that volume = area × length

⇒ 240 = 2 × × length

⇒ length =

⇒ length = 150000 dcm = 15 m

Let the edge of a cube be a cm.

We know that the body diagonal of a cube = 3a cm

∴ According to question,

3 a = 43

a = = 4 cm

As, total surface area of a cube = 6a² sq.cm

= 6×(4)²

= 6×16

= 96 sq. cm

Let the edge of a cube be a cm.

We know that total length of edges of a cube = 12a cm

According to question ,

12a = 60

⇒ a = cm.

As, Volume of a cube is a³

Therefore, volume = 5³ = 125 cubic cm

Let the edge of a cube be a cm.

We know that total surface area of cube is 6a²

According to question,

6a² = 216

a² = = 36

a = √36 = 6 cm

As, volume of cube is a³

Therefore, volume = 6³ = 216 cubic cm.

Let the edge of a cube be a cm.

And the volume be V cubic cm.

Given, rectangular parallelepiped is converted into 2 cubes of equal area.

So, volume of each cube = = 216 sq. cm

As, a³ = volume

a = voulme^1/3

So, a = 216^1/3 = 6 cm.

Original cube:

Let the edge of the original cube be a cm

∴ Volume of the original cube = a³

Cube after 50% decrease in each side:

As edge is decreased by 50%

∴ a =

So, volume of the changed cube

Hence,

Let the ratio be in x.

∴ length = 3x

breadth = 2x

and height = x

Given, volume = 384 cc

⇒ 3x × 2x × x = 384

⇒ 6x³ = 384

⇒ x³ = = 64

⇒ x = cm.

⇒ x = 4 cm

So, length = 3×4 = 12 cm

breadth = 2×4 = 8 cm

height = 4 cm

Therefore, Surface area of cuboidal box = 2(length× breadth + breadth×height + height×length)

surface area = 2(12×8 + 8×4 + 4×12)

= 2(96 + 32 + 48)

= 2×176

= 352 sq. cm.

Given,

The weight of box filled with tea = 52kg 350gm

Weight of empty box = 3kg 750gm

Weight of tea = Weight of box filled with tea- Weight of empty box

⇒ Weight of tea = 52.350-3.750

⇒ Weight of tea =48.600kg

Volume of box=length × breadth × height

⇒ Volume of box=7.5×5.4×1dcm³

⇒ Volume of box=40.50 dcm³

⇒

The weight of 1 cubic dcm. tea. = 1.2kg/dcm³

Given, Weight of brass plate=4725gm

Weight of 1 square dcm brass is 8.4 gm

Area of brass plate=x × 1mm

Therefore,

[1dcm=100mm]

X =56250 dcm

Let the depth of each hole be h.

Given , length of each hole = 14m

breadth of each hole = 8m

total no. of holes = 30

Total quantity of soil = 2520 m³

As, total volume of soil in 30 holes = total quantity of soil

therefore, 30×14×8×h = 2520

h =

h = 0.762 m.

Let the tank can hold V Lts. of water.

Let each bucket can hold B Lts. of water.

Given, edge of cubical tank = 1.2m

total number of buckets = 64

part of water taken out with bucket of total volume of tank

As, volume of cubical tank = 1.2³ m³

therefore,

As, 1m³ = 1000 L

So, B = L = 0.018 L

Given,

Length of one gross match box=2.8dcm

The breadth of one gross match box=0.9dcm

The height of one gross match box=1.5dcm

The volume of one gross match box=2.8×0.9×1.5

⇒ Volume of one gross match box =3.78dcm³

one gross=12dozen

1 dozen =12 units

Therefore, one gross=12×12units

one gross=144units

Therefore,

⇒ [1dcm=100mm]

The volume of one match box=26.25cubic cm

Length of one match box=5cm

The breadth of one match box=3.5 cm

The height of one match box=? cm

The volume of one match box=26.25cubic cm

⇒ 5×3.5×height=26.25

⇒

Height of one cube=1.5cm

Let the rise in level be H m.

