AREA

1. The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares .A) 22 cmB) 24 cmC) 26 cmD) 28 cmAns. B We know perimeter of square = 4(side)So Side of first square = 40/4 = 10 cmSide of second square = 32/4 = 8 cm Area of third Square = 10*10 – 8*8 = 36 cm So side of third square = 6 [because area of square = side*side]Perimeter = 4*Side = 4*6 = 24 cm2. The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas.A) 4:25B) 4:15C) 3:25D) 3:15Ans. ALet the diagonals of the squares be 2x and 5x.Then ratio of their areas will be Area of square=12∗Diagonal212∗2×2 : 12∗5×2 4×2 : 25×2 = 4 : 25 Area of square = 12 ∗ Diagonal2 12 ∗ 2×2: 12 ∗ 5×2 4x 2: 25×2 = 4 : 25 Area of square= 12∗Diagonal2 12 ∗ 2×2 : 12∗ 5×2 4×2 : 25×2 = 4:25 3. A farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garden.A) 10 m * 5 mB) 15 m * 5 mC) 20 m * 5 mD) 25 m * 5 mAns. CFrom the question, 2b+l = 30=> l = 30-2b Area=100m2=l×b=100 =b(30−2b) =100b2−15b+50=0=(b−10)(b−5)=0Area=100m2=l×b=100=b(30−2b)=100b2−15b+50=0=(b−10)(b−5)=0Area=100m2=l×b=100=b(30−2b)=100b2−15b+50=0=(b−10)(b−5)=0b = 10 or b = 5when b = 10 then l = 10when b = 5 then l = 20Since the garden is rectangular so we will take value of breadth 5.So its dimensions are 20 m * 5 m 4. A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is ?A) 25%B) 26%C) 27%D) 28%Ans. DLet original length = xand original width = yDecrease in area will be =xy−(80 x 100 × 90y 100)=(xy−1825xy) = 725xy Decrease = (7xy 25xy × 100) %=28% = xy−(80 x 100 × 90y 100)=(xy−1825xy)=725 xy Decrease = (7xy 25xy × 100) %=28%=xy − (80x 100 × 90y 100)=(xy−1825xy)=725 xy Decrease = (7xy25xy×100)%=28% 5. What will be the cost of gardening 1 meter boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs. 10 per square meter ?A) Rs. 3430B) Rs. 3440C) Rs. 3450D) Rs. 3460 Ans. BWe know Perimeter = 2(l+b), rightSo,2(l+b) = 340As we have to make 1 meter boundary around this, so Area of boundary = ((l+2)+(b+2)-lb)= 2(l+b)+4 = 340+4 = 344 So required cost will be = 344 * 10 = 3440      

1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.A) 120m2120m2120m2B) 130m2130m2130m2C) 140m2140m2140m2D) 150m2150m2150m2 Ans.A We know h2=b2+h2=>Other side =(17)2−(15)2−−−−−−−−−−−√=289−225−−−−−−−−√=64−−√=8meterArea=Length×Breadth=15×8m2=120m2We know h2=b2+h2=>Other side =(17)2−(15)2−−−−−−−−−−−√=289−225−−−−−−−−√=64−−√=8meterArea=Length×Breadth=15×8m2=120m2We know h2=b2+h2=>Other side =(17)2−(15)2−−−−−−−−−−−√=289−225−−−−−−−−√=64−−√=8meterArea=Length×Breadth=15×8m2=120m2 2. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:A) 152600m2152600m2152600m2B) 153500m2153500m2153500m2C) 153600m2153600m2153600m2D) 153800m2153800m2153800m2Ans. C Perimeter = Distance travelled in 8 minutes,= Perimeter = 12000/60 * 8 = 1600 meter. [because Distance = Speed * Time] As per question length is 3x and width is 2xWe know perimeter of rectangle is 2(L+B)So, 2(3x+2x) = 1600 = x = 160So Length = 160*3 = 480 meterand Width = 160*2 = 320 meter Finally, Area = length * breadth= 480 * 320 = 153600 3. The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:A) 32%B) 34%C) 42%D) 44%Ans. DLet