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#### Question 1:

Volume of a cuboid $=\left(Length×Breadth×Height\right)$ cubic units
Total surface area $=2\left(lb+bh+lh\right)$ sq units
Lateral surface area $=\left[2\left(l+b\right)×h\right]$ sq units

(i) Length = 22 cm, breadth = 12 cm, height = 7.5 cm
Volume $=\left(Length×Breadth×Height\right)$ =
Total surface area $=2\left(lb+bh+lh\right)$
Lateral surface area $=\left[2\left(l+b\right)×h\right]$

(ii) Length = 15 m, breadth = 6 m, height = 9 dm = 0.9 m
Volume $=\left(Length×Breadth×Height\right)$ =
Total surface area$=2\left(lb+bh+lh\right)$
Lateral surface area $=\left[2\left(l+b\right)×h\right]$

(iii) Length = 24 m, breadth = 25 cm = 0.25 m, height = 6 m
Volume $=\left(Length×Breadth×Height\right)$ =
Total surface area$=2\left(lb+bh+lh\right)$
Lateral surface area $=\left[2\left(l+b\right)×h\right]$

(iv) Length = 48 cm = 0.48 m, breadth = 6 dm = 0.6 m, height = 1 m
Volume $=\left(Length×Breadth×Height\right)$ =
Total surface area $=2\left(lb+bh+lh\right)$
Lateral surface area $=\left[2\left(l+b\right)×h\right]$

#### Question 2:

Volume of a cuboid $=\left(Length×Breadth×Height\right)$ cubic units
Total surface area $=2\left(lb+bh+lh\right)$ sq units
Lateral surface area $=\left[2\left(l+b\right)×h\right]$ sq units

(i) Length = 22 cm, breadth = 12 cm, height = 7.5 cm
Volume $=\left(Length×Breadth×Height\right)$ =
Total surface area $=2\left(lb+bh+lh\right)$
Lateral surface area $=\left[2\left(l+b\right)×h\right]$

(ii) Length = 15 m, breadth = 6 m, height = 9 dm = 0.9 m
Volume $=\left(Length×Breadth×Height\right)$ =
Total surface area$=2\left(lb+bh+lh\right)$
Lateral surface area $=\left[2\left(l+b\right)×h\right]$

(iii) Length = 24 m, breadth = 25 cm = 0.25 m, height = 6 m
Volume $=\left(Length×Breadth×Height\right)$ =
Total surface area$=2\left(lb+bh+lh\right)$
Lateral surface area $=\left[2\left(l+b\right)×h\right]$

(iv) Length = 48 cm = 0.48 m, breadth = 6 dm = 0.6 m, height = 1 m
Volume $=\left(Length×Breadth×Height\right)$ =
Total surface area $=2\left(lb+bh+lh\right)$
Lateral surface area $=\left[2\left(l+b\right)×h\right]$

Therefore, dimensions of the tank are:

∴ Volume =

Also, $1000c{m}^{3}=1L$

∴ Volume

#### Question 3:

Therefore, dimensions of the tank are:

∴ Volume =

Also, $1000c{m}^{3}=1L$

∴ Volume

$1m=100cm$
∴ Dimensions of the iron piece =
Total volume of the piece of iron
1 cm3 measures 8 gms.
∴Weight of the piece

#### Question 4:

$1m=100cm$
∴ Dimensions of the iron piece =
Total volume of the piece of iron
1 cm3 measures 8 gms.
∴Weight of the piece

Volume of the gravel used
Cost of the gravel is Rs 6.40 per cubic meter.
∴ Total cost Rs 240

#### Question 5:

Volume of the gravel used
Cost of the gravel is Rs 6.40 per cubic meter.
∴ Total cost Rs 240

Total volume of the hall

It is given that  of air is required for each person.
The total number of persons that can be accommodated in that hall

#### Question 6:

Total volume of the hall

It is given that  of air is required for each person.
The total number of persons that can be accommodated in that hall

