Rs Aggarwal 2020 2021 for Class 8 Maths Chapter 20 - Volume And Surface Area Of Solids

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Page No 222:

Answer:

Volume of a cuboid $= (L e n g t h \times B r e a d t h \times H e i g h t)$ cubic units
Total surface area $= 2 (l b + b h + l h)$ sq units
Lateral surface area $= [2 (l + b) \times h]$ sq units

(i) Length = 22 cm, breadth = 12 cm, height = 7.5 cm
Volume $= (L e n g t h \times B r e a d t h \times H e i g h t)$ = $(22 \times 12 \times 7.5) = 1980 c m^{3}$
Total surface area $= 2 (l b + b h + l h)$ $= 2 [(22 \times 12) + (22 \times 7.5) + (12 \times 7.5)] = 2 [264 + 165 + 90] = 1038 c m^{2}$
Lateral surface area $= [2 (l + b) \times h]$ $= 2 (22 + 12) \times 7.5 = 510 c m^{2}$

(ii) Length = 15 m, breadth = 6 m, height = 9 dm = 0.9 m
Volume $= (L e n g t h \times B r e a d t h \times H e i g h t)$ = $(15 \times 6 \times 0.9) = 81 m^{3}$
Total surface area $= 2 (l b + b h + l h)$ $= 2 [(15 \times 6) + (15 \times 0.9) + (6 \times 0.9)] = 2 [90 + 13.5 + 5.4] = 217.8 m^{2}$
Lateral surface area $= [2 (l + b) \times h]$ $= 2 (15 + 6) \times 0.9 = 37.8 m^{2}$

(iii) Length = 24 m, breadth = 25 cm = 0.25 m, height = 6 m
Volume $= (L e n g t h \times B r e a d t h \times H e i g h t)$ = $(24 \times 0.25 \times 6) = 36 m^{3}$
Total surface area $= 2 (l b + b h + l h)$ $= 2 [(24 \times 0.25) + (24 \times 6) + (0.25 \times 6)] = 2 [6 + 144 + 1.5] = 303 m^{2}$
Lateral surface area $= [2 (l + b) \times h]$ $= 2 (24 + 0.25) \times 6 = 291 m^{2}$

(iv) Length = 48 cm = 0.48 m, breadth = 6 dm = 0.6 m, height = 1 m
Volume $= (L e n g t h \times B r e a d t h \times H e i g h t)$ = $(0.48 \times 0.6 \times 1) = 0.288 m^{3}$
Total surface area $= 2 (l b + b h + l h)$ $= 2 [(0.48 \times 0.6) + (0.48 \times 1) + (0.6 \times 1)] = 2 [0.288 + 0.48 + 0.6] = 2.736 m^{2}$
Lateral surface area $= [2 (l + b) \times h]$ $= 2 (0.48 + 0.6) \times 1 = 2.16 m^{2}$

Page No 222:

Answer:

$1 m = 100 c m$
Therefore, dimensions of the tank are:
$2 m 75 c m \times 1 m 80 c m \times 1 m 40 c m = 275 c m \times 180 c m \times 140 c m$
∴ Volume = $Length \times Breadth \times Height = 275 \times 180 \times 140 = 6930000 {cm}^{3}$

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

∴ Volume $= \frac{6930000}{1000} = 6930 L$

Page No 222:

Answer:

$1 m = 100 c m$
∴ Dimensions of the iron piece = $105 cm \times 70 cm \times 1.5 cm$
Total volume of the piece of iron $= (105 \times 70 \times 1.5) = 11025 c m^{3}$
1 cm³ measures 8 gms.
∴Weight of the piece $= 11025 \times 8 = 88200 g = \frac{88200}{1000} = 88.2 k g (b e c a u s e 1 k g = 1000 g)$

Page No 222:

Answer:

$1 c m = 0.01 m$
Volume of the gravel used $= Area \times H eight = (3750 \times 0.01) = 37.5 m^{3}$
Cost of the gravel is Rs 6.40 per cubic meter.
∴ Total cost $= (37.5 \times 6.4) =$ Rs 240

Page No 222:

Answer:

Total volume of the hall $= (16 \times 12.5 \times 4.5) = 900 m^{3}$

It is given that $3.6 m^{3}$ of air is required for each person.
The total number of persons that can be accommodated in that hall $= \frac{Total volume}{Volume required by each person} = \frac{900}{3.6} = 250 people$

Page No 222:

Answer:

Volume of the cardboard box $= (120 \times 72 \times 54) = 466560 c m^{3}$

Volume of each bar of soap $= (6 \times 4.5 \times 4) = 108 c m^{3}$

Total number of bars of soap that can be accommodated in that box $= \frac{Volume of the box}{Volume of each soap} = \frac{466560}{108} = 4320$ bars

Page No 222:

Answer:

Volume occupied by a single matchbox $= (4 \times 2.5 \times 1.5) = 15 c m^{3}$

Volume of a packet containing 144 matchboxes $= (15 \times 144) = 2160 c m^{3}$

Volume of the carton $= (150 \times 84 \times 60) = 756000 c m^{3}$

Total number of packets is a carton $= \frac{Volume of the carton}{Volume of a packet} = \frac{75600}{2160} = 350$ packets

Page No 222:

Answer:

Total volume of the block $= (500 \times 70 \times 32) = 1120000 c m^{3}$

Total volume of each plank $= 200 \times 25 \times 8 = 40000 c m^{3}$ $= 200 \times 25 \times 8 = 40000 c m^{3}$

∴ Total number of planks that can be made $= \frac{Total volume of the block}{Volume of each plank} = \frac{1120000}{40000} = 28$ planks

Page No 222:

Answer:

Volume of the brick $= 25 \times 13.5 \times 6 = 2025 c m^{3}$
Volume of the wall $= 800 \times 540 \times 33 = 14256000 c m^{3}$

Total number of bricks $= \frac{Volume of the wall}{Volume of each brick} = \frac{14256000}{2025} = 7040$ bricks

Page No 222:

Answer:

Volume of the wall $= 1500 \times 30 \times 400 = 18000000 c m^{3}$
Total quantity of mortar $= \frac{1}{12} \times 18000000 = 1500000 c m^{3}$
∴ Volume of the bricks $= 18000000 - 1500000 = 16500000 c m^{3}$

Volume of a single brick $= 22 \times 12.5 \times 7.5 = 2062.5 c m^{3}$

∴ Total number of bricks $= \frac{Total volume of the bricks}{Volume of a single brick} = \frac{16500000}{2062.5} = 8000$ bricks

Page No 222:

Answer:

Volume of the cistern $= 11.2 \times 6 \times 5.8 = 389.76 m^{3} = 389.76 \times 1000 = 389760$ litres

Area of the iron sheet required to make this cistern = Total surface area of the cistern
$= 2 (11.2 \times 6 + 11.2 \times 5.8 + 6 \times 5.8) = 2 (67.2 + 64.96 + 34.8) = 333.92 c m^{2}$

Page No 222:

Answer:

Volume of the block $= 0.5 m^{3}$
We know:
$1 h e c t a r e = 10000 m^{2}$
Thickness $= \frac{Volume}{Area} = \frac{0.5}{10000} = 0.00005 m = 0.005 cm = 0.05 mm$

Page No 222:

Answer:

Rainfall recorded = 5 cm = 0.05 m
Area of the field = 2 hectare = $2 \times 10000 m^{2} = 20000 m^{2}$
Total rain over the field = $Area of the field \times Height of the field = 0.05 \times 20000 = 1000 m^{3}$

Page No 222:

Answer:

Area of the cross-section of river $= 45 \times 2 = 90 m^{2}$

Rate of flow $= 3 \frac{k m}{h r} = \frac{3 \times 1000}{60} = 50 \frac{m}{m i n}$

Volume of water flowing through the cross-section in one minute $= 90 \times 50 = 4500 m^{3}$ per minute

Page No 222:

Answer:

Let the depth of the pit be d m.
$Volume = Length \times w idth \times d epth = 5 m \times 3.5 m \times d m$
But,
Given volume = 14 m³
∴ Depth $= d = \frac{volume}{length \times width} = \frac{14}{5 \times 3.5} = 0.8 m$ = 80 cm

Page No 222:

Answer:

Capacity of the water tank $= 576 litres = 0.576 m^{3}$
Width = 90 cm = 0.9 m
Depth = 40 cm = 0.4 m

Length = $= \frac{capacity}{width×depth} = \frac{0.576}{0.9 \times 0.4} = 1.600 m$

Page No 223:

Answer:

Volume of the beam $= 1.35 m^{3}$

Length = 5 m

Thickness = 36 cm = 0.36 m

Width = $= \frac{volume}{thickness×length} = \frac{1.35}{5 \times 0.36} = 0.75 m = 75 cm$

Page No 223:

Answer:

Volume $= height × area$
Given:
Volume = 378 m³
Area = 84 m²

∴ Height $= \frac{volume}{area} = \frac{378}{84} = 4.5 m$

Page No 223:

Answer:

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

Volume of water in the pool = 54600 cubic metres

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

Page No 223:

Answer:

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

External volume of the box $= 60 \times 45 \times 32 = 86400 {cm}^{3}$

Thickness of wood = 2.5 cm

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

Internal volume of the box $= 55 \times 40 \times 27 = 59400 {cm}^{3}$

Volume of wood = External volume - Internal volume $= 86400 - 59400 = 27000 {cm}^{3}$

Page No 223:

Answer:

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

External volume of the box $= 36 \times 25 \times 16.5 = 14850 {cm}^{3}$

Thickness of iron = 1.5 cm

∴ Internal length $= 36 - (1.5 \times 2) = 33$ cm
Internal width $= 25 - (1.5 \times 2) = 22$ cm
Internal height $= 16.5 - 1.5 = 15$ cm (as the box is open)

