Class 9: Surface Area of Cuboid and Cube

$\displaystyle \textbf{Question 1: } \text{Find the lateral surface area and total surface area of a cuboid}$
$\displaystyle \text{of length }80\text{ cm, breadth }40\text{ cm and height }20\text{ cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Dimensions of the cuboid: Length }(l)=80\text{ cm, Breadth }(b)=40\text{ cm, Height }(h)=20\text{ cm.}$
$\displaystyle \text{Lateral surface area}=2(l+b)\times h$
$\displaystyle =2(80+40)\times20=4800\text{ cm}^2$
$\displaystyle \text{Total surface area}=2(lb+bh+hl)$
$\displaystyle =2(80\times40+40\times20+20\times80)$
$\displaystyle =2(3200+800+1600)=11200\text{ cm}^2$
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$\displaystyle \textbf{Question 2: } \text{Find the lateral surface area and total surface area of a cube}$
$\displaystyle \text{of edge }10\text{ cm.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Edge of the cube }(a)=10\text{ cm.}$
$\displaystyle \text{Lateral surface area}=4a^2=4\times10^2=400\text{ cm}^2$
$\displaystyle \text{Total surface area}=6a^2=6\times10^2=600\text{ cm}^2$
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$\displaystyle \textbf{Question 3: } \text{Find the ratio of the total surface area and lateral surface}$
$\displaystyle \text{area of a cube.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the edge of the cube}=a.$
$\displaystyle \text{Lateral surface area}=4a^2$
$\displaystyle \text{Total surface area}=6a^2$
$\displaystyle \therefore \frac{\text{Total surface area}}{\text{Lateral surface area}}=\frac{6a^2}{4a^2}=\frac{3}{2}$
$\displaystyle \therefore \text{Required ratio}=3:2.$
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$\displaystyle \textbf{Question 4: } \text{Mary wants to decorate her Christmas tree by placing it on a}$
$\displaystyle \text{wooden block covered with coloured paper. If the box has length,}$
$\displaystyle \text{breadth and height }80\text{ cm, }40\text{ cm and }20\text{ cm respectively, find the}$
$\displaystyle \text{number of square sheets of paper of side }40\text{ cm required.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Dimensions of the box: Length }(l)=80\text{ cm, Breadth }(b)=40\text{ cm, Height }(h)=20\text{ cm.}$
$\displaystyle \text{Total surface area}=2(lb+bh+hl)$
$\displaystyle =2(80\times40+40\times20+20\times80)$
$\displaystyle =2(3200+800+1600)=11200\text{ cm}^2$
$\displaystyle \text{Area of one square sheet}=40\times40=1600\text{ cm}^2$
$\displaystyle \therefore \text{Number of sheets}=\frac{11200}{1600}=7$
$\displaystyle \therefore \text{Mary requires }7\text{ square sheets of paper.}$
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$\displaystyle \textbf{Question 5: } \text{The length, breadth and height of a room are }5\text{ m, }4\text{ m and }3\text{ m}$
$\displaystyle \text{respectively. Find the cost of whitewashing the walls of the room and the ceiling}$
$\displaystyle \text{at the rate of Rs. }7.50\text{ per m}^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Dimensions of the room: Length }(l)=5\text{ m, Breadth }(b)=4\text{ m, Height }(h)=3\text{ m.}$
$\displaystyle \text{Area to be whitewashed}=\text{Lateral surface area}+\text{Area of ceiling}$
$\displaystyle =2(l+b)h+lb$
$\displaystyle =2(5+4)\times3+5\times4$
$\displaystyle =54+20=74\text{ m}^2$
$\displaystyle \text{Rate of whitewashing}=\text{Rs. }7.50\text{ per m}^2$
$\displaystyle \therefore \text{Total cost}=74\times7.50=\text{Rs. }555$
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$\displaystyle \textbf{Question 6: } \text{Three equal cubes are placed adjacently in a row. Find the ratio of the total}$