original length = x metres and original breadth = y metres.Original area =xy m2New Length =120100x=65xNew Breadth =120100y=65y=>New Area =65x∗65y=>New Area =3625xyArea Difference=3625xy−xy=1125xyIncrease%=DiffernceActual∗100=11xy25∗1xy∗100==44%Original area =xy m2New Length =120100x=65xNew Breadth =120100y=65yNew Area =65x∗65yNew Area =3625xyArea Difference=3625xy−xy=1125xyIncrease%=DiffernceActual∗100=11xy25∗1xy∗100=44% 4. The area of a rectangle is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the rectangular field ?18 meter20 meter22 meter25 meterAns. B Let breadth =x metres.Then length =(115x/100)metres. =x∗115×100=460×2=(460×100/115)x2=400x = 20 = x∗115×100=460×2=(460×100/115)x2=400x = 20 = x∗115×100 = 460×2 = (460×100/115)x2 = 400x = 20 5. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered.If the area of the field is 680 sq.ft, how many feet of fencing will be required ?A) 88 feetB) 86 feetC) 84 feetD) 82 feetAns. AWe are given with length and area, so we can find the breadth.as Length * Breadth = Area=> 20 * Breadth = 680=> Breadth = 34 feet Area to be fenced = 2B + L = 2*34 + 20 = 88 feet        

1. A stream, which flows at a uniform rate of 2.5 km per hour, is 20 metres wide. If the depth of a certain ferry is 1.2 m, how many litres of water will pass through the ferry in a minute?A) 1000000B) 900000C) 500000D) None of theseAns. AThe total quantity of water = = = 1000 m3 = 1000000 litres 2. The annual rainfall at a place is 43 cm. Find the weight (in metric tonnes) of the annual rainfall on a hectare of land, taking the weight of water to be 1 metric tonne for 1 cubic metre.A) 4000B) 4300C) 5000D) None of theseAns. BHeight of water on 1 sq m area of land annually = 43 cmVolume of water on 10,000 sq m. area of land = × 10000 m3 = 4300 m3 Weight of water = 4300 metric ton 3. A metallic right circular cone of height 9 cm and base radius 7 cm is melted into a cuboid with two sides as 11 cm and 6 cm. What is the third side of the cuboid?A) 5 cmB) 6 cmC) 7 cmD) 9 cmAns. CLet the third side of cuboid = x cm Now, according to the question: 1/3 πr2h = l × b × h⇒ 66x = 462 ⇒ x = 7 cm 4. A garden is 112m long & 78 m wide. It has walking path 25m wide all around it on inside. Find the area of path ?A) 925 m2B) 927 m2C) 929 m2D) 934 m2Ans. AIf path is within the garden then use this formula:2 * B * (L + B – 2 * B)2 * 25 (112 + 78 – 2 * 25)= 5 * 185 = 925 m2 5. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot? A) 274 & 100 m B) 275 & 200 m C) 275 & 100 m D) None of the above Ans. C Let the common ratio between the length and breadth of the rectangular plot be x. Hence, the length and breadth of the rectangular plot will be 11x m and 4x m respectively.Perimeter of the plot = 2(Length + Breadth) It is given that the cost of fencing the plot at the rate of Rs 100 per metre is Rs 75, 000.∴ 100 × Perimeter = 75000100 × 30x = 750003000x = 75000Dividing both sides by 3000, we obtainx = 25Length = 11x m = (11 × 25) m = 275 mBreadth = 4x m = (4 × 25) m = 100 mHence, the dimensions of the plot are 275 m and 100 m respectively.        

