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#### Page No 222:

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 $=\left[2\left(l+b\right)×h\right]$

#### Page No 222:

Therefore, dimensions of the tank are:

∴ Volume =

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

∴ Volume

#### Page No 222:

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

#### Page No 222:

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

#### Page No 222:

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

#### Page No 222:

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

#### Page No 222:

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

#### Page No 222:

Total volume of the block

Total volume of each plank

∴ Total number of planks that can be made planks

#### Page No 222:

Volume of the brick
Volume of the wall

Total number of bricks  bricks

#### Page No 222:

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

Volume of a single brick

∴ Total number of bricks bricks

#### Page No 222:

Volume of the cistern litres

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

#### Page No 222:

Volume of the block
We know:

Thickness

#### Page No 222:

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

#### Page No 222:

Area of the cross-section of river

Rate of flow

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

#### Page No 222:

Let the depth of the pit be d m.

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

#### Page No 222:

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

Length =

#### Page No 223:

Volume of the beam

Length = 5 m

Thickness = 36 cm = 0.36 m

Width =

#### Page No 223:

Volume
Given:
Volume  = 378 m3
Area = 84 m2

∴ Height

#### Page No 223:

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

#### Page No 223:

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

#### Page No 223:

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

#### Page No 223:

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

#### Page No 223:

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

#### Page No 223:

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

#### Page No 223:

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

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

#### Page No 223:

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

∴ Volume

#### Page No 223:

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

Then volume

Also,

∴ Surface area

#### Page No 223:

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

#### Page No 223:

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.

#### Page No 223:

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

#### Page No 227:

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 $=2\mathrm{\pi }r\left(h+r\right)$

#### Page No 227:

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

#### Page No 227:

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

Now, volume ​

#### Page No 227:

Volume
Height = 1 m =100 cm

#### Page No 227:

Diameter = 14 m
Height = 5 m

∴ Area of the metal sheet required = total surface area

#### Page No 227:

Circumference of the base = 88 cm
Height = 60 cm

Area of the curved surface
Circumference
∴ Volume

#### Page No 227:

Length = height = 14 m
Lateral surface area
∴ Volume

#### Page No 227:

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

#### Page No 227:

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

Then
∴ Total surface area

#### Page No 227:

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

#### Page No 227:

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

#### Page No 227:

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

#### Page No 227:

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

#### Page No 227:

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

#### Page No 227:

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:

#### Page No 227:

Diameter = 7 m
​Depth = 20 m

​Volume of the earth dug out
Volume of the earth piled upon the given plot

#### Page No 227:

Inner diameter = 14 m
Depth = 12 m
​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

#### Page No 227:

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

#### Page No 228:

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: ]

#### Page No 228:

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

#### Page No 228:

(b) 17

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

#### Page No 228:

(b)

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

∴ Volume

#### Page No 228:

(c)

Volume

∴ Total surface area

#### Page No 228:

(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

#### Page No 228:

(c) 6400

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

#### Page No 228:

(c) 1000

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

#### Page No 228:

(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

(c) 164 sq cm

Surface area

#### Page No 228:

(c) 36 kg

Volume of the iron beam
∴ Weight

#### Page No 229:

(a) 2 m

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

#### Page No 229:

(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

(b)

Surface area

#### Page No 229:

(d)

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

#### Page No 229:

(b) 486 sq cm

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

#### Page No 229:

(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.

#### Page No 229:

(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.

#### Page No 229:

(a) 12 cm

Total volume
∴ Edge of the new cube

#### Page No 229:

(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

(b) 12 m

Diameter = 14 m
Volume = 1848 m3

(c) 4 : 3

(d) 640

(b) 84 m

(a) 1100 cm3
Volume

#### Page No 230:

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

#### Page No 230:

(b) 396 cm3
Here, curved surface area

#### Page No 230:

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

(c) 20:27

#### Page No 231:

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

#### Page No 231:

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.

#### Page No 231:

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

#### Page No 231:

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.

#### Page No 231:

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

#### Page No 231:

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) 280 cm3
Area = 35 cm2
Height = 8 cm

#### Page No 231:

(a) 28 m
Volume of the cuboid
​Volume of the cylinder

#### Page No 231:

Lateral surface area

#### Page No 231:

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

#### Page No 231:

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

∴ Volume

#### Page No 231:

(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.