Rs Aggarwal 2019 2020 Solutions for Class 8 Math Chapter 20 Volume And Surface Area Of Solids are provided here with simple step-by-step explanations. These solutions for Volume And Surface Area Of Solids are extremely popular among Class 8 students for Math Volume And Surface Area Of Solids Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 8 Math Chapter 20 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 222:

Answer:

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

(i) Length = 22 cm, breadth = 12 cm, height = 7.5 cm
Volume =(Length×Breadth×Height) = (22×12×7.5)=1980 cm3
Total surface area =2(lb+bh+lh)= 222×12+22×7.5+12×7.5=2264+165+90=1038 cm2
Lateral surface area =2l+b×h=222+12×7.5=510 cm2

(ii) Length = 15 m, breadth = 6 m, height = 9 dm = 0.9 m
Volume =(Length×Breadth×Height) = (15×6×0.9)=81 m3
Total surface area=2(lb+bh+lh) = 215×6+15×0.9+6×0.9=290+13.5+5.4=217.8 m2
Lateral surface area =2l+b×h=215+6×0.9=37.8 m2

(iii) Length = 24 m, breadth = 25 cm = 0.25 m, height = 6 m
Volume =(Length×Breadth×Height) = (24×0.25×6)=36 m3
Total surface area=2(lb+bh+lh) = 224×0.25+24×6+0.25×6=26+144+1.5=303 m2
Lateral surface area =2l+b×h=224+0.25×6=291 m2

(iv) Length = 48 cm = 0.48 m, breadth = 6 dm = 0.6 m, height = 1 m
Volume =(Length×Breadth×Height) = (0.48×0.6×1)=0.288 m3
Total surface area =2(lb+bh+lh)= 20.48×0.6+0.48×1+0.6×1=20.288+0.48+0.6=2.736 m2
Lateral surface area =2l+b×h=20.48+0.6×1=2.16 m2

Page No 222:

Answer:

 1 m = 100 cm
Therefore, dimensions of the tank are:
2 m 75 cm× 1 m 80 cm× 1 m 40 cm=275 cm × 180 cm × 140 cm
∴ Volume =  Length × Breadth× Height = 275×180×140=6930000 cm3

Also, 1000cm3=1L

∴ Volume =69300001000=6930 L

Page No 222:

Answer:

1m=100cm
∴ Dimensions of the iron piece = 105 cm×70 cm×1.5 cm
Total volume of the piece of iron =(105×70×1.5)=11025 cm3
1 cm3 measures 8 gms.
∴Weight of the piece =11025 × 8 = 88200 g =882001000 = 88.2 kg                      because 1 kg = 1000 g

Page No 222:

Answer:

1 cm = 0.01 m
Volume of the gravel used = Area × Height = (3750 × 0.01)=37.5 m3
Cost of the gravel is Rs 6.40 per cubic meter.
∴ Total cost =(37.5×6.4)= Rs 240

Page No 222:

Answer:

Total volume of the hall=(16×12.5×4.5)=900 m3

It is given that 3.6 m3 of air is required for each person.
The total number of persons that can be accommodated in that hall =Total volumeVolume required by each person= 9003.6=250 people 

Page No 222:

Answer:

Volume of the cardboard box =(120×72×54)=466560 cm3

Volume of each bar of soap=(6×4.5×4)=108 cm3

Total number of bars of soap that can be accommodated in that box=Volume of the boxVolume of each soap=466560108=4320 bars

Page No 222:

Answer:

Volume occupied by a single matchbox=(4×2.5×1.5)=15 cm3

Volume of a packet containing 144 matchboxes =(15×144)=2160 cm3

Volume of the carton=(150×84×60)=756000 cm3

Total number of packets is a carton=Volume of the cartonVolume of a packet = 756002160=350 packets

Page No 222:

Answer:

Total volume of the block =(500×70×32)=1120000 cm3
 
Total volume of each plank =200×25×8=40000 cm3=200×25×8=40000 cm3

∴ Total number of planks that can be made=Total volume of the blockVolume of each plank = 112000040000=28 planks

Page No 222:

Answer:

