Class 10: Surface Area and Volume

$\displaystyle \textbf{Question 1: }\text{A tent consists of a frustum of a cone capped by a cone. If the radii of the}$
$\displaystyle \text{ends of the frustum are }13\text{ m and }7\text{ m, the height of the frustum is }8\text{ m and the}$
$\displaystyle \text{slant height of the conical cap is }12\text{ m, find the canvas required for the tent.}$
$\displaystyle \text{(Take }\pi=\frac{22}{7}\text{).}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the frustum, }R=13\text{ m},\qquad r=7\text{ m},\qquad h=8\text{ m.}$
$\displaystyle \text{Let the slant height of the frustum be }l\text{ m.}$
$\displaystyle l=\sqrt{h^2+(R-r)^2}$
$\displaystyle =\sqrt{8^2+(13-7)^2}$
$\displaystyle =\sqrt{64+36}=\sqrt{100}=10\text{ m.}$
$\displaystyle \text{Curved surface area of the frustum}=\pi(R+r)l$
$\displaystyle =\frac{22}{7}\times(13+7)\times10$
$\displaystyle =200\pi\text{ m}^2.$
$\displaystyle \text{Radius of the conical cap}=7\text{ m},\qquad \text{slant height}=12\text{ m.}$
$\displaystyle \text{Curved surface area of the conical cap}=\pi rl$
$\displaystyle =\frac{22}{7}\times7\times12$
$\displaystyle =84\pi\text{ m}^2.$
$\displaystyle \text{Canvas required}=200\pi+84\pi$
$\displaystyle =284\pi$
$\displaystyle =284\times\frac{22}{7}=\frac{6248}{7}\text{ m}^2$
$\displaystyle =892\frac{4}{7}\text{ m}^2.$
$\displaystyle \therefore \text{The canvas required for the tent is }892\frac{4}{7}\text{ m}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{A bucket is in the form of a frustum of a cone of height }30\text{ cm with}$
$\displaystyle \text{radii of its lower and upper ends as }10\text{ cm and }20\text{ cm respectively. Find the capacity}$
$\displaystyle \text{and surface area of the bucket. Also, find the cost of milk which can completely fill the}$
$\displaystyle \text{container, at the rate of Rs. }25\text{ per litre. (Use }\pi=3.14\text{).}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the frustum, }R=20\text{ cm},\qquad r=10\text{ cm},\qquad h=30\text{ cm.}$
$\displaystyle \text{Capacity of the bucket}=\frac{1}{3}\pi h(R^2+Rr+r^2)$
$\displaystyle =\frac{1}{3}\times3.14\times30(20^2+20\times10+10^2)$
$\displaystyle =10\times3.14(400+200+100)$
$\displaystyle =10\times3.14\times700$
$\displaystyle =21980\text{ cm}^3.$
$\displaystyle 21980\text{ cm}^3=21.98\text{ litres.}$
$\displaystyle \therefore \text{The capacity of the bucket is }21.98\text{ litres.}$
$\displaystyle \text{Let the slant height of the frustum be }l\text{ cm.}$
$\displaystyle l=\sqrt{h^2+(R-r)^2}$
$\displaystyle =\sqrt{30^2+(20-10)^2}$
$\displaystyle =\sqrt{900+100}=10\sqrt{10}\text{ cm.}$
$\displaystyle \text{Since the bucket is open at the top,}$
$\displaystyle \text{Surface area}=\pi(R+r)l+\pi r^2$
$\displaystyle =3.14(20+10)(10\sqrt{10})+3.14(10)^2$
$\displaystyle =942\sqrt{10}+314$
$\displaystyle \approx3292.87\text{ cm}^2.$
$\displaystyle \therefore \text{The surface area of the bucket is approximately }3292.87\text{ cm}^2.$
$\displaystyle \text{Cost of milk}=21.98\times25$
$\displaystyle =\text{Rs. }549.50.$
$\displaystyle \therefore \text{The cost of milk required to fill the bucket is Rs. }549.50.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{A bucket is in the form of a frustum of a cone with a capacity of }12317.6\text{ cm}^3$
$\displaystyle \text{of water. The radii of the top and bottom circular ends are }20\text{ cm and }12\text{ cm}$
$\displaystyle \text{respectively. Find the height of the bucket and the area of the metal sheet used in its}$
$\displaystyle \text{making. (Use }\pi=3.14\text{).}\hfill\text{[CBSE 2006C, 2016, 19, 20]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the frustum, }R=20\text{ cm},\qquad r=12\text{ cm.}$