Volume of the cardboard box

Volume of each bar of soap

Total number of bars of soap that can be accommodated in that box bars

#### Question 7:

Volume of the cardboard box

Volume of each bar of soap

Total number of bars of soap that can be accommodated in that box bars

Volume occupied by a single matchbox

Volume of a packet containing 144 matchboxes

Volume of the carton

Total number of packets is a carton packets

#### Question 8:

Volume occupied by a single matchbox

Volume of a packet containing 144 matchboxes

Volume of the carton

Total number of packets is a carton packets

Total volume of the block

Total volume of each plank

∴ Total number of planks that can be made planks

#### Question 9:

Total volume of the block

Total volume of each plank

∴ Total number of planks that can be made planks

Volume of the brick
Volume of the wall

Total number of bricks  bricks

#### Question 10:

Volume of the brick
Volume of the wall

Total number of bricks  bricks

Volume of the wall
Total quantity of mortar
∴ Volume of the bricks

Volume of a single brick

∴ Total number of bricks bricks

#### Question 11:

Volume of the wall
Total quantity of mortar
∴ Volume of the bricks

Volume of a single brick

∴ Total number of bricks bricks

Volume of the cistern litres

Area of the iron sheet required to make this cistern = Total surface area of the cistern

#### Question 12:

Volume of the cistern litres

Area of the iron sheet required to make this cistern = Total surface area of the cistern

Volume of the block
We know:

Thickness

#### Question 13:

Volume of the block
We know:

Thickness

Rainfall recorded = 5 cm = 0.05 m
Area of the field = 2 hectare =
Total rain over the field =

#### Question 14:

Rainfall recorded = 5 cm = 0.05 m
Area of the field = 2 hectare =
Total rain over the field =

Area of the cross-section of river

Rate of flow

Volume of water flowing through the cross-section in one minute  per minute

#### Question 15:

Area of the cross-section of river

Rate of flow

Volume of water flowing through the cross-section in one minute  per minute

Let the depth of the pit be d m.

But,
Given volume = 14 m3
∴ Depth  = 80 cm

#### Question 16:

Let the depth of the pit be d m.

But,
Given volume = 14 m3
∴ Depth  = 80 cm

Capacity of the water tank
Width = 90 cm = 0.9 m
Depth = 40 cm = 0.4 m

Length =

#### Question 17:

Capacity of the water tank
Width = 90 cm = 0.9 m
Depth = 40 cm = 0.4 m

Length =

Volume of the beam

Length = 5 m

Thickness = 36 cm = 0.36 m

Width =

#### Question 18:

Volume of the beam

Length = 5 m

Thickness = 36 cm = 0.36 m

Width =

Volume
Given:
Volume  = 378 m3
Area = 84 m2

∴ Height

#### Question 19:

Volume
Given:
Volume  = 378 m3
Area = 84 m2

∴ Height

Length of the pool = 260 m
Width of the pool = 140 m

Volume of water in the pool = 54600 cubic metres

∴ Height of water $=\frac{\text{volume}}{\text{length×width}}=\frac{54600}{260×140}=1.5$ metres

#### Question 20:

Length of the pool = 260 m
Width of the pool = 140 m

Volume of water in the pool = 54600 cubic metres

∴ Height of water $=\frac{\text{volume}}{\text{length×width}}=\frac{54600}{260×140}=1.5$ metres

External length = 60 cm
External width = 45 cm
External height = 32 cm

External volume of the box

Thickness of wood = 2.5 cm

∴  Internal length $=60-\left(2.5×2\right)=55$ cm
Internal width $=45-\left(2.5×2\right)=40$ cm
Internal height $=32-\left(2.5×2\right)=27$ cm

Internal volume of the box

Volume of wood = External volume - Internal volume

#### Question 21:

External length = 60 cm
External width = 45 cm
External height = 32 cm

External volume of the box

Thickness of wood = 2.5 cm

∴  Internal length $=60-\left(2.5×2\right)=55$ cm
Internal width $=45-\left(2.5×2\right)=40$ cm
Internal height $=32-\left(2.5×2\right)=27$ cm

Internal volume of the box

Volume of wood = External volume - Internal volume

External length = 36 cm
External width = 25 cm
External height = 16.5 cm

External volume of the box

Thickness of iron = 1.5 cm

∴ Internal length $=36-\left(1.5×2\right)=33$ cm
Internal width $=25-\left(1.5×2\right)=22$ cm
Internal height  cm  (as the box is open)