1. The volume of the largest right circular cone that can be cut out of a cube of edge 7 cm is:A) 79.8cm379.8cm379.8cm3B) 79.4cm379.4cm379.4cm3C) 89.8cm389.8cm389.8cm3D) 89.4cm389.4cm389.4cm3Ans. CVolume of the largest cone = Volume of the cone with diameter of base 7 and height 7 cmVolume of cone =13pr2h=13*227*3.5*3.5*7=269.53cm3=89.8cm3Volume of cone =13pr2h=13*227*3.5*3.5*7=269.53cm3=89.8cm3Volume of cone =13pr2h=13*227*3.5*3.5*7=269.53cm3=89.8cm3 (radius is taken as 3.5, as diameter is 7 cm) 2. A hollow spherical metallic ball has an external diameter 6 cm and is 1/2 cm thick. The volume of metal used in the metal is:A) 4715cm34715cm34715cm3B) 4735cm34735cm34735cm3C) 4775cm34775cm34775cm3D) 4795cm34795cm34795cm3Ans. B If we are given with external radius and thickness, we can get the internal radius by subtracting them. Then the volume of metal can be obtained by its formula as,External radius = 3 cm,Internal radius = (3-0.5) cm = 2.5 cmVolume of sphere =43pr3=43*227*[32-2.52]cm3=43*227*918cm3=1433cm3=4723cm3Volume of sphere =43pr3=43*227*[32-2.52]cm3=43*227*918cm3=1433cm3=4723cm3Volume of sphere =43pr3=43*227*[32-2.52]cm3=43*227*918cm3=1433cm3=4723cm3 3. The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is:A) 4 cmB) 6 cmC) 8 cmD) 10 cmAns. BCurved surface area of sphere =4pr2Surface area of cylinder =2prh=>4pr2=2prh=>r2=6*122=>r2=36=>r=6Curved surface area of sphere =4pr2Surface area of cylinder =2prh=>4pr2=2prh=>r2=6*122=>r2=36=>r=6Curved surface area of sphere =4pr2Surface area of cylinder =2prh=>4pr2=2prh=>r2=6*122=>r2=36=>r=6Note: Diameter of cylinder is 12 so radius is taken as 6. View Answer Comment 4. 12 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 2 cm height. Find the diameter of each sphere.A) 4 cmB) 6 cmC) 8 cmD) 10 cmAns. AIn this type of question, just equate the two volumes to get the answer as,Volume of cylinder =pr2hVolume of sphere =43pr3=>12*43pr3=pr2h=>12*43pr3=p*8*8*2=>r3=8*8*2*312*4=>r3=8=>r=2cm=>Diameter =2*2=4cmVolume of cylinder =pr2hVolume of sphere =43pr3=>12*43pr3=pr2h=>12*43pr3=p*8*8*2=>r3=8*8*2*312*4=>r3=8=>r=2cm=>Diameter =2*2=4cmVolume of cylinder =pr2hVolume of sphere =43pr3=>12*43pr3=pr2h=>12*43pr3=p*8*8*2=>r3=8*8*2*312*4=>r3=8=>r=2cm=>Diameter =2*2=4cm View Answer 5. A cone of height 9 cm with diameter of its base 18 cm is carved out from a wooden solid sphere of radius 9 cm. The percentage of the wood wasted is :A) 45%B) 56%C) 67%D) 75%Ans. DWe will first subtract the cone volume from wood volume to get the wood wasted.Then we can calculate its percentage.Sphere Volume =43pr3Cone Volume =13pr2hVolume of wood wasted =(43p*9*9*9)-(13p*9*9*9)=p*9*9*9cm3Required Percentage =p*9*9*943p*9*9*9*100%=34*100%=75%Sphere Volume =43pr3Cone Volume =13pr2hVolume of wood wasted =(43p*9*9*9)-(13p*9*9*9)=p*9*9*9cm3Required Percentage =p*9*9*943p*9*9*9*100%=34*100%=75%Sphere Volume =43pr3Cone Volume =13pr2hVolume of wood wasted =(43p*9*9*9)-(13p*9*9*9)=p*9*9*9cm3Required Percentage =p*9*9*943p*9*9*9*100%=34*100%=75% 6. A hemisphere and a cone have equal bases. If their heights are also equal, then the ratio of their curved surface will be :A) 2:12:12:1B) 1:2v1:2v1:2vC) 2v:12v:12v:1D) 3v:13v:13v:1Ans. CLet the radius of hemisphere and cone be R, Height of hemisphere H = R.So the height of the cone = height of the hemisphere = RSlant height of the cone =R2+R2——-v=2vRHemisphere Curved surface areaCone Curved surface area=2pR2p*R*2vR=2v:1=R2+R2——-v=2vRHemisphere Curved surface areaCone Curved surface area=2pR2p*R*2vR=2v:1=R2+R2——-v=2vRHemisphere Curved surface areaCone Curved surface area=2pR2p*R*2vR=2v:1    

