Class 10: Cone and Sphere – Exercise 22(a)

$\displaystyle \textbf{Important Formulae:}$
$\displaystyle \text{Parameters of a cone: Radius of the base }(r),\text{ Height }(h)\text{ and Slant height }(l).$
$\displaystyle \text{Volume of a cone}=\frac{1}{3}\pi r^2h$
$\displaystyle \text{Curved surface area of a cone}=\pi rl$
$\displaystyle \text{Total surface area of a cone}=\pi r^2+\pi rl=\pi r(r+l)$
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$\displaystyle \textbf{Question 1: }\text{Find the volume of a cone whose slant height is }17\text{ cm and radius} \\ \text{of base is }8\text{ cm}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of a cone}=\frac{1}{3}\pi r^2h$
$\displaystyle l=17\text{ cm},\quad r=8\text{ cm}$
$\displaystyle h=\sqrt{l^2-r^2}=\sqrt{17^2-8^2}=\sqrt{289-64}=\sqrt{225}=15\text{ cm}$
$\displaystyle \therefore \text{Volume}=\frac{1}{3}\times\frac{22}{7}\times8^2\times15$
$\displaystyle =\frac{7040}{7}\text{ cm}^3\approx1005.71\text{ cm}^3$
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$\displaystyle \textbf{Question 2: }\text{The curved surface area of a cone is }12320\text{ cm}^2.\text{ If the radius of its} \\ \text{base is }56\text{ cm, find its height.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Curved surface area of a cone}=\pi rl$
$\displaystyle \pi\times56\times l=12320$
$\displaystyle l=\frac{12320}{\frac{22}{7}\times56}=70\text{ cm}$
$\displaystyle h=\sqrt{l^2-r^2}=\sqrt{70^2-56^2}$
$\displaystyle =\sqrt{4900-3136}=\sqrt{1764}=42\text{ cm}$
$\displaystyle \therefore \text{Height of the cone}=42\text{ cm}$
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$\displaystyle \textbf{Question 3: }\text{The circumference of the base of a }12\text{ m high conical tent is }66\text{ m}.\text{ Find} \\ \text{the volume of the air contained in it.}$
$\displaystyle \text{Answer:}$
$\displaystyle h=12\text{ m}$
$\displaystyle \text{Circumference of base}=66\text{ m}$
$\displaystyle 2\pi r=66$
$\displaystyle r=\frac{66}{2\times\frac{22}{7}}=10.5\text{ m}$
$\displaystyle \text{Volume of a cone}=\frac{1}{3}\pi r^2h$
$\displaystyle =\frac{1}{3}\times\frac{22}{7}\times(10.5)^2\times12$
$\displaystyle =1386\text{ m}^3$
$\displaystyle \therefore \text{Volume of air contained in the tent}=1386\text{ m}^3$
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$\displaystyle \textbf{Question 4: }\text{The radius and height of a right circular cone are in the ratio }$
$\displaystyle 5:12\text{ and its volume} \ \text{is }2512\text{ cm}^3. \ \text{Find the radius and slant height of the cone.} \\ \text{Take }\pi=3.14.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }r=5x\text{ and }h=12x$
$\displaystyle \text{Volume of the cone}=2512\text{ cm}^3$
$\displaystyle \frac{1}{3}\pi r^2h=2512$
$\displaystyle \frac{1}{3}\times3.14\times(5x)^2\times12x=2512$
$\displaystyle 3.14\times100x^3=2512$
$\displaystyle 314x^3=2512$
$\displaystyle x^3=8$
$\displaystyle x=2$
$\displaystyle \therefore r=5x=5\times2=10\text{ cm}$
$\displaystyle h=12x=12\times2=24\text{ cm}$
$\displaystyle l=\sqrt{r^2+h^2}$
$\displaystyle =\sqrt{10^2+24^2}=\sqrt{100+576}$
$\displaystyle =\sqrt{676}=26\text{ cm}$
$\displaystyle \therefore \text{Radius}=10\text{ cm and slant height}=26\text{ cm}$