Internal volume of the box

Volume of iron = External volume − Internal volume

Given:

Total weight of the box

#### Question 22:

External length = 36 cm
External width = 25 cm
External height = 16.5 cm

External volume of the box

Thickness of iron = 1.5 cm

∴ Internal length $=36-\left(1.5×2\right)=33$ cm
Internal width $=25-\left(1.5×2\right)=22$ cm
Internal height  cm  (as the box is open)

Internal volume of the box

Volume of iron = External volume − Internal volume

Given:

Total weight of the box

External length = 56 cm
External width = 39 cm
External height = 30 cm

External volume of the box

Thickness of wood = 3 cm

∴ Internal length $=56-\left(3×2\right)=50$ cm
Internal width $=39-\left(3×2\right)=33$ cm
Internal height $=30-\left(3×2\right)=24$ cm

Capacity of the box = Internal volume of the box

Volume of wood = External volume − Internal volume

#### Question 23:

External length = 56 cm
External width = 39 cm
External height = 30 cm

External volume of the box

Thickness of wood = 3 cm

∴ Internal length $=56-\left(3×2\right)=50$ cm
Internal width $=39-\left(3×2\right)=33$ cm
Internal height $=30-\left(3×2\right)=24$ cm

Capacity of the box = Internal volume of the box

Volume of wood = External volume − Internal volume

External length = 62 cm
External width = 30 cm
External height = 18 cm

∴ External volume of the box

Thickness of the wood = 2 cm

Now, internal length $=62-\left(2×2\right)=58$ cm
Internal width $=30-\left(2×2\right)=26$ cm
Internal height $=18-\left(2×2\right)=14$ cm

∴ Capacity of the box = internal volume of the box

#### Question 24:

External length = 62 cm
External width = 30 cm
External height = 18 cm

∴ External volume of the box

Thickness of the wood = 2 cm

Now, internal length $=62-\left(2×2\right)=58$ cm
Internal width $=30-\left(2×2\right)=26$ cm
Internal height $=18-\left(2×2\right)=14$ cm

∴ Capacity of the box = internal volume of the box

External length = 80 cm
External width = 65 cm
External height = 45 cm

∴ External volume of the box

Thickness of the wood = 2.5 cm

Then internal length$=80-\left(2.5×2\right)=75$ cm
Internal width $=65-\left(2.5×2\right)=60$ cm
Internal height $=45-\left(2.5×2\right)=40$ cm

Capacity of the box = internal volume of the box

Volume of the wood = external volume − internal volume

It is given that

∴ Weight of the wood

#### Question 25:

External length = 80 cm
External width = 65 cm
External height = 45 cm

∴ External volume of the box

Thickness of the wood = 2.5 cm

Then internal length$=80-\left(2.5×2\right)=75$ cm
Internal width $=65-\left(2.5×2\right)=60$ cm
Internal height $=45-\left(2.5×2\right)=40$ cm

Capacity of the box = internal volume of the box

Volume of the wood = external volume − internal volume

It is given that

∴ Weight of the wood

(i) Length of the edge of the cube = a = 7 m
Now, we have the following:
â€‹Volume
Lateral surface area
Total Surface area

(ii) Length of the edge of the cube = a = 5.6 cm
â€‹Now, we have the following:
Volume
Lateral surface area
Total Surface area

(iii) Length of the edge of the cube = a = 8 dm 5 cm = 85 cm
â€‹Now, we have the following:
Volume
Lateral surface area
Total Surface area

#### Question 26:

(i) Length of the edge of the cube = a = 7 m
Now, we have the following:
â€‹Volume
Lateral surface area
Total Surface area

(ii) Length of the edge of the cube = a = 5.6 cm
â€‹Now, we have the following:
Volume
Lateral surface area
Total Surface area

(iii) Length of the edge of the cube = a = 8 dm 5 cm = 85 cm
â€‹Now, we have the following:
Volume
Lateral surface area
Total Surface area

Let a be the length of the edge of the cube.
Total surface area

∴ Volume

#### Question 27:

Let a be the length of the edge of the cube.
Total surface area

∴ Volume

Let a be the length of the edge of the cube.