$\displaystyle \text{Let the height of the bucket be }h\text{ cm.}$
$\displaystyle \text{Volume of the frustum}=\frac{1}{3}\pi h(R^2+Rr+r^2)$
$\displaystyle 12317.6=\frac{1}{3}\times3.14\times h(20^2+20\times12+12^2)$
$\displaystyle 12317.6=\frac{3.14h}{3}(400+240+144)$
$\displaystyle 12317.6=\frac{3.14\times784}{3}h$
$\displaystyle h=\frac{3\times12317.6}{3.14\times784}$
$\displaystyle \approx15.01\text{ cm.}$
$\displaystyle \therefore \text{The height of the bucket is approximately }15.01\text{ cm.}$
$\displaystyle \text{Let the slant height of the frustum be }l\text{ cm.}$
$\displaystyle l=\sqrt{h^2+(R-r)^2}$
$\displaystyle =\sqrt{(15.01)^2+(20-12)^2}$
$\displaystyle \approx17.01\text{ cm.}$
$\displaystyle \text{Since the bucket is open at the top,}$
$\displaystyle \text{Area of metal sheet}=\pi(R+r)l+\pi r^2$
$\displaystyle =3.14(20+12)(17.01)+3.14(12)^2$
$\displaystyle \approx2161.27\text{ cm}^2.$
$\displaystyle \therefore \text{The area of the metal sheet used is approximately }2161.27\text{ cm}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{The radii of the circular ends of a solid frustum of a cone are }33\text{ cm}$
$\displaystyle \text{and }27\text{ cm and its slant height is }10\text{ cm. Find its total surface area.}$
$\displaystyle \hfill\text{[CBSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the frustum, }R=33\text{ cm},\qquad r=27\text{ cm},\qquad l=10\text{ cm.}$
$\displaystyle \text{Total surface area}=\pi(R+r)l+\pi R^2+\pi r^2$
$\displaystyle =\pi\{(33+27)\times10+33^2+27^2\}$
$\displaystyle =\pi(600+1089+729)$
$\displaystyle =2418\pi\text{ cm}^2.$
$\displaystyle =2418\times\frac{22}{7}\text{ cm}^2$
$\displaystyle \approx7599.43\text{ cm}^2.$
$\displaystyle \therefore \text{The total surface area of the frustum is }2418\pi\text{ cm}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{A solid is in the shape of a frustum of a cone. The diameters of the two}$
$\displaystyle \text{circular ends are }60\text{ cm and }36\text{ cm and the height is }9\text{ cm. Find the area of its}$
$\displaystyle \text{whole surface and the volume.}\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Outer radius }R=\frac{60}{2}=30\text{ cm},\qquad \text{inner radius }r=\frac{36}{2}=18\text{ cm.}$
$\displaystyle \text{Height of the frustum}=9\text{ cm.}$
$\displaystyle \text{Let the slant height of the frustum be }l\text{ cm.}$
$\displaystyle l=\sqrt{h^2+(R-r)^2}$
$\displaystyle =\sqrt{9^2+(30-18)^2}$
$\displaystyle =\sqrt{81+144}=\sqrt{225}=15\text{ cm.}$
$\displaystyle \text{Whole surface area}=\pi(R+r)l+\pi R^2+\pi r^2$
$\displaystyle =\pi\{(30+18)\times15+30^2+18^2\}$
$\displaystyle =\pi(720+900+324)$
$\displaystyle =1944\pi\text{ cm}^2$
$\displaystyle \approx6109.71\text{ cm}^2.$
$\displaystyle \therefore \text{The whole surface area is }1944\pi\text{ cm}^2.$
$\displaystyle \text{Volume of the frustum}=\frac{1}{3}\pi h(R^2+Rr+r^2)$
$\displaystyle =\frac{1}{3}\pi\times9(30^2+30\times18+18^2)$
$\displaystyle =3\pi(900+540+324)$
$\displaystyle =5292\pi\text{ cm}^3.$
$\displaystyle =5292\times\frac{22}{7}=16632\text{ cm}^3.$
$\displaystyle \therefore \text{The volume of the frustum is }16632\text{ cm}^3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{A bucket, open at the top and made of a metal sheet, is in the form of a}$
$\displaystyle \text{frustum of a cone. The depth of the bucket is }24\text{ cm and the diameters of its upper}$
$\displaystyle \text{and lower circular ends are }30\text{ cm and }10\text{ cm respectively. Find the cost of the}$
$\displaystyle \text{metal sheet used at the rate of Rs. }10\text{ per }100\text{ cm}^2.\text{ (Use }\pi=3.14\text{).}$