Given, length of tank = 2.1m

breadth of tank = 1.5m

volume that can be added = 630 L

As, 1m³ = 1000 L

So, volume that can be added = = 0.63 m³.

According to question,

Capacity of tank to be filled = Volume of water that can be added.

H = 0.35 m

Let the height to be increased = H m

Given, length of rectangular field = 20m.

breadth of rectangular field = 15m.

edge of each cubic hole = 4m

total no. of cubic holes = 4

According to the question,

Total volume of soil dug out of 4 cubical holes = volume of soil deposited on rectangular field.

4×4³ = 20×15×H

⇒ 256 = 300H

H = 0.853 m

Let the depth of hole scooped be H m

Given, Length of field to be scooped = 27m

breadth of field to be scooped = 18.2 m

Length of low land = 48m

breadth of low land = 31.5m

height of low land = 6.5 dcm = 0.65 m

A/Q, Volume of field to be scooped = volume of low land

27×18.2×H = 48×31.5×0.65

H =

H = 2m

Let the ratio be in x m

So, length of cuboidal pot = 4x m

breadth of cuboidal pot = 3x m

Given: Height of cuboidal pot = 7 dcm = 0.7 m

As, 1m³ = 1000 L

Volume of 1^st drum = 800 L = 0.8 m³

Volume of 2^nd drum = 725L = 0.725 m³

Volume of 3^rd drum = 575 L = 0.575 m³

A/Q, Volume from all the 3 drums = volume of cuboidal pot

therefore, 0.8 + 0.725 + 0.575 = 4x × 3x × 0.7

2.1 = 8.4x²

x² = 0.25

x = 0.25

x = 0.5 m

Thus, length of pot = 4x = 4X0.5 = 2 m

breadth of pot = 3x = 3X0.5 = 1.3 m

2^nd part: Given if depth of cuboidal pot = 5dcm = 0.5 m

then volume = length×breadth×height

volume = 2×1.5×0.5

volume = 1.5 m3 = 1.5×1000L

volume = 1500 L

Thus the mentioned volume of 1620 L cannot be kept in that pot .

Given, Length of tank = 2.5m

breadth of tank = 1.6m

Let height of tank be H m

Requirement of 1^st family = 1200L = = 1.2 m³

Requirement of 2^nd family = 1050L = = 1.05 m³

Requirement of 3^rd family = 950L = = 0.95m³

As, 1m³ = 1000 L

According to question,

25% of the requirement is to be stored in a tank.

25% of Requirement of 1^st family = = 0.3 m³

25% of Requirement of 2^nd family = = 0.2625m³

25% of Requirement of 3^rdfamily = = 0.2375m³

As, 25% of the total requirement of the water = volume of the land to be irrigated

0.3 + 0.2625 + 0.2375 = 2.5×1.6×H

0.8 = 4H

H = 0.2m

2^nd part: breadth is increased by 4 dcm = 0.4 m

As, length = 2.5 m

Let the new depth be d m

again, As, 25% of the total requirement of the water = volume of the land to be irrigated

0.8 = 0.4×2.5×d

d

d = 0.8 m

Given, thickness = 5cm = 0.5dm

weight of empty box = 115.5 kg

weight of filled box = 880.5 kg

Length of inner side of box = 12dcm

breadth of inner side of box = 8.5 dcm

According to question, volume of rice inside the box = Inner capacity of box

weight of filled box-weight of empty box = length × breadth height

( 880.5-115.5)kg = 12×8.5×h

dcm³ = 102 h {1 cubic dcm rice is 1.5kg}

h =

h = 5 dcm

As, thickness of the box = 0.5dcm

So, outer length of box (L) = inner length + 2xthickness = 12 + 2×0.5 = 13 dcm

outer breadth of box(B) = inner breadth + 2Xthickness = 8.5 + 2×0.5 = 9.5 dcm

outer height of box(H) = inner height + 2X thickness = 5 + 2×0.5 = 6 dcm

As,Outer surface area = 2(LB + BH + LH)