Then volume

Also,

∴ Surface area

#### Question 28:

Let a be the length of the edge of the cube.

Then volume

Also,

∴ Surface area

1 m = 100 cm
Volume of the original block

Length of the edge of one cube = 45 cm
Then volume of one cube

∴ Total number of blocks that can be cast

#### Question 29:

1 m = 100 cm
Volume of the original block

Length of the edge of one cube = 45 cm
Then volume of one cube

∴ Total number of blocks that can be cast

Let a be the length of the edge of a cube.
Volume of the cube$={a}^{3}$
Total surface area$=6{a}^{2}$

If the length is doubled, then the new length becomes 2a.
Now, new volume $=\left(2a{\right)}^{3}=8{a}^{3}$
Also, new surface area=$=6\left(2a{\right)}^{2}=6×4{a}^{2}=24{a}^{2}$
∴ The volume is increased by a factor of 8, while the surface area increases by a factor of 4.

#### Question 30:

Let a be the length of the edge of a cube.
Volume of the cube$={a}^{3}$
Total surface area$=6{a}^{2}$

If the length is doubled, then the new length becomes 2a.
Now, new volume $=\left(2a{\right)}^{3}=8{a}^{3}$
Also, new surface area=$=6\left(2a{\right)}^{2}=6×4{a}^{2}=24{a}^{2}$
∴ The volume is increased by a factor of 8, while the surface area increases by a factor of 4.

Cost of wood = Rs $500/{\mathrm{m}}^{3}$

Cost of the given block = Rs 256

∴ Volume of the given block

Also, length of its edge = a  = 80 cm

#### Question 1:

Cost of wood = Rs $500/{\mathrm{m}}^{3}$

Cost of the given block = Rs 256

∴ Volume of the given block

Also, length of its edge = a  = 80 cm

Volume of a cylinder =
Lateral surface$=2\mathrm{\pi }rh$
Total surface area $=2\mathrm{\pi }r\left(h+r\right)$

(i) Base radius = 7 cm; height = 50 cm
Now, we have the following:
Volume
Lateral surface area$=2\mathrm{\pi }rh$
Total surface area $=2\mathrm{\pi }r\left(h+r\right)$

(ii) Base radius = 5.6 m; height = 1.25 m
Now, we have the following:
Volume
Lateral surface area$=2\mathrm{\pi }rh$
Total surface area $=2\mathrm{\pi }r\left(h+r\right)$

(iii) Base radius = 14 dm = 1.4 m, height = 15 m
Now, we have the following:
Volume
Lateral surface area$=2\mathrm{\pi }rh$
Total surface area $=2\mathrm{\pi }r\left(h+r\right)$

#### Question 2:

Volume of a cylinder =
Lateral surface$=2\mathrm{\pi }rh$
Total surface area $=2\mathrm{\pi }r\left(h+r\right)$

(i) Base radius = 7 cm; height = 50 cm
Now, we have the following:
Volume
Lateral surface area$=2\mathrm{\pi }rh$
Total surface area $=2\mathrm{\pi }r\left(h+r\right)$

(ii) Base radius = 5.6 m; height = 1.25 m
Now, we have the following:
Volume
Lateral surface area$=2\mathrm{\pi }rh$
Total surface area $=2\mathrm{\pi }r\left(h+r\right)$

(iii) Base radius = 14 dm = 1.4 m, height = 15 m
Now, we have the following:
Volume
Lateral surface area$=2\mathrm{\pi }rh$
Total surface area $=2\mathrm{\pi }r\left(h+r\right)$

#### Question 3:

Height = 7 m
Radius = 10 cm = 0.1 m
Volume
Weight of wood = 225 kg/m3
∴ Weight of the pole

#### Question 4:

Height = 7 m
Radius = 10 cm = 0.1 m
Volume
Weight of wood = 225 kg/m3
∴ Weight of the pole