1. If a right circular cone of height 24 cm has a volume of 1232 cm cube, then the area of its curved surface is :A) 450cm2450cm2450cm2B) 550cm2550cm2550cm2C) 650cm2650cm2650cm2D) 750cm2750cm2750cm2Ans. BVolume is given, we can calculate the radius from it, then by calculating slant height, we can get curved surface area. 13*p*r2*h=123213*227*r2*24=1232r2=1232*7*322*24=49r=7Now, r = 7cm and h = 24 cm l=r2+h2——v=72+242——-v=25cmCurved surface area =prl=227*7*25=550cm213*p*r2*h=123213*227*r2*24=1232r2=1232*7*322*24=49r=7Now, r = 7cm and h = 24 cm l=r2+h2——v=72+242——-v=25cmCurved surface area =prl=227*7*25=550cm213*p*r2*h=123213*227*r2*24=1232r2=1232*7*322*24=49r=7Now, r = 7cm and h = 24 cm l=r2+h2——v=72+242——-v=25cmCurved surface area =prl=227*7*25=550cm2 2. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m cube) is:A) 4120 m cubeB) 4140 m cubeC) 5140 m cubeD) 5120 m cubeAns. DExplanation:l = (48 – 16)m = 32 m, [because 8+8 = 16]b = (36 -16)m = 20 m,h = 8 m.Volume of the box = (32 x 20 x 8) m cube = 5120 m cube. 3. The maximum length of a pencil that can he kept is a rectangular box of dimensions 8 cm x 6 cm x 2 cm, isA) 217–v217–v217–vB) 216–v216–v216–vC) 226–v226–v226–vD) 224–v224–v224–vAns. CIn this question we need to calculate the diagonal of cuboid,which is = l2+b2+h2———-v=82+62+22———-v=104—v=226–vl2+b2+h2———-v=82+62+22———-v=104—v=226–vl2+b2+h2———-v=82+62+22———-v=104—v=226–v 4. The slant height of a conical mountain is 2.5 km and the area of its base is 1.54 km square. The height of mountain is :A) 2.3 kmB) 2.4 kmC) 2.5 kmD) 2.6 kmAns. BLet the radius of the base be r km. Then, pr2=1.54r2=1.54*722=0.49=0.7kmNow l=2.5 km, r = 0.7 kmh=2.52-0.72———vkm=6.25-0.49———v=5.76—-vkm=2.4kmpr2=1.54r2=1.54*722=0.49=0.7kmNow l=2.5 km, r = 0.7 kmh=2.52-0.72———vkm=6.25-0.49———v=5.76—-vkm=2.4kmpr2=1.54r2=1.54*722=0.49=0.7kmNow l=2.5 km, r = 0.7 kmh=2.52-0.72———vkm=6.25-0.49———v=5.76—-vkm=2.4km 5. The radii of two cones are in ratio 2:1, their volumes are equal. Find the ratio of their heights.A) 1:4B) 1:3C) 1:2D) 1:5Ans. ALet their radii be 2x, x and their heights be h and H resp. Then,Volume of cone =13pr2h13*p*(2x)2*h13*p*x2*H=>hH=14=>h:H=1:4Volume of cone =13pr2h13*p*(2x)2*h13*p*x2*H=>hH=14=>h:H=1:4Volume of cone =13pr2h13*p*(2x)2*h13*p*x2*H=>hH=14=>h:H=1:4      

1. Two right circular cylinders of equal volumes have their heights in the ratio 1:2. Find the ratio of their radii.A) 3v:13v:13v:1B) 7v:17v:17v:1C) 2v:12v:12v:1D) 2:12:12:1Ans. CLet their heights be h and 2h and radii be r and R respectively then.pr2h=pR2(2h)=>r2R2=2hh=21=>rR=2v1=>r:R=2v:1pr2h=pR2(2h)=>r2R2=2hh=21=>rR=2v1=>r:R=2v:1pr2h=pR2(2h)=>r2R2=2hh=21=>rR=2v1=>r:R=2v:1 2. A cylindrical tank of diameter 35 cm is full of water. If 11 litres of water is drawn off, the water level in the tank will drop by:A) 1137cm1137cm1137cmB) 1127cm1127cm1127cmC) 1117cm1117cm1117cmD) 11cm11cm11cmAns. ALet the drop in the water level be h cm, then,Volume of cylinder= pr2h=>227*352*352*h=11000=>h=11000*7*422*35*35cm=807cm=1137cmVolume of cylinder= pr2h=>227*352*352*h=11000=>h=11000*7*422*35*35cm=807cm=1137cmVolume of cylinder= pr2h=>227*352*352*h=11000=>h=11000*7*422*35*35cm=807cm=1137cm 3. 