$\displaystyle \text{surface area of the new cuboid to the sum of the surface areas of the three cubes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the edge of each cube}=l.$
$\displaystyle \text{Total surface area of one cube}=6l^2$
$\displaystyle \text{Sum of the total surface areas of three cubes}=3\times6l^2=18l^2$
$\displaystyle \text{Dimensions of the new cuboid: Length}=3l,\ \text{Breadth}=l,\ \text{Height}=l.$
$\displaystyle \text{Total surface area of the cuboid}=2(lb+bh+hl)$
$\displaystyle =2(3l\times l+l\times l+l\times3l)$
$\displaystyle =2(3l^2+l^2+3l^2)=14l^2$
$\displaystyle \therefore \frac{\text{Total surface area of cuboid}}{\text{Sum of total surface areas of three cubes}}=\frac{14l^2}{18l^2}=\frac{7}{9}$
$\displaystyle \therefore \text{Required ratio}=7:9.$
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$\displaystyle \textbf{Question 7: } \text{A }4\text{ cm cube is cut into }1\text{ cm cubes. Calculate the total surface area}$
$\displaystyle \text{of all the small cubes.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Edge of the big cube}=4\text{ cm.}$
$\displaystyle \text{Volume of the big cube}=4^3=64\text{ cm}^3$
$\displaystyle \text{Edge of each small cube}=1\text{ cm.}$
$\displaystyle \text{Volume of each small cube}=1^3=1\text{ cm}^3$
$\displaystyle \therefore \text{Number of small cubes}=\frac{64}{1}=64$
$\displaystyle \text{Total surface area of each small cube}=6(1)^2=6\text{ cm}^2$
$\displaystyle \therefore \text{Total surface area of }64\text{ small cubes}=64\times6=384\text{ cm}^2$
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$\displaystyle \textbf{Question 8: } \text{The length of a hall is }18\text{ m and the width is }12\text{ m. The sum of the areas}$
$\displaystyle \text{of the floor and the flat roof is equal to the sum of the areas of the four walls.}$
$\displaystyle \text{Find the height of the hall.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length }(l)=18\text{ m, Breadth }(b)=12\text{ m and Height }=h\text{ m.}$
$\displaystyle \text{Area of floor}=18\times12=216\text{ m}^2$
$\displaystyle \text{Area of roof}=18\times12=216\text{ m}^2$
$\displaystyle \text{Area of the four walls}=2(l+b)h$
$\displaystyle =2(18+12)h=60h$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 60h=216+216=432$
$\displaystyle \therefore h=\frac{432}{60}=\frac{36}{5}=7.2\text{ m}$
$\displaystyle \therefore \text{Height of the hall}=7.2\text{ m.}$
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$\displaystyle \textbf{Question 9: } \text{Hameed has built a cubical water tank with a lid for his house, with each edge}$
$\displaystyle \text{being }1.5\text{ m long. He gets the outer surface of the tank, excluding the base, covered}$
$\displaystyle \text{with square tiles of side }25\text{ cm. Find how much he would spend on the tiles,}$
$\displaystyle \text{if the cost of the tiles is Rs. }360\text{ per dozen.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Edge of the cubical tank }(a)=1.5\text{ m.}$
$\displaystyle \text{Area to be tiled}=\text{Lateral surface area}+\text{Area of top}$
$\displaystyle =4a^2+a^2=5a^2$
$\displaystyle =5(1.5)^2=11.25\text{ m}^2$
$\displaystyle \text{Side of each square tile}=25\text{ cm}=0.25\text{ m}$
$\displaystyle \text{Area of one square tile}=0.25\times0.25=0.0625\text{ m}^2$
$\displaystyle \therefore \text{Number of tiles}=\frac{11.25}{0.0625}=180$
$\displaystyle \text{Number of dozens}=\frac{180}{12}=15$
$\displaystyle \therefore \text{Total cost}=15\times360=\text{Rs. }5400$