Diameter = 2r = 140 cm
i.e., radius, r = 70 cm = 0.7 m

Now, volume â€‹

#### Question 5:

Diameter = 2r = 140 cm
i.e., radius, r = 70 cm = 0.7 m

Now, volume â€‹

Volume
Height = 1 m =100 cm

#### Question 6:

Volume
Height = 1 m =100 cm

Diameter = 14 m
Height = 5 m

∴ Area of the metal sheet required = total surface area

#### Question 7:

Diameter = 14 m
Height = 5 m

∴ Area of the metal sheet required = total surface area

Circumference of the base = 88 cm
Height = 60 cm

Area of the curved surface
Circumference
∴ Volume

#### Question 8:

Circumference of the base = 88 cm
Height = 60 cm

Area of the curved surface
Circumference
∴ Volume

Length = height = 14 m
Lateral surface area
∴ Volume

#### Question 9:

Length = height = 14 m
Lateral surface area
∴ Volume

Height = 8 cm
Volume
Now, radius$=r=\sqrt{\frac{1232}{\mathrm{\pi h}}}=\sqrt{\frac{1232×7}{22×8}}=\sqrt{49}=7cm$
Also, curved surface area
∴ Total surface area

#### Question 10:

Height = 8 cm
Volume
Now, radius$=r=\sqrt{\frac{1232}{\mathrm{\pi h}}}=\sqrt{\frac{1232×7}{22×8}}=\sqrt{49}=7cm$
Also, curved surface area
∴ Total surface area

We have: $\frac{radius}{height}=\frac{7}{2}$
i.e., $r=\frac{7}{2}h$
Now, volume

Then
∴ Total surface area

#### Question 11:

We have: $\frac{radius}{height}=\frac{7}{2}$
i.e., $r=\frac{7}{2}h$
Now, volume

Then
∴ Total surface area

Curved surface area
Circumference
Now, height

Also, radius, $r=\frac{4400}{2\mathrm{\pi h}}=\frac{4400×7}{2×22×40}=\frac{35}{2}$

∴ Volume

#### Question 12:

Curved surface area
Circumference
Now, height

Also, radius, $r=\frac{4400}{2\mathrm{\pi h}}=\frac{4400×7}{2×22×40}=\frac{35}{2}$

∴ Volume

For the cubic pack:
Length of the side, a = 5 cm
Height = 14 cm
Volume

For the cylindrical pack:
Height = 12 cm
Volume

We can see that the pack with a circular base has a greater capacity than the pack with a square base.
Also, difference in volume

#### Question 13:

For the cubic pack:
Length of the side, a = 5 cm
Height = 14 cm
Volume

For the cylindrical pack:
Height = 12 cm
Volume

We can see that the pack with a circular base has a greater capacity than the pack with a square base.
Also, difference in volume

Diameter = 48 cm
Radius = 24 cm = 0.24 m
Height = 7 m

Now, we have:
Lateral surface area of one pillar
Surface area to be painted = total surface area of 15 pillars
∴ Total cost

#### Question 14:

Diameter = 48 cm
Radius = 24 cm = 0.24 m
Height = 7 m

Now, we have:
Lateral surface area of one pillar
Surface area to be painted = total surface area of 15 pillars
∴ Total cost

Volume of the rectangular vessel
Radius of the cylindrical vessel = 8 cm
Volume$={\mathrm{\pi r}}^{2}\mathrm{h}$

As the water is poured from the rectangular vessel to the cylindrical vessel, we have:
Volume of the rectangular vessel  = volume of the cylindrical vessel

∴ Height of the water in the cylindrical vessel

#### Question 15:

Volume of the rectangular vessel
Radius of the cylindrical vessel = 8 cm
Volume$={\mathrm{\pi r}}^{2}\mathrm{h}$

As the water is poured from the rectangular vessel to the cylindrical vessel, we have:
Volume of the rectangular vessel  = volume of the cylindrical vessel

∴ Height of the water in the cylindrical vessel

Diameter of the given wire = 1 cm
Length = 11 cm
Now, volume
The volumes of the two cylinders would be the same.
Now, diameter of the new wire = 1 mm = 0.1 cm
∴ New length  ≅ 11 m

#### Question 16:

Diameter of the given wire = 1 cm
Length = 11 cm
Now, volume
The volumes of the two cylinders would be the same.
Now, diameter of the new wire = 1 mm = 0.1 cm
∴ New length  ≅ 11 m