$\displaystyle \hfill\text{[CBSE 2006, 13, 18]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Upper radius }R=\frac{30}{2}=15\text{ cm},\qquad \text{lower radius }r=\frac{10}{2}=5\text{ cm.}$
$\displaystyle \text{Height of the frustum}=24\text{ cm.}$
$\displaystyle \text{Let the slant height of the frustum be }l\text{ cm.}$
$\displaystyle l=\sqrt{h^2+(R-r)^2}$
$\displaystyle =\sqrt{24^2+(15-5)^2}$
$\displaystyle =\sqrt{576+100}=\sqrt{676}=26\text{ cm.}$
$\displaystyle \text{Since the bucket is open at the top,}$
$\displaystyle \text{Area of metal sheet}=\pi(R+r)l+\pi r^2$
$\displaystyle =3.14(15+5)\times26+3.14(5)^2$
$\displaystyle =1632.8+78.5$
$\displaystyle =1711.3\text{ cm}^2.$
$\displaystyle \text{Cost of }100\text{ cm}^2\text{ of metal sheet}=\text{Rs. }10.$
$\displaystyle \text{Cost of }1711.3\text{ cm}^2\text{ of metal sheet}=\frac{1711.3}{100}\times10$
$\displaystyle =\text{Rs. }171.13.$
$\displaystyle \therefore \text{The cost of the metal sheet used is Rs. }171.13.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{A bucket, made of metal sheet, is in the form of a frustum of a cone whose}$
$\displaystyle \text{height is }35\text{ cm and radii of its circular ends are }30\text{ cm and }12\text{ cm. How many}$
$\displaystyle \text{litres of milk does it contain if it is full to the brim? If the milk is sold at Rs. }40$
$\displaystyle \text{per litre, find the amount received by the person.}\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the frustum, }R=30\text{ cm},\qquad r=12\text{ cm},\qquad h=35\text{ cm.}$
$\displaystyle \text{Volume of the bucket}=\frac{1}{3}\pi h(R^2+Rr+r^2)$
$\displaystyle =\frac{1}{3}\times\frac{22}{7}\times35(30^2+30\times12+12^2)$
$\displaystyle =\frac{110}{3}(900+360+144)$
$\displaystyle =\frac{110}{3}\times1404$
$\displaystyle =51480\text{ cm}^3.$
$\displaystyle 1000\text{ cm}^3=1\text{ litre.}$
$\displaystyle \therefore 51480\text{ cm}^3=51.48\text{ litres.}$
$\displaystyle \therefore \text{The bucket can contain }51.48\text{ litres of milk.}$
$\displaystyle \text{Amount received}=51.48\times40$
$\displaystyle =\text{Rs. }2059.20.$
$\displaystyle \therefore \text{The amount received by the person is Rs. }2059.20.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{A milk container is made of metal sheet in the shape of a frustum of a}$
$\displaystyle \text{cone whose volume is }10459\frac{3}{7}\text{ cm}^3.\text{ The radii of its lower and upper circular}$
$\displaystyle \text{ends are }8\text{ cm and }20\text{ cm respectively. Find the cost of metal sheet used in making}$
$\displaystyle \text{the container at the rate of Rs. }1.40\text{ per cm}^2.\text{ (Use }\pi=\frac{22}{7}\text{).}\hfill\text{[CBSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the frustum, }R=20\text{ cm},\qquad r=8\text{ cm.}$
$\displaystyle \text{Let the height of the frustum be }h\text{ cm.}$
$\displaystyle \text{Volume of the frustum}=\frac{1}{3}\pi h(R^2+Rr+r^2)$
$\displaystyle 10459\frac{3}{7}=\frac{1}{3}\times\frac{22}{7}\times h(20^2+20\times8+8^2)$
$\displaystyle \frac{73216}{7}=\frac{22h}{21}(400+160+64)$
$\displaystyle \frac{73216}{7}=\frac{4576h}{7}$
$\displaystyle h=\frac{73216}{4576}=16\text{ cm.}$
$\displaystyle \text{Let the slant height of the frustum be }l\text{ cm.}$
$\displaystyle l=\sqrt{h^2+(R-r)^2}$
$\displaystyle =\sqrt{16^2+(20-8)^2}$
$\displaystyle =\sqrt{256+144}=\sqrt{400}=20\text{ cm.}$
$\displaystyle \text{Area of metal sheet used}=\text{Curved surface area}+\text{Area of larger base}$
$\displaystyle =\pi(R+r)l+\pi R^2$
$\displaystyle =\frac{22}{7}\{(20+8)\times20+20^2\}$
$\displaystyle =\frac{22}{7}(560+400)$
$\displaystyle =\frac{21120}{7}\text{ cm}^2.$