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$\displaystyle \textbf{Question 10: } \text{Each edge of a cube is increased by }50\%\text{. Find the percentage increase}$
$\displaystyle \text{in the surface area of the cube.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the initial edge of the cube}=l.$
$\displaystyle \text{New edge}=l+50\%\text{ of }l=1.5l$
$\displaystyle \text{Initial total surface area}=6l^2$
$\displaystyle \text{New total surface area}=6(1.5l)^2=13.5l^2$
$\displaystyle \text{Increase in surface area}=13.5l^2-6l^2=7.5l^2$
$\displaystyle \therefore \text{Percentage increase}=\frac{7.5l^2}{6l^2}\times100=125\%$
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$\displaystyle \textbf{Question 11: } \text{The dimensions of a rectangular box are in the ratio }2:3:4\text{ and the difference}$
$\displaystyle \text{between the cost of covering it with a sheet of paper at the rates of Rs. }8\text{ and}$
$\displaystyle \text{Rs. }9.50\text{ per m}^2\text{ is Rs. }1248\text{. Find the dimensions of the box.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the dimensions of the box be }2x\text{ m, }3x\text{ m and }4x\text{ m.}$
$\displaystyle \text{Total surface area}=2(lb+bh+hl)$
$\displaystyle =2(6x^2+12x^2+8x^2)=52x^2$
$\displaystyle \text{Cost at Rs. }8\text{ per m}^2=52x^2\times8=416x^2$
$\displaystyle \text{Cost at Rs. }9.50\text{ per m}^2=52x^2\times9.5=494x^2$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 494x^2-416x^2=1248$
$\displaystyle 78x^2=1248$
$\displaystyle x^2=16$
$\displaystyle \therefore x=4$
$\displaystyle \therefore \text{Dimensions of the box}=8\text{ m, }12\text{ m and }16\text{ m.}$
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$\displaystyle \textbf{Question 12: } \text{A closed iron tank }12\text{ m long, }9\text{ m wide and }4\text{ m deep is to be made.}$
$\displaystyle \text{Determine the cost of iron sheet used at the rate of Rs. }5\text{ per metre, the sheet being}$
$\displaystyle 2\text{ m wide.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length }(l)=12\text{ m, Breadth }(b)=9\text{ m, Height }(h)=4\text{ m.}$
$\displaystyle \text{Total surface area}=2(lb+bh+hl)$
$\displaystyle =2(12\times9+9\times4+4\times12)$
$\displaystyle =2(108+36+48)=384\text{ m}^2$
$\displaystyle \text{Width of the iron sheet}=2\text{ m}$
$\displaystyle \therefore \text{Length of sheet required}=\frac{384}{2}=192\text{ m}$
$\displaystyle \text{Cost of iron sheet}=\text{Rs. }5\text{ per metre}$
$\displaystyle \therefore \text{Total cost}=192\times5=\text{Rs. }960$
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$\displaystyle \textbf{Question 13: } \text{Ravish wanted to make a temporary shelter for his car by making a box-like}$
$\displaystyle \text{structure with tarpaulin that covers all the four sides and the top of the car, with the}$
$\displaystyle \text{front face as a flap which can be rolled up. Assuming that the stitching margins are}$
$\displaystyle \text{negligible, how much tarpaulin would be required to make the shelter of height }2.5\text{ m}$
$\displaystyle \text{with base dimensions }4\text{ m}\times3\text{ m?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length }(l)=4\text{ m, Breadth }(b)=3\text{ m, Height }(h)=2.5\text{ m.}$
$\displaystyle \text{Area of tarpaulin required}=\text{Lateral surface area}+\text{Area of top}$
$\displaystyle =2(l+b)h+lb$
$\displaystyle =2(4+3)\times2.5+4\times3$
$\displaystyle =35+12=47\text{ m}^2$
$\displaystyle \therefore \text{Tarpaulin required}=47\text{ m}^2.$
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$\displaystyle \textbf{Question 14: } \text{An open box is made of wood }3\text{ cm thick. Its external length, breadth and}$