Length of the edge, a = 2.2 cm
Volume of the cube
Volume of the wire$={\mathrm{\pi r}}^{2}\mathrm{h}$
Radius = 1 mm = 0.1 cm
As volume of cube = volume of wire, we have:

#### Question 17:

Length of the edge, a = 2.2 cm
Volume of the cube
Volume of the wire$={\mathrm{\pi r}}^{2}\mathrm{h}$
Radius = 1 mm = 0.1 cm
As volume of cube = volume of wire, we have:

Diameter = 7 m
â€‹Depth = 20 m

â€‹Volume of the earth dug out
Volume of the earth piled upon the given plot

#### Question 18:

Diameter = 7 m
â€‹Depth = 20 m

â€‹Volume of the earth dug out
Volume of the earth piled upon the given plot

Inner diameter = 14 m
Depth = 12 m
â€‹Volume of the earth dug out

Width of embankment = 7 m

Since volume of embankment = volume of earth dug out, we have:

∴ Height of the embankment = 4 m

#### Question 19:

Inner diameter = 14 m
Depth = 12 m
â€‹Volume of the earth dug out

Width of embankment = 7 m

Since volume of embankment = volume of earth dug out, we have:

∴ Height of the embankment = 4 m

Diameter = 84 cm
Length = 1 m = 100 cm
Now, lateral surface area

#### Question 20:

Diameter = 84 cm
Length = 1 m = 100 cm
Now, lateral surface area

Thickness of the cylinder = 1.5 cm
External diameter = 12 cm
also, internal radius = 4.5 cm
Height = 84 cm

Now, we have the following:
Total volume
Inner volume
Now, volume of the metal = total volume − inner volume
∴ Weight of iron    [Given: ]

#### Question 21:

Thickness of the cylinder = 1.5 cm
External diameter = 12 cm
also, internal radius = 4.5 cm
Height = 84 cm

Now, we have the following:
Total volume
Inner volume
Now, volume of the metal = total volume − inner volume
∴ Weight of iron    [Given: ]

Length = 1 m = 100 cm
Inner diameter = 12 cm
Now, inner volume
Thickness = 1 cm

Now, we have the following:
Total volume
Volume of the tube
Density of the tube = 7.7 g/cm3
∴ Weight of the tube

#### Question 1:

Length = 1 m = 100 cm
Inner diameter = 12 cm
Now, inner volume
Thickness = 1 cm

Now, we have the following:
Total volume
Volume of the tube
Density of the tube = 7.7 g/cm3
∴ Weight of the tube

(b) 17

Length of the diagonal of a cuboid $=\sqrt{{l}^{2}+{b}^{2}+{h}^{2}}$

#### Question 2:

(b) 17

Length of the diagonal of a cuboid $=\sqrt{{l}^{2}+{b}^{2}+{h}^{2}}$

(b)

Total surface area , where a is the length of the edge of the cube.
$⇒6{a}^{2}=150$

∴ Volume

#### Question 3:

(b)

Total surface area , where a is the length of the edge of the cube.
$⇒6{a}^{2}=150$

∴ Volume

(c)

Volume

∴ Total surface area

#### Question 4:

(c)

Volume

∴ Total surface area

(b)

Rate of painting = 10 paise per sq cm = Rs 0.1/cm2
Total cost = Rs 264.60
Now, total surface area
Also, length of edge, a

#### Question 5:

(b)

Rate of painting = 10 paise per sq cm = Rs 0.1/cm2
Total cost = Rs 264.60
Now, total surface area
Also, length of edge, a

(c) 6400

Volume of each brick
Volume of the wall
∴  No. of bricks =

#### Question 6:

(c) 6400

Volume of each brick
Volume of the wall
∴  No. of bricks =

(c) 1000

Volume of the smaller cube
Volume of box     [1 m = 100 cm]
∴ Total no. of cubes $=\frac{100×100×100}{10×10×10}=1000$

#### Question 7:

(c) 1000

Volume of the smaller cube
Volume of box     [1 m = 100 cm]
∴ Total no. of cubes $=\frac{100×100×100}{10×10×10}=1000$

(a)

Let a be the length of the smallest edge.
Then the edges are in the proportion a : 2a : 3a.
Now, surface area
$⇒a=\sqrt{\frac{88}{22}}=\sqrt{4}=2$
Also, 2a = 4 and 3a = 6
∴ Volume