Internal volume of the box $= 33 \times 22 \times 15 = 10890 {cm}^{3}$

Volume of iron = External volume − Internal volume $= 14850 - 10890 = 3960 {cm}^{3}$

Given:
${1 cm}^{3} of iron = 8.5 grams$

Total weight of the box $= 3960 \times 8.5 = 33660 grams = 33.66 kilograms$

Page No 223:

Answer:

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

External volume of the box $= 56 \times 39 \times 30 = 65520 {cm}^{3}$

Thickness of wood = 3 cm

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

Capacity of the box = Internal volume of the box $= 50 \times 33 \times 24 = 39600 {cm}^{3}$

Volume of wood = External volume − Internal volume $= 65520 - 39600 = 25920 {cm}^{3}$

Page No 223:

Answer:

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

∴ External volume of the box $= 62 \times 30 \times 18 = 33480 c m^{3}$

Thickness of the wood = 2 cm

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

∴ Capacity of the box = internal volume of the box $= (58 \times 26 \times 14) c m^{3} = 21112 c m^{3}$

Page No 223:

Answer:

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

∴ External volume of the box $= 80 \times 65 \times 45 = 234000 c m^{3}$

Thickness of the wood = 2.5 cm

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

Capacity of the box = internal volume of the box $= (75 \times 60 \times 40) c m^{3} = 180000 c m^{3}$

Volume of the wood = external volume − internal volume $= (234000 - 180000) c m^{3} = 54000 c m^{3}$

It is given that $100 {cm}^{3} of wood weighs 8 g .$

∴ Weight of the wood $= \frac{54000}{100} \times 8 g = 4320 g = 4.32 kg$

Page No 223:

Answer:

(i) Length of the edge of the cube = a = 7 m
Now, we have the following:
Volume $= a^{3} = 7^{3} = 343 m^{3}$
Lateral surface area $= 4 a^{2} = 4 \times 7 \times 7 = 196 m^{2}$
Total Surface area $= 6 a^{2} = 6 \times 7 \times 7 = 294 m^{2}$

(ii) Length of the edge of the cube = a = 5.6 cm
Now, we have the following:
Volume $= a^{3} = 5 . 6^{3} = 175.616 c m^{3}$
Lateral surface area $= 4 a^{2} = 4 \times 5.6 \times 5.6 = 125.44 c m^{2}$
Total Surface area $= 6 a^{2} = 6 \times 5.6 \times 5.6 = 188.16 c m^{2}$

(iii) Length of the edge of the cube = a = 8 dm 5 cm = 85 cm
Now, we have the following:
Volume $= a^{3} = 85^{3} = 614125 c m^{3}$
Lateral surface area $= 4 a^{2} = 4 \times 85 \times 85 = 28900 c m^{2}$
Total Surface area $= 6 a^{2} = 6 \times 85 \times 85 = 43350 c m^{2}$

Page No 223:

Answer:

Let a be the length of the edge of the cube.
Total surface area $= 6 a^{2} = 1176 c m^{2}$
$\Rightarrow a = \sqrt{\frac{1176}{6}} = \sqrt{196} = 14 c m$
∴ Volume $= a^{3} = 14^{3} = 2744 c m^{3}$

Page No 223:

Answer:

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

Then volume $= a^{3} = 729 c m^{3}$

Also, $a = \sqrt[3]{729} = 9 c m$

∴ Surface area $= 6 a^{2} = 6 \times 9 \times 9 = 486 c m^{2}$

Page No 223:

Answer:

1 m = 100 cm
Volume of the original block $= 225 \times 150 \times 27 = 911250 c m^{3}$

Length of the edge of one cube = 45 cm
Then volume of one cube $= 45^{3} = 91125 c m^{3}$

∴ Total number of blocks that can be cast $= \frac{volume of the block}{volume of one cube} = \frac{911250}{91125} = 10$

Page No 223:

Answer:

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 $= (2 a)^{3} = 8 a^{3}$
Also, new surface area= $= 6 (2 a)^{2} = 6 \times 4 a^{2} = 24 a^{2}$
∴ The volume is increased by a factor of 8, while the surface area increases by a factor of 4.

Page No 223:

Answer:

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

Cost of the given block = Rs 256

∴ Volume of the given block $= a^{3} = \frac{256}{500} = 0.512 m^{3} = 512000 {cm}^{3}$

Also, length of its edge = a $= \sqrt[3]{0.512} = 0.8 m$ = 80 cm

Page No 227:

Answer:

Volume of a cylinder = $π r^{2} h$
Lateral surface $= 2 π r h$
Total surface area $= 2 π r (h + r)$

(i) Base radius = 7 cm; height = 50 cm
Now, we have the following:
Volume $= \frac{22}{7} \times 7 \times 7 \times 50 = 7700 c m^{3}$
Lateral surface area $= 2 π r h$ $= 2 \times \frac{22}{7} \times 7 \times 50 = 2200 c m^{2}$
Total surface area $= 2 π r (h + r)$ $= 2 \times \frac{22}{7} \times 7 (50 + 7) = 2508 c m^{2}$