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length if the wire in meters will be:A) 76 mB) 80 mC) 84 mD) 88 mAns. CLet the length of the wire be h Radius=12mm=120cmpr2h=66227*120*120*h=66=>h=66*20*20*722=8400cm=84mRadius=12mm=120cmpr2h=66227*120*120*h=66=>h=66*20*20*722=8400cm=84mRadius=12mm=120cmpr2h=66227*120*120*h=66=>h=66*20*20*722=8400cm=84m 4. A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weights 8g/cm cube, then find the weight of the pipe.A) 3.696 kgB) 3.686 kgC) 2.696 kgD) 2.686 kgAns. AExternal radius = 4 cmInternal radius = 3 cm [because thickness of pipe is 1 cm]Volume of iron =pr2h=227*[42-32]*21cm3=227*1*21cm3=462cm3Volume of iron =pr2h=227*[42-32]*21cm3=227*1*21cm3=462cm3Volume of iron =pr2h=227*[42-32]*21cm3=227*1*21cm3=462cm3 Weight of iron = 462*8 = 3696 gm = 3.696 kg 5. The curved surface of a right circular cone of height 15 cm and base diameter 16 cm is:A) 116pcm2116pcm2116pcm2B) 122pcm2122pcm2122pcm2C) 124pcm2124pcm2124pcm2D) 136pcm2136pcm2136pcm2Ans. DCurved surface area of cone=prll=r2+h2——vl=82+152——-v=17cmCurved surface area =prl=p*8*17=136pcm2Curved surface area of cone=prll=r2+h2——vl=82+152——-v=17cmCurved surface area =prl=p*8*17=136pcm2Curved surface area of cone=prll=r2+h2——vl=82+152——-v=17cmCurved surface area =prl=p*8*17=136pcm2      

1. The perimeter of one face of a cube is 20 cm. Its volume will be:A) 125cm3125cm3125cm3B) 400cm3400cm3400cm3C) 250cm3250cm3250cm3D) 625cm3625cm3625cm3Ans. AEdge of cude = 20/4 = 5 cm Volume = a*a*a = 5*5*5 = 125 cm cube View Answer 2. The cost of the paint is Rs. 36.50 per kg. If 1 kg of paint covers 16 square feet, how much will it cost to paint outside of a cube having 8 feet each side.A) Rs. 850B) Rs. 860C) Rs. 876D) Rs. 886Ans. CSurface area =6a2=6*82=384sq feetQuantity required =38416=24kgCost of painting =36.50*24=Rs.876Surface area =6a2=6*82=384sq feetQuantity required =38416=24kgCost of painting =36.50*24=Rs.876Surface area =6a2=6*82=384sq feetQuantity required =38416=24kgCost of painting =36.50*24=Rs.876 3. How many cubes of 10 cm edge can be put in a cubical box of 1 m edge.A) 10000 cubesB) 1000 cubesC) 100 cubesD) 50 cubesAns. BNumber of cubes =100*100*10010*10*10=1000Number of cubes =100*100*10010*10*10=1000Number of cubes =100*100*10010*10*10=1000 Note: 1 m = 100 cm 4. If the volume of two cubes are in the ratio 27:1, the ratio of their edges is:A) 3:1B) 3:2C) 3:5D) 3:7Ans.ALet the edges be a and b of two cubes, then a3b3=271=>(ab)3=(31)3ab=31=>a:b=3:1a3b3=271=>(ab)()3=(31)3ab=31=>a:b=3:1a3b3=271=>(ab)()3=(31)3ab=31=>a:b=3:1 5. A circular well with a diameter of 2 meters, is dug to a depth of 14 meters. What is the volume of the earth dug out.A) 40m340m340m3B) 42m342m342m3C) 44m344m344m3D) 46m346m346m3Ans. CVolume=pr2hVolume=(227*1*1*14)m3=44m3Volume=pr2hVolume=(227*1*1*14)m3=44m3Volume=pr2hVolume=(227*1*1*14)m3=44m3    

1. Find the surface area of a 10cm*4cm*3cm brick.A) 154 cm squareB) 156 cm squareC) 160 cm squareD) 164 cm squareAns. DSurface area of a cuboid = 2(lb+bh+hl) cm squareSo, Surface area of a brick = 2(10*4+4*3+3*10) cm square = 2(82) cm square = 164 cm square 2. A cistern 6 m long and 4 m wide contains water up to a breadth of 1 m 25 cm. Find the total area of the wet surface.A) 42 m sqaureB) 49 m sqaureC) 52 m sqaureD) 64 m sqaureAns. BArea of the wet surface = 2[lb+bh+hl] – lb = 2 [bh+hl] + lb= 2[(4*1.25+6*1.25)]+6*4 = 49 m square 3. A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets into it. The mass of the man is :A) 50 kgB) 60 kgC) 70 kgD) 80 kgAns. BIn this type of question, first we will calculate the volume of water displaces then will multiply with the density of water.Volume of water displaced = 3*2*0.01 = 0.06 m cube Mass of Man = Volume of water displaced * Density of