#### Question 8:

(a)

Let a be the length of the smallest edge.
Then the edges are in the proportion a : 2a : 3a.
Now, surface area
$⇒a=\sqrt{\frac{88}{22}}=\sqrt{4}=2$
Also, 2a = 4 and 3a = 6
∴ Volume

(b) 1: 9

(b) 1: 9

(c) 164 sq cm

Surface area

#### Question 10:

(c) 164 sq cm

Surface area

(c) 36 kg

Volume of the iron beam
∴ Weight

#### Question 11:

(c) 36 kg

Volume of the iron beam
∴ Weight

(a) 2 m

42000 L = 42 m3
$\mathrm{Volume}=lbh$

#### Question 12:

(a) 2 m

42000 L = 42 m3
$\mathrm{Volume}=lbh$

(b) 88

Volume of the room
One person requires 3 m3.
∴ Total no. of people that can be accommodated$=\frac{264}{3}=88$

#### Question 13:

(b) 88

Volume of the room
One person requires 3 m3.
∴ Total no. of people that can be accommodated$=\frac{264}{3}=88$

(a) 30000

(a) 30000

(b)

Surface area

#### Question 15:

(b)

Surface area

(d)

Diagonal of the cube
i.e., a = 4 cm
∴ Volume

#### Question 16:

(d)

Diagonal of the cube
i.e., a = 4 cm
∴ Volume

(b) 486 sq cm

Diagonal
i.e., a = 9
∴ Total surface area

#### Question 17:

(b) 486 sq cm

Diagonal
i.e., a = 9
∴ Total surface area

(d) If each side of the cube is doubled, its volume becomes 8 times the original volume.

Let the original side be a units.
Then original volume = a3 cubic units
Now, new side  = 2a units
Then new volume = (2a)3 sq units = 8 a3cubic units
Thus, the volume becomes 8 times the original volume.

#### Question 18:

(d) If each side of the cube is doubled, its volume becomes 8 times the original volume.

Let the original side be a units.
Then original volume = a3 cubic units
Now, new side  = 2a units
Then new volume = (2a)3 sq units = 8 a3cubic units
Thus, the volume becomes 8 times the original volume.

(b) becomes 4 times.

Let the side of the cube be a units.
Surface area = 6a2 sq units
Now, new side = 2a units
New surface area = 6(2a2 ) sq units = 24a2 sq units.
Thus, the surface area becomes 4 times the original area.

#### Question 19:

(b) becomes 4 times.

Let the side of the cube be a units.
Surface area = 6a2 sq units
Now, new side = 2a units
New surface area = 6(2a2 ) sq units = 24a2 sq units.
Thus, the surface area becomes 4 times the original area.

(a) 12 cm

Total volume
∴ Edge of the new cube

#### Question 20:

(a) 12 cm

Total volume
∴ Edge of the new cube

(d)

Length of the cuboid so formed = 25 cm
Breadth of the cuboid = 5 cm
Height of the cuboid = 5 cm
∴ Volume of cuboid

#### Question 21:

(d)

Length of the cuboid so formed = 25 cm
Breadth of the cuboid = 5 cm
Height of the cuboid = 5 cm
∴ Volume of cuboid

(d) 44 m3

Diameter = 2 m
Height = 14 m

(d) 44 m3

Diameter = 2 m
Height = 14 m

(b) 12 m

Diameter = 14 m
Volume = 1848 m3

(b) 12 m

Diameter = 14 m
Volume = 1848 m3

(c) 4 : 3

(c) 4 : 3

(d) 640

(d) 640

(b) 84 m

(b) 84 m

(a) 1100 cm3
Volume

#### Question 27:

(a) 1100 cm3
Volume

(a) 1837 cm2
Diameter = 7 cm
Height = 80 cm
∴ Total surface area

#### Question 28:

(a) 1837 cm2
Diameter = 7 cm
Height = 80 cm
∴ Total surface area

(b) 396 cm3
Here, curved surface area

#### Question 29:

(b) 396 cm3
Here, curved surface area

(a) 770 cm3
Diameter = 14 cm
Now, curved surface area

#### Question 30:

(a) 770 cm3
Diameter = 14 cm
Now, curved surface area

(c) 20:27

#### Question 1:

(c) 20:27

Total surface area$=6{a}^{2}$
$⇒6{a}^{2}=384$

#### Question 2:

Total surface area$=6{a}^{2}$
$⇒6{a}^{2}=384$

Volume of a soap cake
Volume of the box

No. of soap cakes

∴ 640 cakes of soap can be placed in a box of the given size.