$\displaystyle \text{Cost of metal sheet}=\frac{21120}{7}\times1.40$
$\displaystyle =\text{Rs. }4224.$
$\displaystyle \therefore \text{The cost of the metal sheet used in making the container is Rs. }4224.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{A solid cone of base radius }10\text{ cm is cut into two parts through the}$
$\displaystyle \text{mid-point of its height by a plane parallel to its base. Find the ratio of the volumes of}$
$\displaystyle \text{the two parts of the cone.}\hfill\text{[CBSE 2013, 17]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the height and base radius of the original cone be }h\text{ and }r\text{ respectively.}$
$\displaystyle \text{The cone is cut by a plane through the mid-point of its height.}$
$\displaystyle \therefore \text{Height of the smaller cone}=\frac{h}{2}.$
$\displaystyle \text{The smaller cone and the original cone are similar.}$
$\displaystyle \therefore \frac{\text{Radius of smaller cone}}{r}=\frac{h/2}{h}=\frac{1}{2}.$
$\displaystyle \therefore \text{Radius of the smaller cone}=\frac{r}{2}.$
$\displaystyle \frac{\text{Volume of smaller cone}}{\text{Volume of original cone}}$
$\displaystyle =\left(\frac{1}{2}\right)^3=\frac{1}{8}.$
$\displaystyle \therefore \text{Volume of smaller cone}:\text{Volume of original cone}=1:8.$
$\displaystyle \text{Volume of remaining frustum}=8-1=7\text{ parts.}$
$\displaystyle \therefore \text{Volume of smaller cone}:\text{Volume of frustum}=1:7.$
$\displaystyle \therefore \text{The ratio of the volumes of the two parts is }1:7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{In figure, from the top of a solid cone of height }12\text{ cm and}$
$\displaystyle \text{base radius }6\text{ cm, a cone of height }4\text{ cm is removed by a plane parallel to the base.}$
$\displaystyle \text{Find the total surface area of the remaining solid. (Use }\pi=\frac{22}{7}\text{ and }\sqrt{5}=2.236\text{).}$
$\displaystyle \hfill\text{[CBSE 2015]}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{Height of the original cone}=12\text{ cm},\qquad \text{base radius}=6\text{ cm.}$
$\displaystyle \text{Height of the smaller cone removed}=4\text{ cm.}$
$\displaystyle \text{Let the radius of the smaller cone be }r\text{ cm.}$
$\displaystyle \text{Since the smaller cone and the original cone are similar,}$
$\displaystyle \frac{r}{6}=\frac{4}{12}$
$\displaystyle \therefore r=2\text{ cm.}$
$\displaystyle \text{Height of the remaining frustum}=12-4=8\text{ cm.}$
$\displaystyle \text{For the frustum, }R=6\text{ cm},\qquad r=2\text{ cm},\qquad h=8\text{ cm.}$
$\displaystyle \text{Let the slant height of the frustum be }l\text{ cm.}$
$\displaystyle l=\sqrt{h^2+(R-r)^2}$
$\displaystyle =\sqrt{8^2+(6-2)^2}$
$\displaystyle =\sqrt{64+16}=\sqrt{80}=4\sqrt{5}$
$\displaystyle =4\times2.236=8.944\text{ cm.}$
$\displaystyle \text{Total surface area of the remaining solid}$
$\displaystyle =\pi(R+r)l+\pi R^2+\pi r^2$
$\displaystyle =\frac{22}{7}\{(6+2)\times8.944+6^2+2^2\}$
$\displaystyle =\frac{22}{7}(71.552+36+4)$
$\displaystyle =\frac{22}{7}\times111.552$
$\displaystyle \approx350.88\text{ cm}^2.$
$\displaystyle \therefore \text{The total surface area of the remaining solid is approximately }350.88\text{ cm}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{The height of a cone is }10\text{ cm. The cone is divided into two parts using a}$
$\displaystyle \text{plane parallel to its base at the middle of its height. Find the ratio of the volumes of the}$
$\displaystyle \text{two parts.}\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The plane divides the height of the cone into two equal parts.}$
$\displaystyle \therefore \text{Height of the smaller cone}=\frac{10}{2}=5\text{ cm.}$