water = 0.06 * 1000 = 60 kg 4. How many bricks, each measuring 25cm*11.25cm*6cm, will be needed to build a wall 8m*6m*22.5mA) 6100B) 6200C) 6300D) 6400Ans. DTo solve this type of question, simply divide the volume of wall with the volume of brick to get the numbers of required bricksSo lets solve this Number of bricks = Volume of wallVolume of 1 brick=800*600*22.525*11.25*6=6400Volume of wallVolume of 1 brick=800*600*22.525*11.25*6=6400Volume of wallVolume of 1 brick=800*600*22.525*11.25*6=6400 5. A swimming pool 9 m wide and 12 m long is 1 m deep on the shallow side and 4 m deep on the deeper side. Its volume is:A) 260B) 262C) 270D) 272Ans. CVolume will be length * breadth * height, but in this case two heights are given so we will take average, Volume=(12*9*(1+42))m312*9*2.5m3=270m3Volume=(12*9*(1+42))m312*9*2.5m3=270m3Volume=(12*9*(1+42))m312*9*2.5m3=270  

Q.1 The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding. Solution : Diameter of the sphere = 7 m. Therefore, radius is 3.5 m. So, the ridingspace available for the motorcyclist is the surface area of the ‘sphere’ which isgiven by4pr2 = 4 ×22 / 7× 3.5 × 3.5 m2= 154 m2 Q.2 A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 m, find the cost of painting it, given the cost of painting is ` 5 per 100 cm2. Solution : Since only the rounded surface of the dome is to be painted, we would need to find the curved surface area of the hemisphere to know the extent of painting that needs to be done. Now, circumference of the dome = 17.6 m. Therefore, 17.6 = 2pr. So, the radius of the dome = 17.6 ×7/ 2 ??22 m = 2.8 mThe curved surface area of the dome = 2?r2= 2 × 22/ 7 × 2.8 × 2.8 m2= 49.28 m2Now, cost of painting 100 cm2 is ` 5.So, cost of painting 1 m2 = ` 500Therefore, cost of painting the whole dome= ` 500 × 49.28= ` 24640 Q.3 A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm × 12 cm × 8 cm, how many bricks would berequired? Solution : Since the wall with all its bricks makes up the space occupied by it, weneed to find the volume of the wall, which is nothing but a cuboid.Here, Length = 10 m = 1000 cmThickness = 24 cmHeight = 4 m = 400 cmTherefore, Volume of the wall = length × thickness × height= 1000 × 24 × 400 cm3Now, each brick is a cuboid with length = 24 cm, breadth = 12 cm and height = 8 cmSo, volume of each brick = length × breadth × height= 24 × 12 × 8 cm3So, number of bricks required =volume of the wallvolume of each brick=1000 × 24 × 40024× 12 × 8= 4166.6So, the wall requires 4167 bricks Q.4 A child playing with building blocks, which are of the shape of cubes, has built a structure. If the edge of each cube is 3 cm, find the volume of the structure built by the child. Solution : Volume of each cube = edge × edge × edge= 3 × 3 × 3 cm3 = 27 cm3Number of cubes in the structure = 15Therefore, volume of the structure = 27 × 15 cm3= 405 cm3 Q.5 The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10 m, how much concrete mixture would be required to build 14 such pillars? Solution : Since the concrete mixture that is to beused to build up the pillars is going to occupy theentire space of the pillar, what we need to find hereis the volume of the cylinders.Radius of base of a cylinder = 20 cmHeight of the cylindrical pillar = 10 m = 1000 cmSo, volume of each cylinder = ?r2h=22/7 * 20* 20 *1000?cm3=8800000/7 cm3=8.8 / 7 m3 (Since 1000000 cm3 = 1m3)Therefore, volume of 14 pillars = volume of each cylinder × 14=8.8/7 * 14= 17.6 m3So, 14 pillars would need 17.6 m3 of concrete mixture.      