(ii) Base radius = 5.6 m; height = 1.25 m
Now, we have the following:
Volume $= \frac{22}{7} \times 5.6 \times 5.6 \times 1.25 = 123.2 m^{3}$
Lateral surface area $= 2 π r h$ $= 2 \times \frac{22}{7} \times 5.6 \times 1.25 = 44 m^{2}$
Total surface area $= 2 π r (h + r)$ $= 2 \times \frac{22}{7} \times 5.6 (1.25 + 5.6) = 241.12 m^{2}$

(iii) Base radius = 14 dm = 1.4 m, height = 15 m
Now, we have the following:
Volume $= \frac{22}{7} \times 1.4 \times 1.4 \times 15 = 92.4 m^{3}$
Lateral surface area $= 2 π r h$ $= 2 \times \frac{22}{7} \times 1.4 \times 15 = 132 m^{2}$
Total surface area $= 2 π r (h + r)$ $= 2 \times \frac{22}{7} \times 1.4 (15 + 1.4) = 144.32 c m^{2}$

Page No 227:

Answer:

$r = 1.5 m h = 10.5 m Capacity of the tank = volume of the tank = π r^{2} h = \frac{22}{7} \times 1.5 \times 1.5 \times 10.5 = 74.25 m^{3} We know that 1 m^{3} = 1000 L ∴ 74.25 m^{3} = 74250 L$

Page No 227:

Answer:

Height = 7 m
Radius = 10 cm = 0.1 m
Volume $= {πr}^{2} h = \frac{22}{7} \times 0.1 \times 0.1 \times 7 = 0.22 m^{3}$
Weight of wood = 225 kg/m³
∴ Weight of the pole $= 0.22 \times 225 = 49.5 k g$

Page No 227:

Answer:

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

Now, volume $= {πr}^{2} h = 1.54 m^{3}$

$\Rightarrow \frac{22}{7} \times 0.7 \times 0.7 \times h = 1.54 ∴ h = \frac{1.54 \times 7}{0.7 \times 0.7 \times 22} = \frac{154 \times 7}{154 \times 7} = 1 m$

Page No 227:

Answer:

Volume $= {πr}^{2} h = 3850 {cm}^{3}$
Height = 1 m =100 cm

Now, radius, $r = \sqrt{\frac{3850}{π \times h}} = \sqrt{\frac{3850 \times 7}{22 \times 100}} = 1.75 \times 7 = 3.5 c m$
∴ Diameter =2(radius) $= 2 \times 3.5 = 7 c m$

Page No 227:

Answer:

Diameter = 14 m
Radius $= \frac{14}{2} = 7 m$
Height = 5 m

∴ Area of the metal sheet required = total surface area

$= 2 πr (h + r) = 2 \times \frac{22}{7} \times 7 (5 + 7) m^{2} = 44 \times 12 m^{2} = 528 m^{2}$

Page No 227:

Answer:

Circumference of the base = 88 cm
Height = 60 cm

Area of the curved surface $= c i r c u m f e r e n c e \times h e i g h t = 88 \times 60 = 5280 c m^{2}$
Circumference $= 2 πr = 88 c m$
Then radius $= r = \frac{88}{2 π} = \frac{88 \times 7}{2 \times 22} = 14 c m$
∴ Volume $= {πr}^{2} h = \frac{22}{7} \times 14 \times 14 \times 60 = 36960 {cm}^{3}$

Page No 227:

Answer:

Length = height = 14 m
Lateral surface area $= 2 πrh = 220 m^{2}$
Radius $= r = \frac{220}{2 πh} = \frac{220 \times 7}{2 \times 22 \times 14} = \frac{10}{4} = 2.5 m$
∴ Volume $= {πr}^{2} h = \frac{22}{7} \times 2.5 \times 2.5 \times 14 = 275 m^{3}$

Page No 227:

Answer:

Height = 8 cm
Volume $= {πr}^{2} h = 1232 {cm}^{3}$
Now, radius $= r = \sqrt{\frac{1232}{πh}} = \sqrt{\frac{1232 \times 7}{22 \times 8}} = \sqrt{49} = 7 c m$
Also, curved surface area $= 2 πrh = 2 \times \frac{22}{7} \times 7 \times 8 = 352 {cm}^{2}$
∴ Total surface area $= 2 πr (h + r) = (2 \times \frac{22}{7} \times 7 \times 8) + (2 \times \frac{22}{7} \times {(7)}^{2}) = 352 + 308 = 660 {cm}^{2}$

Page No 227:

Answer:

We have: $\frac{r a d i u s}{h e i g h t} = \frac{7}{2}$
i.e., $r = \frac{7}{2} h$
Now, volume $= {πr}^{2} h = π {(\frac{7}{2} h)}^{2} h = 8316 {cm}^{3}$
$\Rightarrow \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times h^{3} = 8316 \Rightarrow h^{3} = \frac{8316 \times 2}{11 \times 7} = 108 \times 2 = 216 \Rightarrow h = \sqrt[3]{216} = 6 c m$