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$\displaystyle \textbf{Question 5: }\text{Two right circular cones }X\text{ and }Y\text{ are made. Cone }X\text{ has three times}$
$\displaystyle \text{the radius of cone }Y, \ \text{and cone }Y\text{ has half the volume of cone }X.\text{ Find the ratio of their heights.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius of cone }Y=x$
$\displaystyle \therefore \text{Radius of cone }X=3x$
$\displaystyle \text{Let the heights of cones }X\text{ and }Y\text{ be }h_x\text{ and }h_y\text{ respectively.}$
$\displaystyle \text{Given, }V_y=\frac{1}{2}V_x$
$\displaystyle \frac{1}{3}\pi x^2h_y=\frac{1}{2}\left(\frac{1}{3}\pi(3x)^2h_x\right)$
$\displaystyle x^2h_y=\frac{1}{2}\times9x^2h_x$
$\displaystyle 2h_y=9h_x$
$\displaystyle \frac{h_x}{h_y}=\frac{2}{9}$
$\displaystyle \therefore h_x:h_y=2:9$
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$\displaystyle \textbf{Question 6: }\text{The diameters of two cones are equal. If their slant heights are} \\ \text{in the ratio } 5:4,\text{ find the ratio of their curved surface areas.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the common radius of the two cones be }r$
$\displaystyle \text{Let their slant heights be }l_1\text{ and }l_2\text{ respectively.}$
$\displaystyle \text{Given, }l_1:l_2=5:4$
$\displaystyle \text{Curved surface area of a cone}=\pi rl$
$\displaystyle \frac{\text{Curved surface area of Cone 1}}{\text{Curved surface area of Cone 2}}=\frac{\pi rl_1}{\pi rl_2}$
$\displaystyle =\frac{l_1}{l_2}=\frac{5}{4}$
$\displaystyle \therefore \text{The ratio of the curved surface areas is }5:4.$
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$\displaystyle \textbf{Question 7: }\text{There are two cones. The curved surface area of one is twice that of the other.}$
$\displaystyle \text{The slant height of the latter is twice that of the former. Find the ratio of their radii.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the curved surface areas of the two cones be }S_1\text{ and }S_2\text{ respectively.}$
$\displaystyle \text{Let their radii be }r_1,r_2\text{ and slant heights be }l_1,l_2\text{ respectively.}$
$\displaystyle \text{Given, }S_1=2S_2\text{ and }l_2=2l_1$
$\displaystyle \frac{S_1}{S_2}=\frac{\pi r_1l_1}{\pi r_2l_2}=2$
$\displaystyle \frac{r_1l_1}{r_2(2l_1)}=2$
$\displaystyle \frac{r_1}{2r_2}=2$
$\displaystyle \frac{r_1}{r_2}=4$
$\displaystyle \therefore r_1:r_2=4:1$
$\\$

$\displaystyle \textbf{Question 8: }\text{A heap of wheat is in the form of a cone of diameter }16.8\text{ m and height }$
$\displaystyle 3.5\text{ m}. \ \text{Find its volume. How much cloth is required to just cover the heap?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Diameter}=16.8\text{ m}$
$\displaystyle r=\frac{16.8}{2}=8.4\text{ m},\quad h=3.5\text{ m}$
$\displaystyle \text{Volume of the cone}=\frac{1}{3}\pi r^2h$
$\displaystyle =\frac{1}{3}\times\frac{22}{7}\times(8.4)^2\times3.5$
$\displaystyle =258.72\text{ m}^3$
$\displaystyle \therefore \text{Volume of the heap}=258.72\text{ m}^3$
$\displaystyle l=\sqrt{r^2+h^2}$
$\displaystyle =\sqrt{(8.4)^2+(3.5)^2}$
$\displaystyle =\sqrt{70.56+12.25}=\sqrt{82.81}=9.1\text{ m}$
$\displaystyle \text{Cloth required}=\text{curved surface area of the cone}$
$\displaystyle =\pi rl$
$\displaystyle =\frac{22}{7}\times8.4\times9.1$
$\displaystyle =240.24\text{ m}^2$