$\displaystyle \text{The smaller cone and the original cone are similar.}$
$\displaystyle \therefore \frac{\text{Height of smaller cone}}{\text{Height of original cone}}=\frac{5}{10}=\frac{1}{2}.$
$\displaystyle \text{For similar solids, the ratio of volumes is the cube of the ratio of corresponding lengths.}$
$\displaystyle \therefore \frac{\text{Volume of smaller cone}}{\text{Volume of original cone}}=\left(\frac{1}{2}\right)^3=\frac{1}{8}.$
$\displaystyle \therefore \text{Volume of smaller cone}:\text{Volume of original cone}=1:8.$
$\displaystyle \text{Volume of the remaining frustum}=8-1=7\text{ parts.}$
$\displaystyle \therefore \text{Volume of smaller cone}:\text{Volume of frustum}=1:7.$
$\displaystyle \therefore \text{The ratio of the volumes of the two parts is }1:7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{A reservoir in the form of the frustum of a right circular cone contains}$
$\displaystyle 44\times10^7\text{ litres of water which fills it completely. The radii of the bottom and top}$
$\displaystyle \text{of the reservoir are }50\text{ m and }100\text{ m respectively. Find the depth of water and the}$
$\displaystyle \text{lateral surface area of the reservoir. (Take }\pi=\frac{22}{7}\text{).}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Radius of the top }R=100\text{ m},\qquad \text{radius of the bottom }r=50\text{ m.}$
$\displaystyle 44\times10^7\text{ litres}=\frac{44\times10^7}{1000}\text{ m}^3$
$\displaystyle =440000\text{ m}^3.$
$\displaystyle \text{Let the depth of the reservoir be }h\text{ m.}$
$\displaystyle \text{Volume of the frustum}=\frac{1}{3}\pi h(R^2+Rr+r^2)$
$\displaystyle 440000=\frac{1}{3}\times\frac{22}{7}\times h(100^2+100\times50+50^2)$
$\displaystyle 440000=\frac{1}{3}\times\frac{22}{7}\times h(10000+5000+2500)$
$\displaystyle 440000=\frac{1}{3}\times\frac{22}{7}\times17500h$
$\displaystyle 440000=\frac{55000}{3}h$
$\displaystyle h=\frac{440000\times3}{55000}=24\text{ m.}$
$\displaystyle \therefore \text{The depth of water is }24\text{ m.}$
$\displaystyle \text{Let the slant height of the frustum be }l\text{ m.}$
$\displaystyle l=\sqrt{h^2+(R-r)^2}$
$\displaystyle =\sqrt{24^2+(100-50)^2}$
$\displaystyle =\sqrt{576+2500}=\sqrt{3076}$
$\displaystyle \approx55.46\text{ m.}$
$\displaystyle \text{Lateral surface area of the reservoir}=\pi(R+r)l$
$\displaystyle =\frac{22}{7}\times(100+50)\times55.46$
$\displaystyle \approx26146.23\text{ m}^2.$
$\displaystyle \therefore \text{The lateral surface area of the reservoir is approximately }26146.23\text{ m}^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{A cone of radius }4\text{ cm is divided into two parts by drawing a plane}$
$\displaystyle \text{through the mid-point of its axis and parallel to its base. Compare the volumes of the}$
$\displaystyle \text{two parts.}\hfill\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The plane passes through the mid-point of the axis of the cone.}$
$\displaystyle \therefore \text{Height of the smaller cone}:\text{Height of the original cone}=1:2.$
$\displaystyle \text{The smaller cone and the original cone are similar.}$
$\displaystyle \therefore \text{Their corresponding radii are also in the ratio }1:2.$
$\displaystyle \text{For similar cones, the ratio of volumes is the cube of the ratio of corresponding lengths.}$
$\displaystyle \therefore \text{Volume of smaller cone}:\text{Volume of original cone}$
$\displaystyle =1^3:2^3=1:8.$
$\displaystyle \text{Volume of the remaining frustum}=8-1=7\text{ parts.}$
$\displaystyle \therefore \text{Volume of smaller cone}:\text{Volume of frustum}=1:7.$
$\displaystyle \therefore \text{The volumes of the two parts are in the ratio }1:7.$
$\displaystyle \\$