Then $r = \frac{7}{2} h = \frac{7}{2} \times 6 = 21 c m$
∴ Total surface area $= 2 πr (h + r) = 2 \times \frac{22}{7} \times 21 \times (6 + 21) = 3564 {cm}^{2}$

Page No 227:

Answer:

Curved surface area $= 2 π r h = 4400 c m^{2}$
Circumference $= 2 πr = 110 cm$
Now, height $= h = \frac{c u r v e d s u r f a c e a r e a}{c i r c u m f e r e n c e} = \frac{4400}{110} = 40 c m$

Also, radius, $r = \frac{4400}{2 πh} = \frac{4400 \times 7}{2 \times 22 \times 40} = \frac{35}{2}$

∴ Volume $= {πr}^{2} h = \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \times 40 = 22 \times 5 \times 35 \times 10 = 38500 {cm}^{3}$

Page No 227:

Answer:

For the cubic pack:
Length of the side, a = 5 cm
Height = 14 cm
Volume $= a^{2} h = 5 \times 5 \times 14 = 350 c m^{3}$

For the cylindrical pack:
Base radius $= r = 3.5 c m$
Height = 12 cm
Volume $= {πr}^{2} h = \frac{22}{7} \times 3.5 \times 3.5 \times 12 = 462 c m^{3}$

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 $= 462 - 350 = 112 c m^{3}$

Page No 227:

Answer:

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

Now, we have:
Lateral surface area of one pillar $= πdh = \frac{22}{7} \times 0.48 \times 7 = 10.56 m^{2}$
Surface area to be painted = total surface area of 15 pillars $= 10.56 \times 15 = 158.4 m^{2}$
∴ Total cost $= Rs (158.4 \times 2.5) = Rs 396$

Page No 227:

Answer:

Volume of the rectangular vessel $= 22 \times 16 \times 14 = 4928 c m^{3}$
Radius of the cylindrical vessel = 8 cm
Volume $= {πr}^{2} h$

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 $= \frac{v o l u m e}{{πr}^{2}} = \frac{4928 \times 7}{22 \times 8 \times 8} = \frac{28 \times 7}{8} = \frac{49}{2} = 24.5 c m$

Page No 227:

Answer:

Diameter of the given wire = 1 cm
Radius = 0.5 cm
Length = 11 cm
Now, volume $= {πr}^{2} h = \frac{22}{7} \times 0.5 \times 0.5 \times 11 = 8.643 c m^{3}$
The volumes of the two cylinders would be the same.
Now, diameter of the new wire = 1 mm = 0.1 cm
Radius = 0.05 cm
∴ New length $= \frac{volume}{{πr}^{2}} = \frac{8.643 \times 7}{22 \times 0.05 \times 0.05} = 1100.02 c m$ ≅ 11 m

Page No 227:

Answer:

Length of the edge, a = 2.2 cm
Volume of the cube $= a^{3} = {(2.2)}^{3} = 10.648 c m^{3}$
Volume of the wire $= {πr}^{2} h$
Radius = 1 mm = 0.1 cm
As volume of cube = volume of wire, we have:

$h = \frac{v o l u m e}{{πr}^{2}} = \frac{10.648 \times 7}{22 \times 0.1 \times 0.1} = 338.8 c m$

Page No 227:

Answer:

Diameter = 7 m
Radius = 3.5 m
Depth = 20 m

Volume of the earth dug out $= {πr}^{2} h = \frac{22}{7} \times 3.5 \times 3.5 \times 20 = 770 m^{3}$
Volume of the earth piled upon the given plot $= 28 \times 11 \times h = 770 m^{3}$

$∴ h = \frac{770}{28 \times 11} = \frac{70}{28} = 2.5 m$

Page No 227:

Answer:

Inner diameter = 14 m
i.e., radius = 7 m
Depth = 12 m
Volume of the earth dug out $= {πr}^{2} h = \frac{22}{7} \times 7 \times 7 \times 12 = 1848 m^{3}$

Width of embankment = 7 m
Now, total radius $= 7 + 7 = 14 m$

$Volume of the embankment = total volume - inner volume = {πr}_{o}^{2} h - {πr}_{i}^{2} h = πh ({r_{o}}^{2} - {r_{i}}^{2}) = \frac{22}{7} h (14^{2} - 7^{2}) = \frac{22}{7} h (196 - 49) = \frac{22}{7} h \times 147 = 21 \times 22 h = 462 \times h m^{3}$

Since volume of embankment = volume of earth dug out, we have:
$1848 = 462 h$
$\Rightarrow h = \frac{1848}{462} = 4 m$
∴ Height of the embankment = 4 m

Page No 227:

Answer:

Diameter = 84 cm
i.e., radius = 42 cm
Length = 1 m = 100 cm
Now, lateral surface area $= 2 πrh = 2 \times \frac{22}{7} \times 42 \times 100 = 26400 {cm}^{2}$
∴ Area of the road $= lateral surface area \times no . of rotations = 26400 \times 750 = 19800000 {cm}^{2} = 1980 m^{2}$