= 2(13×9.5 + 9.5×6 + 6×13)

= 2(123.5 + 57 + 78)

= 2×285.5

= 517 dcm²

According to question, cost of 1 dcm² = Rs.1.5

So, cost of 517 dcm² = Rs.(1.5×517)

= Rs 775.5

Given, length of cuboidal pond = 20m

breadth of cuboidal pond = 18.5 m

height of cuboidal pond = 3.2m

Thus volume of cuboidal pond = length × breadth × height

= 20×18.5×3.2

= 1184 m³

According to question, time required to irrigate upto 160 L = 1 hr

As, 1m3 = 1000L

160L = = 0.16 m³

So, time required to irrigate 1184 m³ hr.

= 7400 hr

2^nd part: Length of field = 59.2m

breadth of field = 40 m

Let height of field = H m

According to question,

Volume of field to be poured with water = volume of water

length ×breadth × height = 1184 m³

59.2 × 40 × H = 1184

H =

H = 0.5 m

volume = area × height

⇒ 440 = 88 × height

⇒ height = = 5 cm

Volume of cuboidal hole = Total Volume of planks

⇒ 40 × 12 × 16 = n × 5 × 4 × 2

⇒ 7680 = n × 40

⇒ n = = 192

Surface area of cube = 6 edge²

edge² = = 64 sq.cm

edge = √64 = 8 m

Volume = edge³ = 8³ = 512 m³

Let the edges of two cubes be a cm and b cm.

ratio of volumes of cube =

ratio of surface area of cube =

Let the edge be of a cm

then total surface area of cube = s = 6a²

a =

and body diagonal = d = 3a

a =

Thus,

Squaring both sides,

(i) FALSE

Let the side of 1^st cube be a cm.

volume of 1^st cube = a³

Now, as length of 2^nd cube = 2a

Then , volume becomes = (2a)³ = 8a³ = a³ + 7a³

i.e. its 7 times more than that of 1^st cube.

(ii) FALSE

Area = 2 hectares = 2 × 100 acres = 200 × 100 sq m = 20000 sq m

height = 5 cm = 0.005 m

Thus,

volume of rain water it can hold = area×height

= 20000 × 0.005 = 100 m³

(i) total number of diagonals of a cuboid =

surface diagonals + body diagonals

= 6 + 4 = 10

(ii) Length of surface diagonal = (a² + a²) = (2a²) = 2 a

(iii) When the length of rectangular parallelepiped are equal then it becomes a CUBE.

In cuboid –number of surfaces = 6 = x

number of edges = 12 = y

number of vertices = 8 = z

number of diagonal = surface diagonal + body diagonal = 6 + 4 = 10 = p

Thus, x-y + z + p = 6-12 + 8 + 10 = 12.

Let the edge of cube be a

Surface area = 6a²

According to question, edge increased by 50%

=

new surface area =

Therefore, ratio of surface areas

volume of 1^st cube = 3³ = 27 cm³

volume of 2^nd cube = 4³ = 64 cm³

volume of 3^rd cube = 5³ = 125 cm³

As, a new solid is made by melting these three solid cubes, So, volume of new cube is the sum of volume of all the 3 cubes.

Thus, volume of new cube = 27 + 64 + 125 = 216 cm³

new edge³ = 6³

So, new edge = 6 cm

As, the length of 2 adjacent walls of room are 12m and 8m = length and breadth of the floor

So, area of the floor = 12×8 = 96 sq m.

volume of 1^st cuboid = length breadth × height

= 4 × 6 × 4 = 96 cubic unit

Volume of 2^nd cuboid = length breadth × height =

8 × (2h-1) × 2 = 32h-16

Given, volumes are equal.

So, 32h-16 = 96

⇒ 32h = 96 + 16 = 112

⇒ h =

h = 3.5 units.