Volume of the brick =25×13.5×6=2025 cm3
Volume of the wall =800×540×33=14256000 cm3

Total number of bricks =Volume of the wallVolume of each brick=142560002025=7040 bricks  

Page No 222:

Answer:

Volume of the wall=1500×30×400=18000000 cm3
Total quantity of mortar=112×18000000=1500000 cm3
∴ Volume of the bricks=18000000-1500000=16500000 cm3

Volume of a single brick=22×12.5×7.5=2062.5 cm3

∴ Total number of bricks=Total volume of the bricksVolume of a single brick=165000002062.5=8000 bricks

Page No 222:

Answer:

Volume of the cistern=11.2×6×5.8=389.76 m3=389.76×1000=389760 litres

Area of the iron sheet required to make this cistern = Total surface area of the cistern
=2(11.2×6+11.2×5.8+6×5.8)=2(67.2+64.96+34.8)=333.92 cm2

Page No 222:

Answer:

Volume of the block=0.5 m3
We know:
 1 hectare = 10000 m2
Thickness=VolumeArea=0.510000= 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 =  2×10000 m2  = 20000 m2
Total rain over the field = Area of the field × Height of the field = 0.05× 20000 = 1000 m3

Page No 222:

Answer:

Area of the cross-section of river =45×2=90 m2

Rate of flow=3 kmhr=3×100060=50 mmin

Volume of water flowing through the cross-section in one minute =90×50=4500 m3 per minute

Page No 222:

Answer:

Let the depth of the pit be d m.
Volume = Length × width × depth = 5 m × 3.5 m× d m
But,
Given volume = 14 m3
∴ Depth = d =volumelength × width=145×3.5=0.8 m = 80 cm

Page No 222:

Answer:

Capacity of the water tank =576 litres=0.576 m3
Width = 90 cm = 0.9 m
Depth = 40 cm = 0.4 m

Length = =capacitywidth×depth=0.5760.9×0.4=1.600 m



Page No 223:

Answer:

Volume of the beam=1.35 m3

Length = 5 m

Thickness = 36 cm = 0.36 m

Width = =volumethickness×length=1.355×0.36=0.75 m=75 cm

Page No 223:

Answer:

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

∴ Height =volumearea=37884=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 =volumelength×width=54600260×140=1.5 metres

Page No 223:

Answer:

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

External volume of the box=60×45×32=86400 cm3

Thickness of wood = 2.5 cm

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

Internal volume of the box= 55 × 40 × 27 = 59400 cm3

Volume of wood = External volume - Internal volume= 86400 - 59400 = 27000 cm3

Page No 223:

Answer:

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

External volume of the box= 36 × 25 × 16.5 = 14850 cm3

Thickness of iron = 1.5 cm

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

Internal volume of the box= 33 × 22 × 15 = 10890 cm3

Volume of iron = External volume − Internal volume= 14850 - 10890 = 3960 cm3

Given: 
1 cm3 of iron = 8.5 grams

Total weight of the box = 3960 × 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 × 39 × 30 = 65520 cm3

Thickness of wood = 3 cm

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

Capacity of the box = Internal volume of the box= 50 × 33 × 24 = 39600 cm3

Volume of wood = External volume − Internal volume= 65520 - 39600 = 25920 cm3

Page No 223:

Answer:

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

∴ External volume of the box=62×30×18=33480 cm3

Thickness of the wood = 2 cm

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

∴ Capacity of the box = internal volume of the box=(58×26×14) cm3=21112 cm3

Page No 223:

Answer:

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

∴ External volume of the box=80×65×45=234000 cm3

Thickness of the wood = 2.5 cm

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

Capacity of the box = internal volume of the box=(75×60×40) cm3=180000 cm3

Volume of the wood = external volume − internal volume=(234000-180000) cm3=54000 cm3

It is given that 100 cm3 of wood  weighs 8 g.