$\displaystyle \text{height are }1.48\text{ m, }1.16\text{ m and }8.3\text{ dm respectively. Find the cost of painting}$
$\displaystyle \text{the inner surface at Rs. }50\text{ per m}^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Thickness of wood}=3\text{ cm}=0.03\text{ m.}$
$\displaystyle \text{External length}=1.48\text{ m, External breadth}=1.16\text{ m.}$
$\displaystyle \text{External height}=8.3\text{ dm}=0.83\text{ m.}$
$\displaystyle \text{Inner length}=1.48-2(0.03)=1.42\text{ m}$
$\displaystyle \text{Inner breadth}=1.16-2(0.03)=1.10\text{ m}$
$\displaystyle \text{Inner height}=0.83-0.03=0.80\text{ m}$
$\displaystyle \text{Inner surface area}=2(l+b)h+lb$
$\displaystyle =2(1.42+1.10)\times0.80+1.42\times1.10$
$\displaystyle =4.032+1.562=5.594\text{ m}^2$
$\displaystyle \therefore \text{Cost of painting}=5.594\times50=\text{Rs. }279.70$
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$\displaystyle \textbf{Question 15: } \text{The cost of preparing the walls of a room }12\text{ m long at the rate of}$
$\displaystyle \text{Rs. }1.35\text{ per m}^2\text{ is Rs. }340.20\text{ and the cost of matting the floor at }85\text{ paise}$
$\displaystyle \text{per m}^2\text{ is Rs. }91.80\text{. Find the height of the room.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of the room }(l)=12\text{ m, Breadth }=b\text{ m, Height }=h\text{ m.}$
$\displaystyle \text{Area of the four walls}=2(l+b)h=2(12+b)h$
$\displaystyle \text{Rate of preparing the walls}=\text{Rs. }1.35\text{ per m}^2$
$\displaystyle \therefore 340.20=2(12+b)h\times1.35$
$\displaystyle 340.20=2.70(12+b)h\qquad\ldots\text{(i)}$
$\displaystyle \text{Rate of matting the floor}=85\text{ paise per m}^2=\text{Rs. }0.85\text{ per m}^2$
$\displaystyle \therefore 91.80=lb\times0.85$
$\displaystyle 91.80=12b\times0.85$
$\displaystyle \therefore b=\frac{91.80}{12\times0.85}=9\text{ m}$
$\displaystyle \text{Substituting }b=9\text{ in (i),}$
$\displaystyle 340.20=2.70(12+9)h$
$\displaystyle 340.20=2.70\times21h$
$\displaystyle \therefore h=\frac{340.20}{2.70\times21}=6\text{ m}$
$\displaystyle \therefore \text{Height of the room}=6\text{ m.}$
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$\displaystyle \textbf{Question 16: } \text{The dimensions of a room are }12.5\text{ m}\times9\text{ m}\times7\text{ m. There are }2\text{ doors}$
$\displaystyle \text{and }4\text{ windows in the room; each door measures }2.5\text{ m}\times1.2\text{ m and each window}$
$\displaystyle \text{measures }1.5\text{ m}\times1\text{ m. Find the cost of painting the walls at Rs. }3.50\text{ per m}^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length }(l)=12.5\text{ m, Breadth }(b)=9\text{ m, Height }(h)=7\text{ m.}$
$\displaystyle \text{Area of the four walls}=2(l+b)h$
$\displaystyle =2(12.5+9)\times7=301\text{ m}^2$
$\displaystyle \text{Area of }2\text{ doors}=2\times2.5\times1.2=6\text{ m}^2$
$\displaystyle \text{Area of }4\text{ windows}=4\times1.5\times1=6\text{ m}^2$
$\displaystyle \text{Total area of doors and windows}=6+6=12\text{ m}^2$
$\displaystyle \therefore \text{Area to be painted}=301-12=289\text{ m}^2$
$\displaystyle \text{Rate of painting}=\text{Rs. }3.50\text{ per m}^2$
$\displaystyle \therefore \text{Cost of painting}=289\times3.50=\text{Rs. }1011.50$
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$\displaystyle \textbf{Question 17: } \text{The length and breadth of a hall are in the ratio }4:3\text{ and its height is }5.5\text{ m.}$
$\displaystyle \text{The cost of decorating its walls, including doors and windows, at Rs. }6.60\text{ per m}^2$