Q.1 Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and produce different crops. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get for their crops? Solution : Let ABCD be the field.Perimeter = 400 mSo, each side = 400 m ??4 = 100 m.i.e. AB = AD = 100 m.Let diagonal BD = 160 m.Then semi-perimeter s of ??ABD is given bys =100 + 100 +160 /2m = 180 mTherefore, area of ??ABD = 180(180 ??100) (180 – 100) (180 – 160)= 180 ??80 ??80 ??20 m2 = 4800 m2Therefore, each of them will get an area of 4800 m2 . Q.2 Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden box covered with coloured paper with picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively how many square sheets of paper of side 40 cm would she require? Solution : Since Mary wants to paste the paper on the outer surface of the box; the quantity of paperrequired would be equal to the surface area of the box which is of the shape of a cuboid. The dimensionsof the box are: Length =80 cm, Breadth = 40 cm, Height = 20 cm.The surface area of the box = 2(lb + bh + hl)= 2[(80 × 40) + (40 × 20) + (20 × 80)] cm2= 2[3200 + 800 + 1600] cm2= 2 × 5600 cm2 = 11200 cm2The area of each sheet of the paper = 40 × 40 cm2= 1600 cm2Therefore, number of sheets required =surface area of boxarea of one sheet of paper=11200/ 1600= 7So, she would require 7 sheets. Q.3 Hameed has built a cubical water tank with lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the tiles, if the cost of the tiles is ` 360 per dozen. Solution : Since Hameed is getting the five outer faces of the tank covered with tiles, the would need to know the surface area of the tank, to decide on the number of tiles required.Edge of the cubical tank = 1.5 m = 150 cm (= a)So, surface area of the tank = 5 × 150 × 150 cm2Area of each square tile = side × side = 25 × 25 cm2So, the number of tiles required =surface area of the tankarea of each tile=5 ×150×15025× 25= 180Cost of 1 dozen tiles, i.e., cost of 12 tiles = ` 360Therefore, cost of one tile = `36012= ` 30So, the cost of 180 tiles = 180 × ` 30 = ` 5400 Q.4 Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope. (see Fig 13.10). What would be the area of chart paper required by her, if she wanted to make a kaleidoscope of length 25 cm with a 3.5 cm radius? Solution : Radius of the base of the cylindrical kaleidoscope (r) = 3.5 cm.Height (length) of kaleidoscope (h) = 25 cm.Area of chart paper required = curved surface area of the kaleidoscope= 2??rh= 2 * 22/7 * 3.5 * 25= 550 cm2 Q.5 A corn cob , shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm and length (height) as 20 cm. If each 1 cm2 of the surface of the cob carries an average of four grains, find how many grains you would find on the entire cob. Solution : Since the grains of corn are found only on the curved surface of the corn cob, we would need to know the curved surface area of the corn cob to find the total number of grains on it. In this question, we are given the height of the cone, so we need to find its slant height.Here, l = r 2 ??h2 = (2.1)2 ??202 cm= 404.41 cm = 20.11 cmTherefore, the curved surface area of the corn cob = ?rl= 22/7 × 2.1 × 20.11 cm2 = 132.726 cm2 = 132.73 cm2 (approx.)Number of grains of corn on 1 cm2 of the surface of the corn cob = 4Therefore, number of grains on the entire curved surface of the cob= 132.73 × 4 = 530.92 = 531 (approx.)So, there would be approximately 531 grains of corn on the cob.     iibm