Page No 228:

Answer:

Thickness of the cylinder = 1.5 cm
External diameter = 12 cm
i.e., radius = 6 cm
also, internal radius = 4.5 cm
Height = 84 cm

Now, we have the following:
Total volume $= {πr}^{2} h = \frac{22}{7} \times 6 \times 6 \times 84 = 9504 {cm}^{3}$
Inner volume $= {πr}^{2} h = \frac{22}{7} \times 4.5 \times 4.5 \times 84 = 5346 {cm}^{3}$
Now, volume of the metal = total volume − inner volume $= 9504 - 5346 = 4158 c m^{3}$
∴ Weight of iron $= 4158 \times 7.5 = 31185 g = 31.185 kg$ [Given: $1 c m^{3} = 7.5 g$ ]

Page No 228:

Answer:

Length = 1 m = 100 cm
Inner diameter = 12 cm
Radius = 6 cm
Now, inner volume $= {πr}^{2} h = \frac{22}{7} \times 6 \times 6 \times 100 = 11314.286 {cm}^{3}$
Thickness = 1 cm
Total radius = 7 cm

Now, we have the following:
Total volume $= {πr}^{2} h = \frac{22}{7} \times 7 \times 7 \times 100 = 15400 {cm}^{3}$
Volume of the tube $= total volume - inner volume = 15400 - 11314.286 = 4085.714 c m^{3}$
Density of the tube = 7.7 g/cm³
∴ Weight of the tube $= volume \times density = 4085.714 \times 7.7 = 31459.9978 g = 31.459 kg$

Page No 228:

Answer:

(b) 17

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

∴ $\sqrt{l^{2} + b^{2} + h^{2}} = \sqrt{12^{2} + 9^{2} + 8^{2}} = \sqrt{144 + 81 + 64} = \sqrt{289} = 17 c m$

Page No 228:

Answer:

(b) $125 c m^{3}$

Total surface area $= 6 a^{2} = 150 c m^{2}$ , where a is the length of the edge of the cube.
$\Rightarrow 6 a^{2} = 150$
$\Rightarrow a = \sqrt{\frac{150}{6}} = \sqrt{25} = 5 c m$
∴ Volume $= a^{3} = 5^{3} = 125 c m^{3}$

Page No 228:

Answer:

(c) $294 c m^{2}$

Volume $= a^{3} = 343 c m^{3}$
$\Rightarrow a = \sqrt[3]{343} = 7 c m$
∴ Total surface area $= 6 a^{2} = 6 \times 7 \times 7 = 294 c m^{2}$

Page No 228:

Answer:

(b) $9261 c m^{3}$

Rate of painting = 10 paise per sq cm = Rs 0.1/cm²
Total cost = Rs 264.60
Now, total surface area $= \frac{264.6}{0.1} = 2646 c m^{2}$
Also, length of edge, a $= \sqrt{\frac{2646}{6}} = \sqrt{441} = 21 c m$
$∴ Volume = 21^{3} = 9261 c m^{3}$

Page No 228:

Answer:

(c) 6400

Volume of each brick $= 25 \times 11.25 \times 6 = 1687.5 c m^{3}$
Volume of the wall $= 800 \times 600 \times 22.5 = 10800000 c m^{3}$
∴ No. of bricks = $\frac{10800000}{1687.5} = 6400$

Page No 228:

Answer:

(c) 1000

Volume of the smaller cube $= (10 c m)^{3} = 1000 c m^{3}$
Volume of box $= (100 c m)^{3} = 1000000 c m^{3}$ [1 m = 100 cm]
∴ Total no. of cubes $= \frac{100 \times 100 \times 100}{10 \times 10 \times 10} = 1000$

Page No 228:

Answer:

(a) $48 c m^{3}$

Let a be the length of the smallest edge.
Then the edges are in the proportion a : 2a : 3a.
Now, surface area $= 2 (a \times 2 a + a \times 3 a + 2 a \times 3 a) = 2 (2 a^{2} + 3 a^{2} + 6 a^{2}) = 22 a^{2} = 88 c m^{2}$
$\Rightarrow a = \sqrt{\frac{88}{22}} = \sqrt{4} = 2$
Also, 2a = 4 and 3a = 6
∴ Volume $= a \times 2 a \times 3 a = 2 \times 4 \times 6 = 48 c m^{3}$

Page No 228:

Answer:

(b) 1: 9

$\frac{Volume 1}{Volume 2} = \frac{1}{27} = \frac{a^{3}}{b^{3}} \Rightarrow a = \frac{b}{\sqrt[3]{27}} = \frac{b}{3} or b = 3 a or \frac{b}{a} = 3$
$Now, \frac{surface area 1}{surface area 2} = \frac{6 a^{2}}{6 b^{2}} = \frac{a^{2}}{b^{2}} = \frac{(b / 3)^{2}}{b^{2}} = \frac{1}{9} ∴ Ratio of the surface areas = 1 : 9$