∴ Weight of the wood =54000100×8 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=a3=73=343 m3
Lateral surface area =4a2=4×7×7=196 m2
Total Surface area=6a2=6×7×7=294 m2

(ii) Length of the edge of the cube = a = 5.6 cm
​Now, we have the following:
Volume=a3=5.63=175.616 cm3
Lateral surface area =4a2=4×5.6×5.6=125.44 cm2
Total Surface area=6a2=6×5.6×5.6=188.16 cm2

(iii) Length of the edge of the cube = a = 8 dm 5 cm = 85 cm
​Now, we have the following:
Volume=a3=853=614125 cm3
Lateral surface area =4a2=4×85×85=28900 cm2
Total Surface area=6a2=6×85×85=43350 cm2

Page No 223:

Answer:

Let a be the length of the edge of the cube.
Total surface area=6a2=1176 cm2
a=11766=196=14 cm
 ∴ Volume=a3=143=2744 cm3

Page No 223:

Answer:

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

Then volume =a3=729 cm3

Also, a=7293=9 cm

∴ Surface area=6a2=6×9×9=486 cm2

Page No 223:

Answer:

1 m = 100 cm
Volume of the original block =225×150×27=911250 cm3

Length of the edge of one cube = 45 cm
Then volume of one cube=453=91125 cm3

∴ Total number of blocks that can be cast =volume of the blockvolume of one cube=91125091125=10

Page No 223:

Answer:

Let a be the length of the edge of a cube.
Volume of the cube=a3
Total surface area=6a2

If the length is doubled, then the new length becomes 2a.
Now, new volume =(2a)3=8a3
Also, new surface area==6(2a)2=6×4a2=24a2 
∴ 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/m3

Cost of the given block = Rs 256

∴ Volume of the given block =a3=256500=0.512 m3 = 512000 cm3

Also, length of its edge = a =0.5123=0.8 m = 80 cm



Page No 227:

Answer:

Volume of a cylinder = πr2 h
Lateral surface=2πrh
Total surface area =2πr(h+r)

(i) Base radius = 7 cm; height = 50 cm
Now, we have the following:
Volume=227×7×7×50=7700 cm3
Lateral surface area=2πrh=2×227×7×50=2200 cm2
Total surface area =2πr(h+r)=2×227×750+7=2508 cm2

(ii) Base radius = 5.6 m; height = 1.25 m
Now, we have the following:
Volume=227×5.6×5.6×1.25=123.2 m3
Lateral surface area=2πrh=2×227×5.6×1.25=44 m2
Total surface area =2πr(h+r)=2×227×5.61.25+5.6=241.12 m2

(iii) Base radius = 14 dm = 1.4 m, height = 15 m
Now, we have the following:
Volume=227×1.4×1.4×15=92.4 m3
Lateral surface area=2πrh=2×227×1.4×15=132 m2
Total surface area =2πr(h+r)=2×227×1.415+1.4=144.32 cm2

Page No 227:

Answer:

r = 1.5 mh = 10.5 mCapacity of the tank=volume of the tank = πr2h = 227×1.5×1.5×10.5=74.25 m3We know that 1 m3=1000 L 74.25 m3=74250 L

Page No 227:

Answer:

Height = 7 m
Radius = 10 cm = 0.1 m
Volume=πr2h=227×0.1×0.1×7=0.22 m3
Weight of wood = 225 kg/m3
∴ Weight of the pole=0.22×225=49.5 kg

Page No 227:

Answer:

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

Now, volume ​=πr2h=1.54 m3

227×0.7×0.7×h=1.54 h=1.54×70.7×0.7×22=154×7154×7=1 m

Page No 227:

Answer:

Volume=πr2h=3850 cm3
Height = 1 m =100 cm

Now, radius, r=3850π×h=3850×722×100=1.75×7=3.5 cm
∴ Diameter =2(radius) =2×3.5=7 cm

Page No 227:

Answer:

Diameter = 14 m
Radius =142=7 m
Height = 5 m

∴ Area of the metal sheet required = total surface area

 =2πr(h+r)=2×227×7(5+7) m2=44×12 m2=528 m2

Page No 227:

Answer:

Circumference of the base = 88 cm
Height = 60 cm

Area of the curved surface =circumference×height=88×60=5280 cm2
Circumference =2πr=88 cm
Then radius=r=882π=88×72×22=14 cm
∴ Volume=πr2h=227×14×14×60=36960 cm3

Page No 227:

Answer:

Length = height = 14 m
Lateral surface area=2πrh=220 m2
Radius =r=2202πh=220×72×22×14=104=2.5 m
∴ Volume=πr2h=227×2.5×2.5×14=275 m3  

Page No 227:

Answer:

Height = 8 cm
Volume=πr2h=1232 cm3
Now, radius=r=1232πh=1232×722×8=49=7cm
Also, curved surface area =2πrh=2×227×7×8=352 cm2
∴ Total surface area =2πr(h+r)=2×227×7×8+2×227×72=352+308=660 cm2

Page No 227:

Answer:

We have: radiusheight=72
i.e., r=72h
Now, volume =πr2h=π72h2h=8316 cm3
227×72×72×h3=8316h3=8316×211×7=108×2=216h=2163=6 cm

Then r=72h=72×6=21 cm
∴ Total surface area =2πr(h+r)=2×227×21×(6+21)=3564 cm2

Page No 227:

Answer:

Curved surface area =2πrh=4400 cm2
Circumference =2πr=110 cm
Now, height=h=curved surface areacircumference=4400110=40 cm

Also, radius, r=44002πh=4400×72×22×40=352

∴ Volume=πr2h=227×352×352×40=22×5×35×10=38500 cm3

Page No 227:

Answer:

For the cubic pack:
Length of the side, a = 5 cm
Height = 14 cm
Volume=a2h=5×5×14=350 cm3

For the cylindrical pack:
Base radius =r=3.5 cm
Height = 12 cm
Volume=πr2h=227×3.5×3.5×12=462 cm3

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 cm3

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=227×0.48×7=10.56 m2
Surface area to be painted = total surface area of 15 pillars =10.56×15=158.4 m2
∴ Total cost=Rs (158.4×2.5)=Rs 396

Page No 227:

Answer:

Volume of the rectangular vessel =22×16×14=4928 cm3
Radius of the cylindrical vessel = 8 cm
Volume=πr2h

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=volumeπr2=4928×722×8×8=28×78=492=24.5 cm

Page No 227:

Answer:

Diameter of the given wire = 1 cm
Radius = 0.5 cm
Length = 11 cm
Now, volume=πr2h=227×0.5×0.5×11=8.643 cm3
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 =volumeπr2=8.643×722×0.05×0.05=1100.02 cm ≅ 11 m

Page No 227:

Answer:

Length of the edge, a = 2.2 cm
Volume of the cube =a3=2.23=10.648 cm3
Volume of the wire=πr2h
Radius = 1 mm = 0.1 cm
As volume of cube = volume of wire, we have:

h=volumeπr2=10.648×722×0.1×0.1=338.8 cm

Page No 227:

Answer:

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

​Volume of the earth dug out =πr2h=227×3.5×3.5×20=770 m3
Volume of the earth piled upon the given plot=28×11×h=770 m3

 h=77028×11=7028=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=πr2h=227×7×7×12=1848 m3

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

Volume of the embankment=total volume - inner volume=πro2h-πri2h=πhro2-ri2=227h142-72=227h196-49=227h×147=21×22h=462×h m3

Since volume of embankment = volume of earth dug out, we have:
1848=462 h
h=1848462=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 =2πrh=2×227×42×100=26400 cm2
∴ Area of the road =lateral surface area × no. of rotations=26400×750=19800000 cm2=1980 m2



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=πr2h=227×6×6×84=9504 cm3
Inner volume =πr2h=227×4.5×4.5×84=5346 cm3
Now, volume of the metal = total volume − inner volume   =9504-5346=4158 cm3
∴ Weight of iron =4158×7.5= 31185 g= 31.185 kg   [Given: 1 cm3=7.5g]

Page No 228:

Answer:

Length = 1 m = 100 cm
Inner diameter = 12 cm
Radius = 6 cm
Now, inner volume=πr2h=227×6×6×100=11314.286 cm3
Thickness = 1 cm
Total radius = 7 cm

Now, we have the following:
Total volume=πr2h=227×7×7×100=15400 cm3
Volume of the tube =total volume-inner volume= 15400-11314.286=4085.714 cm3
Density of the tube = 7.7 g/cm3
∴ Weight of the tube =volume×density=4085.714×7.7=31459.9978 g=31.459 kg