$\displaystyle \text{is Rs. }5082\text{. Find the length and breadth of the hall.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length }(l)=4x\text{ m and breadth }(b)=3x\text{ m.}$
$\displaystyle \text{Height }(h)=5.5\text{ m.}$
$\displaystyle \text{Area of the four walls}=2(l+b)h$
$\displaystyle =2(4x+3x)\times5.5=77x\text{ m}^2$
$\displaystyle \text{Cost of decorating the walls}=\text{Rs. }5082$
$\displaystyle \therefore 77x\times6.60=5082$
$\displaystyle x=\frac{5082}{77\times6.60}=10$
$\displaystyle \therefore \text{Length}=4x=40\text{ m}$
$\displaystyle \text{Breadth}=3x=30\text{ m}$
$\displaystyle \therefore \text{The length and breadth of the hall are }40\text{ m and }30\text{ m respectively.}$
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$\displaystyle \textbf{Question 18: } \text{A wooden bookshelf has external dimensions as follows: Height }=110\text{ cm,}$
$\displaystyle \text{Depth }=25\text{ cm and Breadth }=85\text{ cm. The thickness of the plank is }5\text{ cm}$
$\displaystyle \text{everywhere. The external faces are to be polished and the inner faces are to be}$
$\displaystyle \text{painted. If the rate of polishing is }20\text{ paise per cm}^2\text{ and the rate of painting is}$
$\displaystyle 10\text{ paise per cm}^2\text{, find the total expenses required for polishing and painting}$
$\displaystyle \text{the surface of the bookshelf.}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{External breadth}=85\text{ cm, Depth}=25\text{ cm, Height}=110\text{ cm.}$
$\displaystyle \text{Thickness of each plank}=5\text{ cm.}$
$\displaystyle \text{Internal breadth}=85-2(5)=75\text{ cm}$
$\displaystyle \text{Internal depth}=25-5=20\text{ cm}$
$\displaystyle \text{Internal height of each compartment}=\frac{110-4(5)}{3}=30\text{ cm}$
$\displaystyle \text{External area to be polished}$
$\displaystyle =2(25\times110)+2(85\times25)+(85\times110)+2(5\times110)+4(75\times5)$
$\displaystyle =5500+4250+9350+1100+1500$
$\displaystyle =21700\text{ cm}^2$
$\displaystyle \text{Rate of polishing}=20\text{ paise per cm}^2=\text{Rs. }0.20\text{ per cm}^2$
$\displaystyle \therefore \text{Cost of polishing}=21700\times0.20=\text{Rs. }4340$
$\displaystyle \text{Internal area to be painted}=3[2(75+30)\times20+75\times30]$
$\displaystyle =3[4200+2250]=3\times6450=19350\text{ cm}^2$
$\displaystyle \text{Rate of painting}=10\text{ paise per cm}^2=\text{Rs. }0.10\text{ per cm}^2$
$\displaystyle \therefore \text{Cost of painting}=19350\times0.10=\text{Rs. }1935$
$\displaystyle \therefore \text{Total expense}=4340+1935=\text{Rs. }6275$
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$\displaystyle \textbf{Question 19: } \text{The paint in a certain container is sufficient to paint an area equal to }9.375\text{ m}^2.$
$\displaystyle \text{How many bricks of dimensions }22.5\text{ cm}\times10\text{ cm}\times7.5\text{ cm can be painted}$
$\displaystyle \text{out of this container?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length }(l)=22.5\text{ cm, Breadth }(b)=10\text{ cm, Height }(h)=7.5\text{ cm.}$
$\displaystyle \text{Total surface area of one brick}=2(lb+bh+hl)$
$\displaystyle =2(22.5\times10+10\times7.5+7.5\times22.5)$
$\displaystyle =2(225+75+168.75)=937.5\text{ cm}^2$
$\displaystyle \text{Total area that can be painted}=9.375\text{ m}^2$
$\displaystyle =9.375\times10000=93750\text{ cm}^2$
$\displaystyle \therefore \text{Number of bricks}=\frac{93750}{937.5}=100$
$\displaystyle \therefore \text{Number of bricks that can be painted}=100.$
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