$\displaystyle \therefore \text{Cloth required}=240.24\text{ m}^2$
$\\$

$\displaystyle \textbf{Question 9: }\text{Find what length of canvas, }1.5\text{ m wide, is required to make a conical tent }$
$\displaystyle 48\text{ m in diameter} \ \text{and }7\text{ m high. If }10\%\text{ of the canvas is used in folds and} \\ \text{stitching, find the cost at Rs. }24\text{ per metre.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Diameter}=48\text{ m}\Rightarrow r=24\text{ m},\quad h=7\text{ m}$
$\displaystyle l=\sqrt{r^2+h^2}=\sqrt{24^2+7^2}=\sqrt{625}=25\text{ m}$
$\displaystyle \text{Curved surface area}=\pi rl=\frac{22}{7}\times24\times25=1885.714\text{ m}^2$
$\displaystyle \text{Length of }1.5\text{ m wide canvas required}=\frac{1885.714}{1.5}=1257.143\text{ m}$
$\displaystyle \text{Including }10\%\text{ extra for folds and stitching,}$
$\displaystyle \text{Total canvas length}=1257.143\times1.1=1382.857\text{ m}$
$\displaystyle \text{Cost of the canvas}=24\times1382.857=\text{Rs. }33188.57$
$\displaystyle \therefore \text{Required canvas length}=1382.857\text{ m and cost}=\text{Rs. }33188.57$
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$\displaystyle \textbf{Question 10: }\text{A solid cone of height }8\text{ cm and base radius }6\text{ cm is melted and}$
$\displaystyle \text{recast into identical cones, each of height }2\text{ cm and diameter 1 cm. Find the number} \\ \text{of cones formed.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the large cone, }r_1=6\text{ cm},\quad h_1=8\text{ cm}$
$\displaystyle \text{For each small cone, }h_2=2\text{ cm}$
$\displaystyle \text{Diameter}=1\text{ cm}\Rightarrow r_2=\frac{1}{2}=0.5\text{ cm}$
$\displaystyle \text{Let the number of small cones formed be }n$
$\displaystyle n=\frac{\text{Volume of the large cone}}{\text{Volume of one small cone}}$
$\displaystyle =\frac{\frac{1}{3}\pi r_1^2h_1}{\frac{1}{3}\pi r_2^2h_2}$
$\displaystyle =\frac{6^2\times8}{(0.5)^2\times2}$
$\displaystyle =\frac{36\times8}{0.25\times2}$
$\displaystyle =\frac{288}{0.5}=576$
$\displaystyle \therefore \text{The number of cones formed is }576.$
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$\displaystyle \textbf{Question 11: }\text{The total surface area of a right circular cone of slant height }13\text{ cm is }90\pi\text{ cm}^2.$
$\displaystyle \text{Calculate: (i) its radius in cm (ii) its volume in cm}^3.\text{ [Take }\pi=3.14\text{]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the radius be }r\text{ cm and slant height }l=13\text{ cm}$
$\displaystyle \text{Total surface area}=\pi r^2+\pi rl=90\pi$
$\displaystyle r^2+13r=90$
$\displaystyle r^2+18r-5r-90=0$
$\displaystyle r(r+18)-5(r+18)=0$
$\displaystyle (r-5)(r+18)=0$
$\displaystyle r=5\text{ cm}\quad(\because r>0)$
$\displaystyle \text{(i) Radius}=5\text{ cm}$
$\displaystyle h=\sqrt{l^2-r^2}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\text{ cm}$
$\displaystyle \text{Volume}=\frac{1}{3}\pi r^2h$
$\displaystyle =\frac{1}{3}\times3.14\times5^2\times12$
$\displaystyle =314\text{ cm}^3$
$\displaystyle \text{(ii) Volume}=314\text{ cm}^3$
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$\displaystyle \textbf{Question 12: }\text{The area of the base of a conical solid is }38.5\text{ cm}^2\text{ and its volume is }$