Page No 228:

Answer:

Page No 228:

Answer:

Page No 229:

Answer:

(a) 2 m

42000 L = 42 m³
$Volume = l b h$
$∴ Height (h) = \frac{volume}{l b} = \frac{42}{6 \times 3.5} = \frac{6}{6 \times 0.5} = 2 m$

Page No 229:

Answer:

(b) 88

Volume of the room $= 10 \times 8 \times 3.3 = 264 m^{3}$
One person requires 3 m³.
∴ Total no. of people that can be accommodated $= \frac{264}{3} = 88$

Page No 229:

Answer:

(a) 30000

$Volume = 3 \times 2 \times 5 = 30 m^{3} = 30000 L$

Page No 229:

Answer:

(b) $1390 c m^{2}$

Surface area $= 2 (25 \times 15 + 15 \times 8 + 25 \times 8) = 2 (375 + 120 + 200) = 1390 c m^{2}$

Page No 229:

Answer:

(d) $64 c m^{2}$

Diagonal of the cube $= a \sqrt{3} = 4 \sqrt{3} c m$
i.e., a = 4 cm
∴ Volume $= a^{3} = 4^{3} = 64 c m^{3}$

Page No 229:

Answer:

(b) 486 sq cm

Diagonal $= \sqrt{3} a c m = 9 \sqrt{3} c m$
i.e., a = 9
∴ Total surface area $= 6 a^{2} = 6 \times 81 = 486 c m^{2}$

Page No 229:

Answer:

(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)³sq units = 8 a³cubic units
Thus, the volume becomes 8 times the original volume.

Page No 229:

Answer:

(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:

Answer:

(a) 12 cm

Total volume $= 6^{3} + 8^{3} + 10^{3} = 216 + 512 + 1000 = 1728 c m^{3}$
∴ Edge of the new cube $= \sqrt[3]{1728} = 12 c m$

Page No 229:

Answer:

(d) $625 c m^{3}$

Length of the cuboid so formed = 25 cm
Breadth of the cuboid = 5 cm
Height of the cuboid = 5 cm
∴ Volume of cuboid $= 25 \times 5 \times 5 = 625 c m^{3}$

Page No 229:

Answer:

(d) 44 m³

Diameter = 2 m
Radius = 1 m
Height = 14 m
$∴ Volume = {πr}^{2} h = \frac{22}{7} \times 1 \times 1 \times 14 = 44 m^{3}$

Page No 229:

Answer:

(b) 12 m

Diameter = 14 m
Radius = 7 m
Volume = 1848 m³

$Now, volume = {πr}^{2} h = \frac{22}{7} \times 7 \times 7 \times h = 1848 m^{3} ∴ h = \frac{1848}{22 \times 7} = 12 m$

Page No 229:

Answer:

(c) 4 : 3

$Here, \frac{Total surface area}{Lateral surface area} = \frac{2 π r (h + r)}{2 π r h} = \frac{h + r}{h} = \frac{20 + 60}{60} = \frac{4}{3} = 4 : 3$

Page No 229:

Answer:

(d) 640
$Total no . of coins = \frac{volume of cylinder}{volume of each coin} = \frac{π \times 3 \times 3 \times 8}{π \times 0.75 \times 0.75 \times 0.2} = 640$

Page No 230:

Answer:

(b) 84 m
$Length = \frac{volume}{π r^{2} 2} = \frac{66 \times 7}{22 \times 0.05 \times 0.05} = 8400 c m = 84 m$

Page No 230:

Answer:

(a) 1100 cm³
Volume $= {πr}^{2} h = \frac{22}{7} \times 5 \times 5 \times 14 = 1100 {cm}^{3}$

Page No 230:

Answer:

(a) 1837 cm²
Diameter = 7 cm
Radius =3.5 cm
Height = 80 cm
∴ Total surface area $= 2 πr (r + h) = 2 \times \frac{22}{7} \times 3.5 (3.5 + 80) = 22 (83.5) = 1837 {cm}^{2}$

Page No 230:

Answer:

(b) 396 cm³
Here, curved surface area $= 2 πrh = 264 {cm}^{3}$
$\Rightarrow r = \frac{264 \times 7}{2 \times 22 \times 14} = 3 c m$
$∴ Volume = {πr}^{2} h = \frac{22}{7} \times 3 \times 3 \times 14 = 396 {cm}^{3}$

Page No 230:

Answer:

(a) 770 cm³
Diameter = 14 cm
Radius = 7 cm
Now, curved surface area $= 2 πrh = 220 {cm}^{2}$
$\Rightarrow h = \frac{220 \times 7}{2 \times 22 \times 7} = 5 c m$
$∴ Volume = {πr}^{2} h = \frac{22}{7} \times 7 \times 7 \times 5 = 770 {cm}^{3}$

Page No 230:

Answer:

(c) 20:27

$We have the following : \frac{r_{1}}{r_{2}} = \frac{2}{3} \frac{h_{1}}{h_{2}} = \frac{5}{3} ∴ \frac{V_{1}}{V_{2}} = \frac{{πr}_{1}^{2} h_{1}}{{πr}_{2}^{2} h_{2}} = \frac{20}{27}$

Page No 231:

Answer:

Total surface area $= 6 a^{2}$
$\Rightarrow 6 a^{2} = 384$
$\Rightarrow a = \sqrt{\frac{384}{6}} = 8 c m$
$∴ Volume = a^{3} = 512 c m^{3}$

Page No 231:

Answer:

Volume of a soap cake $= 7 \times 5 \times 2.5 = 87.5 c m^{3}$
Volume of the box $= 56 \times 40 \times 25 = 56000 c m^{3}$

No. of soap cakes $= \frac{56000}{87.5} = 640 u n i t s$

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

Page No 231:

Answer:

$\frac{Radius}{height} = \frac{r}{h} = \frac{5}{7} \Rightarrow r = \frac{5}{7} h$
Now, volume $= {πr}^{2} h = \frac{22}{7} \times \frac{5}{7} h \times \frac{5}{7} h \times h = 550 c m^{3}$
$∴ h = \sqrt[3]{\frac{550 \times 7 \times 7 \times 7}{22 \times 5 \times 5}} = 7 c m Also, r = \frac{5}{7} h = 5 c m$

Page No 231:

Answer:

Volume of the coin $= {πr}^{2} h = \frac{22}{7} \times 0.75 \times 0.75 \times 0.2$
Volume of the cylinder $= {πr}^{2} h = \frac{22}{7} \times 2.25 \times 2.25 \times 10$
No. of coins $= \frac{volume of cylinder}{volume of coin} = \frac{2.25 \times 2.25 \times 10}{0.75 \times 0.75 \times 0.2} = 450 coins$

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

Page No 231:

Answer:

Length = 18 cm
Breadth = 10 cm
Height = 8 cm
∴ Total surface area $= 2 (l b + l h + b h) = 2 (18 \times 10 + 18 \times 8 + 10 \times 8) = 2 (180 + 144 + 80) = 808 c m^{2}$

Page No 231:

Answer:

Curved surface area $= 2 πrh = 264 m^{2}$
$∴ r = \frac{264}{2 πh} = \frac{132}{πh} m$

Volume $= {πr}^{2} h = π \times \frac{132}{πh} \times \frac{132}{πh} \times h = 924 m^{3}$
$∴ h = \frac{132 \times 132 \times 7}{22 \times 924} = 6 m$

Now, $r = \frac{132}{πh} = \frac{132 \times 7}{22 \times 6} = 7 m$

i.e., diameter of the pillar, $d = 7 \times 2 = 14 m$

Page No 231:

Answer:

(b) 2310 cm³

Height = 15 cm
Circumference $= 2 πr = 44 cm$
$∴ r = \frac{44 \times 7}{2 \times 22} = 7 c m$
∴ Volume $= {πr}^{2} h = \frac{22}{7} \times 7 \times 7 \times 15 = 2310 {cm}^{3}$

Page No 231:

Answer:

(b) 280 cm³
Area = 35 cm²
Height = 8 cm
$∴ Volume = base area \times height = 35 \times 8 = 280 {cm}^{3}$

Page No 231:

Answer:

(a) 28 m
Volume of the cuboid $= 16 \times 11 \times 8 = 1408 m^{3}$
Volume of the cylinder $= {πr}^{2} h = 1408 m^{3}$
$∴ h = \frac{1408 \times 7}{22 \times 4 \times 4} = 28 m$

Page No 231:

Answer:

Lateral surface area $= 2 ((l + b) \times h) = 2 ((8 + 6) \times 4) = 2 (56) = 112 m^{2}$

Page No 231:

Answer:

(c) 432 sq cm
Volume $= l b h = 3 x \times 4 x \times 6 x = 72 x^{3} = 576 c m^{3}$
$\Rightarrow x = \sqrt[3]{\frac{576}{72}} = 2$
∴ Total surface area $= 2 (l b + b h + l h) = 2 (3 x 4 x + 4 x 6 x + 3 x 6 x) = 2 (48 + 96 + 72) = 432 {cm}^{2}$

Page No 231:

Answer:

(a) 512 cm³
Surface area $= 6 a^{2}$
$\Rightarrow 6 a^{2} = 384$
$\Rightarrow a = \sqrt{\frac{384}{6}} = \sqrt{64} = 8 cm$
∴ Volume $= a^{3} = 8^{3} = 512 {cm}^{3}$

Page No 231:

Answer:

(i) If l, b and h are the length, breadth and height of a cuboid, respectively, then its whole surface area is equal to $2 (l b + l h + b h)$ 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 $2 ((l + b) \times h)$ 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 $π r^{2} h$ cubic units.
(v) If r and h are the radius of the base and height of a cylinder, then its lateral surface area is $2 π r h$ sq units.

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