#### Question 3:

Volume of a soap cake
Volume of the box

No. of soap cakes

∴ 640 cakes of soap can be placed in a box of the given size.

$\frac{\mathrm{Radius}}{\mathrm{height}}=\frac{r}{h}=\frac{5}{7}\phantom{\rule{0ex}{0ex}}⇒r=\frac{5}{7}h$
Now, volume

#### Question 4:

$\frac{\mathrm{Radius}}{\mathrm{height}}=\frac{r}{h}=\frac{5}{7}\phantom{\rule{0ex}{0ex}}⇒r=\frac{5}{7}h$
Now, volume

Volume of the coin$={\mathrm{\pi r}}^{2}\mathrm{h}=\frac{22}{7}×0.75×0.75×0.2$
Volume of the cylinder $={\mathrm{\pi r}}^{2}\mathrm{h}=\frac{22}{7}×2.25×2.25×10$
No. of coins

∴ 450 coins must be melted to form the required cylinder.

#### Question 5:

Volume of the coin$={\mathrm{\pi r}}^{2}\mathrm{h}=\frac{22}{7}×0.75×0.75×0.2$
Volume of the cylinder $={\mathrm{\pi r}}^{2}\mathrm{h}=\frac{22}{7}×2.25×2.25×10$
No. of coins

∴ 450 coins must be melted to form the required cylinder.

Length = 18 cm
Height = 8 cm
∴ Total surface area

#### Question 6:

Length = 18 cm
Height = 8 cm
∴ Total surface area

Curved surface area

Volume

Now, $r=\frac{132}{\mathrm{\pi h}}=\frac{132×7}{22×6}=7m$

i.e., diameter of the pillar,

#### Question 7:

Curved surface area

Volume

Now, $r=\frac{132}{\mathrm{\pi h}}=\frac{132×7}{22×6}=7m$

i.e., diameter of the pillar,

(b) 2310 cm3

Height = 15 cm
Circumference

∴ Volume

(b) 2310 cm3

Height = 15 cm
Circumference

∴ Volume

(b) 280 cm3
Area = 35 cm2
Height = 8 cm

#### Question 9:

(b) 280 cm3
Area = 35 cm2
Height = 8 cm

(a) 28 m
Volume of the cuboid
â€‹Volume of the cylinder

#### Question 10:

(a) 28 m
Volume of the cuboid
â€‹Volume of the cylinder

Lateral surface area

#### Question 11:

Lateral surface area

(c) 432 sq cm
Volume
$⇒x=\sqrt[3]{\frac{576}{72}}=2$
∴ Total surface area

#### Question 12:

(c) 432 sq cm
Volume
$⇒x=\sqrt[3]{\frac{576}{72}}=2$
∴ Total surface area

(a) 512 cm3
Surface area$=6{a}^{2}$
$⇒6{a}^{2}=384$

∴ Volume

#### Question 13:

(a) 512 cm3
Surface area$=6{a}^{2}$
$⇒6{a}^{2}=384$

∴ Volume

(i) If l, b and h are the length, breadth and height of a cuboid, respectively, then its whole surface area is equal to $2\left(lb+lh+bh\right)$ sq units.
(ii) If l, b and h are the length, breadth and height of a cuboid, respectively, then its lateral surface area is equal to $2\left(\left(l+b\right)×h\right)$ sq units.
(iii) If each side of a cube is a, then the lateral surface area is $4{a}^{2}$ sq units.
(iv) If r and h are the radius of the base and height of a cylinder, respectively, then its volume is $\mathrm{\pi }{r}^{2}h$ cubic units.
(v) If r and h are the radius of the base and height of a cylinder, then its lateral surface area is $2\mathrm{\pi }rh$ sq units.