$\displaystyle 154\text{ cm}^3. \ \text{Find the curved surface area of the solid.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of the base}=\pi r^2=38.5\text{ cm}^2$
$\displaystyle r=\sqrt{\frac{38.5\times7}{22}}=3.5\text{ cm}$
$\displaystyle \text{Volume of the cone}=154\text{ cm}^3$
$\displaystyle \frac{1}{3}\pi r^2h=154$
$\displaystyle h=\frac{154\times3\times7}{22\times(3.5)^2}=12\text{ cm}$
$\displaystyle l=\sqrt{r^2+h^2}=\sqrt{(3.5)^2+12^2}$
$\displaystyle =\sqrt{12.25+144}=\sqrt{156.25}=12.5\text{ cm}$
$\displaystyle \text{Curved surface area}=\pi rl$
$\displaystyle =\frac{22}{7}\times3.5\times12.5=137.5\text{ cm}^2$
$\displaystyle \therefore \text{The curved surface area of the solid is }137.5\text{ cm}^2.$
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$\displaystyle \textbf{Question 13: }\text{A vessel in the form of an inverted cone is filled with water to the brim. Its height is }$
$\displaystyle 32\text{ cm} \ \text{and the diameter of its base is }25.2\text{ cm. Six equal solid cones are dropped into it}$
$\displaystyle \text{and are fully submerged. As a result, one-fourth of the water in the vessel overflows. Find the volume} \\ \text{of each solid cone.}$
$\displaystyle \text{Answer:}$
$\displaystyle h=32\text{ cm},\quad d=25.2\text{ cm}$
$\displaystyle r=\frac{25.2}{2}=12.6\text{ cm}$
$\displaystyle \text{Let the volume of each solid cone be }V\text{ cm}^3$
$\displaystyle \text{Volume of water overflowed}=\frac{1}{4}\times\text{Volume of the vessel}$
$\displaystyle 6V=\frac{1}{4}\left(\frac{1}{3}\pi r^2h\right)$
$\displaystyle 6V=\frac{1}{4}\left(\frac{1}{3}\times\frac{22}{7}\times(12.6)^2\times32\right)$
$\displaystyle 6V=1330.56$
$\displaystyle V=\frac{1330.56}{6}=221.76\text{ cm}^3$
$\displaystyle \therefore \text{The volume of each solid cone is }221.76\text{ cm}^3.$
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$\displaystyle \textbf{Question 14: }\text{The volume of a conical tent is }1232\text{ m}^3\text{ and the area of its base floor is }$
$\displaystyle 154\text{ m}^2. \ \text{Calculate: (i) the radius of the floor (ii) the height of the tent (iii) the length of canvas }$
$\displaystyle \text{required to cover the tent, if the width of the canvas is }2\text{ m}.\hfill \text{[ICSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Volume of the tent}=1232\text{ m}^3$
$\displaystyle \text{Area of the base}=154\text{ m}^2$
$\displaystyle \text{(i) }\pi r^2=154$
$\displaystyle r^2=\frac{154\times7}{22}=49$
$\displaystyle r=7\text{ m}$
$\displaystyle \therefore \text{Radius of the floor}=7\text{ m}$
$\displaystyle \text{(ii) }\frac{1}{3}\pi r^2h=1232$
$\displaystyle \frac{1}{3}\times\frac{22}{7}\times7^2\times h=1232$
$\displaystyle h=\frac{1232\times3}{\frac{22}{7}\times7^2}=24\text{ m}$
$\displaystyle \therefore \text{Height of the tent}=24\text{ m}$
$\displaystyle \text{(iii) }l=\sqrt{r^2+h^2}$
$\displaystyle =\sqrt{7^2+24^2}=\sqrt{49+576}=\sqrt{625}=25\text{ m}$
$\displaystyle \text{Curved surface area}=\pi rl$
$\displaystyle =\frac{22}{7}\times7\times25=550\text{ m}^2$
$\displaystyle \text{Length of canvas required}=\frac{\text{Area of canvas}}{\text{Width of canvas}}$
$\displaystyle =\frac{550}{2}=275\text{ m}$
$\displaystyle \therefore \text{Length of canvas required}=275\text{ m}$
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