Page No 228:

Answer:

(b) 17
  
Length of the diagonal of a cuboid =l2+b2+h2

l2+b2+h2=122+92+82=144+81+64=289=17 cm

Page No 228:

Answer:

(b) 125 cm3

Total surface area =6a2=150 cm2, where a is the length of the edge of the cube.
6a2=150
a=1506=25=5 cm
∴ Volume=a3=53=125 cm3

Page No 228:

Answer:

(c) 294 cm2

Volume=a3=343 cm3
a=3433=7 cm
∴ Total surface area=6a2=6×7×7=294 cm2

Page No 228:

Answer:

(b) 9261 cm3

Rate of painting = 10 paise per sq cm = Rs 0.1/cm2
Total cost = Rs 264.60
Now, total surface area  =264.60.1=2646 cm2
Also, length of edge, a =26466=441=21 cm
 Volume= 213=9261 cm3

Page No 228:

Answer:

(c) 6400

Volume of each brick=25×11.25×6=1687.5 cm3
Volume of the wall=800×600×22.5=10800000 cm3
∴  No. of bricks =108000001687.5=6400 

Page No 228:

Answer:

(c) 1000

Volume of the smaller cube=(10 cm)3=1000 cm3
Volume of box=(100 cm)3=1000000 cm3     [1 m = 100 cm]
∴ Total no. of cubes =100×100×10010×10×10=1000

Page No 228:

Answer:

(a) 48 cm3

Let a be the length of the smallest edge.
Then the edges are in the proportion a : 2a : 3a.
Now, surface area=2(a×2a+a×3a+2a×3a)=2(2a2+3a2+6a2)=22a2=88 cm2
a=8822=4=2
Also, 2a = 4 and 3a = 6
∴ Volume=a×2a×3a=2×4×6=48 cm3

Page No 228:

Answer:

(b) 1: 9

Volume 1Volume 2=127=a3b3a=b273=b3 or b = 3a or ba=3
Now, surface area 1surface area 2=6a26b2=a2b2=(b/3)2b2= 19 Ratio of the surface areas = 1 : 9

Page No 228:

Answer:

(c) 164 sq cm  

Surface area =2(10×4+10×3+4×3)=2(40+30+12)=164 cm2

Page No 228:

Answer:

(c) 36 kg

Volume of the iron beam =9×0.4×0.2=0.72 m3
∴ Weight=0.72×50=36 kg



Page No 229:

Answer:

(a) 2 m

42000 L = 42 m3
Volume=lbh
 Height (h)=volumelb=426×3.5=66×0.5=2 m

Page No 229:

Answer:

(b) 88  

Volume of the room=10×8×3.3=264 m3
One person requires 3 m3. 
∴ Total no. of people that can be accommodated=2643=88

Page No 229:

Answer:

(a) 30000

Volume=3×2×5=30 m3=30000 L

Page No 229:

Answer:

(b) 1390 cm2

Surface area=2(25×15+15×8+25×8)=2(375+120+200)=1390 cm2

Page No 229:

Answer:

(d) 64 cm2

Diagonal of the cube=a3=43 cm
i.e., a = 4 cm
∴ Volume=a3=43=64 cm3

Page No 229:

Answer:

(b) 486 sq cm

Diagonal =3a cm = 93cm
i.e., a = 9
∴ Total surface area =6a2=6×81=486 cm2

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)3 sq 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 =63+83+103=216+512+1000=1728 cm3
∴ Edge of the new cube=17283=12 cm

Page No 229:

Answer:

(d) 625 cm3

Length of the cuboid so formed = 25 cm
Breadth of the cuboid = 5 cm
Height of the cuboid = 5 cm
∴ Volume of cuboid=25×5×5=625 cm3

Page No 229:

Answer:

(d) 44 m3

Diameter = 2 m
Radius = 1 m
Height = 14 m
 Volume=πr2h=227×1×1×14=44 m3

Page No 229:

Answer:

(b) 12 m

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

Now, volume=πr2h=227×7×7×h=1848 m3 h = 184822×7=12 m

Page No 229:

Answer:

(c) 4 : 3

Here, Total surface areaLateral surface area=2πr(h+r)2πrh=h+rh=20+6060=43= 4:3

Page No 229:

Answer:

(d) 640 
Total no. of coins =volume of cylindervolume of each coin=π×3×3×8π×0.75×0.75×0.2=640 



Page No 230:

Answer:

(b) 84 m
Length=volumeπr22=66×722×0.05×0.05=8400 cm = 84 m

Page No 230:

Answer:

(a) 1100 cm3
Volume=πr2h=227×5×5×14=1100 cm3

Page No 230:

Answer:

(a) 1837 cm2
Diameter = 7 cm
Radius  =3.5 cm
Height = 80 cm
∴ Total surface area=2πr(r+h)=2×227×3.5(3.5+80)=22(83.5)=1837 cm2

Page No 230:

Answer:

(b) 396 cm3
Here, curved surface area=2πrh=264 cm3
r=264×72×22×14=3 cm
 Volume=πr2h=227×3×3×14=396 cm3

Page No 230:

Answer:

(a) 770 cm3
Diameter = 14 cm
Radius = 7 cm
Now, curved surface area=2πrh=220 cm2
h=220×72×22×7=5 cm
 Volume=πr2h=227×7×7×5=770 cm3

Page No 230:

Answer:

(c) 20:27

We have the following:r1r2=23h1h2=53V1V2=πr12h1πr22h2=2027



Page No 231:

Answer:

Total surface area=6a2
6a2=384
a=3846=8 cm
 Volume=a3=512 cm3

Page No 231:

Answer:

Volume of a soap cake =7×5×2.5=87.5 cm3
Volume of the box=56×40×25=56000 cm3

No. of soap cakes=5600087.5=640 units

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

Page No 231:

Answer:

Radiusheight=rh=57r=57h
Now, volume=πr2h=227×57h×57h×h=550 cm3
 h=550×7×7×722×5×53=7 cmAlso, r=57h=5 cm

Page No 231:

Answer:

Volume of the coin=πr2h=227×0.75×0.75×0.2
Volume of the cylinder =πr2h=227×2.25×2.25×10
No. of coins=volume of cylindervolume of coin=2.25×2.25×100.75×0.75×0.2= 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 =2lb+lh+bh=218×10+18×8+10×8=2180+144+80=808 cm2  

Page No 231:

Answer:

Curved surface area =2πrh=264 m2
 r=2642πh=132πhm

Volume =πr2h=π×132πh×132πh×h=924 m3
 h=132×132×722×924=6 m

Now, r=132πh=132×722×6=7m

i.e., diameter of the pillar, d=7×2=14 m

Page No 231:

Answer:

(b) 2310 cm3

Height = 15 cm
Circumference=2πr=44 cm
 r=44×72×22=7 cm
∴ Volume=πr2h=227×7×7×15=2310 cm3

Page No 231:

Answer:

(b) 280 cm3
Area = 35 cm2
Height = 8 cm
 Volume = base area × height = 35×8=280 cm3

Page No 231:

Answer:

(a) 28 m 
Volume of the cuboid=16×11×8=1408 m3
​Volume of the cylinder =πr2h=1408 m3
 h=1408×722×4×4=28 m

Page No 231:

Answer:

Lateral surface area =2l+b×h=28+6×4=256=112 m2

Page No 231:

Answer:

(c) 432 sq cm
Volume=lbh=3x×4x×6x=72x3 =576 cm3
x=576723=2
∴ Total surface area=2lb+bh+lh=2(3x4x+4x6x+3x6x)=2(48+96+72)=432 cm2

Page No 231:

Answer:

(a) 512 cm3
Surface area=6a2
6a2=384
a=3846=64=8 cm
∴ Volume=a3=83=512 cm3

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(lb+lh+bh) 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)×h) sq units.
(iii) If each side of a cube is a, then the lateral surface area is 4a2 sq units.
(iv) If r and h are the radius of the base and height of a cylinder, respectively, then its volume is πr2h cubic units.
(v) If r and h are the radius of the base and height of a cylinder, then its lateral surface area is 2πrh sq units.



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