Class 10: Cone and Sphere – Exercise 22(a) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Notes: Important formulas to be kept in mind:

Parameters of a Cone: Radius of the base ( $\displaystyle r$ ), Height of the cone ( $\displaystyle h$ ) and Slant Height of a Cone ( $\displaystyle l$ )

$\displaystyle \text{Volume of a Cone } = \frac{1}{3} \pi r^2 h$

Curved Surface area of a Cone $\displaystyle = \pi r l$

Total Surface area of a Cone $\displaystyle = \pi r^2+ \pi r l$

Question 1: Find the volume of a cone whose slant height is $\displaystyle 17 \text{ cm }$ and radius of base is $\displaystyle 8 \text{ cm } .$

Answer:

$\displaystyle \text{Volume of a Cone } = \frac{1}{3} \pi r^2 h$

$\displaystyle l = 17 \text{ cm,} r = 8 \text{ cm }$

$\displaystyle \text{Therefore } h = \sqrt{l^2 - r^2} = \sqrt{17^2-8^2} = 15 \text{ cm }$

$\displaystyle \text{Therefore Volume } = \frac{1}{3} \times \frac{22}{7} \times (8)^2 \times (15) = 1005.71 \text{ cm}^3$

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Question 2: The curved surface area of a cone is $\displaystyle 12320 \text{ cm}^2 .$ If the radius of its base is $\displaystyle 56 \text{ cm } ,$ find its height.

Answer:

$\displaystyle \text{Curved surface area of the cone } = 12320 \text{ cm}^2$

$\displaystyle r = 56 \text{ cm }$

$\displaystyle \pi r l = 12320 \Rightarrow l = \frac{12320}{\pi . 56} = 70 \text{ cm }$

$\displaystyle \text{Therefore } h = \sqrt{l^2 - r^2} = \sqrt{70^2-56^2} = 1542 \text{ cm }$

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Question 3: The circumference of the base of a $\displaystyle 12\text{ m }$ high conical tent is $\displaystyle 66\text{ m}.$ Find the volume of the air contained in it.

Answer:

$\displaystyle h = 12\text{ m }$

$\displaystyle \text{Circumference of the base } = 66\text{ m }$

$\displaystyle \text{Therefore } 2 \pi r = 66 \Rightarrow r = \frac{33}{\pi}$

$\displaystyle \text{Volume of a Cone } = \frac{1}{3} \pi r^2 h$

$\displaystyle = \frac{1}{3} \times \pi \times (\frac{33}{\pi})^2 \times (12) = 1386\text{ m}^3$

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Question 4: The radius and the height of a right circular cone are in the ratio $\displaystyle 5:12$ and its volume is $\displaystyle 2512 \text{ cm}^3 .$ Find the radius and slant height of the cone. (Take $\displaystyle \pi = 3.14$ )

Answer:

Given: Radius and the height of a right circular cone are in the ratio $\displaystyle 5:12$

$\displaystyle \text{Let } r = 5x$ and $\displaystyle h = 12 x$

$\displaystyle \text{Volume } = 2512 \text{ cm}^3$

$\displaystyle \text{Therefore } \frac{1}{3} \pi r^2 h = 2512$

$\displaystyle \Rightarrow \frac{1}{3} \times 3.14 \times (5x)^2 \times (12x) = 2512$

$\displaystyle \Rightarrow x^3 = 8\text{ or } x = 2$

$\displaystyle \text{Therefore Radius } = 5 (2) = 10 \text{ cm } \text{ and } \text{Height } = 12(2) = 24 \text{ cm }$

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Question 5: Two right circular cones $\displaystyle X$ and, $\displaystyle Y$ are made. $\displaystyle X$ having three times the radius of $\displaystyle Y$ and $\displaystyle Y$ having half the volume of $\displaystyle X .$ Calculate the ratio between the heights of $\displaystyle X$ and $\displaystyle Y .$

Answer:

$\displaystyle \text{Let the radius of cone Y is } x \text{ Therefore the radius of cone X } = 3x$

$\displaystyle \text{Let the height of cone Y } = h_y \text{ and the height of cone X } = h_x$

$\displaystyle \text{Given: } V_y = \frac{1}{2} V_x$

$\displaystyle \Rightarrow \frac{1}{3} \pi (x)^2 (h_y) = \frac{1}{2} \{ \frac{1}{3} \pi (3x)^2 (h_x) \}$

$\displaystyle \Rightarrow \frac{h_x}{h_y} = \frac{x^2}{(3x)^2}$

$\displaystyle \Rightarrow \frac{h_x}{h_y} = \frac{2}{9}$

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Question 6: The diameters of two cones are equal. If their slant heights are in the ratio $\displaystyle 5 : 4 ,$ find the ratio of their curved surface areas.

Answer:

Let the radius of Cone 1 and Cone 2 $\displaystyle = x$

Let the slant height of the cones be $\displaystyle l_1$ and $\displaystyle l_2$ respectively.

Curved Surface Area of a cone $\displaystyle = \pi r l$

Therefore

$\displaystyle \frac{\text{Curved Surface Area of Cone 1}}{\text{Curved Surface Area of Cone 2}} = \frac{\pi \times x \times l_1}{\pi \times x \times l_2} = \frac{5}{4}$

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Question 7: There are two cones. The curved surface area of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of their radii.

Answer:

Let the curved surface area of the cones be $\displaystyle S_1$ and $\displaystyle S_2$ respectively.

Similarly, let the slant height of the two cones be $\displaystyle l_1$ and $\displaystyle l_2$ respectively.

Similarly, let the radius of the two cones be $\displaystyle r_1$ and $\displaystyle r_2$ respectively.

$\displaystyle \text{Given: } S_1 = 2 S_2$

$\displaystyle 2 l_1 = l_2$

$\displaystyle \text{Therefore } \frac{\pi . r_1. l_1}{\pi .r_2.l_2} = 2$

$\displaystyle \Rightarrow \frac{r_1}{r_2} = \frac{4}{1}\text{ or } r_1:r_2= 4:1$

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Question 8: A heap of wheat is in the form of a cone of $\displaystyle \text{Diameter } 16.8\text{ m }$ and $\displaystyle \text{Height } 3.5\text{ m }$ . Find its volume. How much cloth is required to just cover the heap?

Answer:

$\displaystyle \text{Diameter } = 16.8\text{ m }$

Radius $\displaystyle = r = 8.4\text{ m }$

$\displaystyle \text{Height } = h = 3.5\text{ m }$

$\displaystyle \text{Volume of a Cone } = \frac{1}{3} \pi r^2 h$

$\displaystyle = \frac{1}{3} \times \pi \times (8.4)^2 \times (3.5) = 258.72\text{ m}^3$

Curved Surface area of a Cone $\displaystyle = \pi r l$

$\displaystyle = \pi \times (8.4) \times (\sqrt{3.5^2+8.4^2} = 240.02\text{ m}^2$

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Question 9: Find what length of canvas, $\displaystyle 1.5\text{ m }$ in width, is required to make a conical tent $\displaystyle 48\text{ m }$ in diameter and $\displaystyle 7\text{ m }$ in height. Given that $\displaystyle 10\%$ of the canvas is used in folds and stitching. Also, find the cost of the canvas at the rate of $\displaystyle \text{ Rs. } 24$ per meter.

Answer:

$\displaystyle \text{Diameter } = 48 \text{ m } \Rightarrow \text{ Radius } (r) = 24\text{ m }$

$\displaystyle \text{Height } (h) = 7\text{ m }$

$\displaystyle \text{Curved Surface Area } = \pi r l = \frac{22}{7} \times 24 \times \sqrt{7^2+24^2} = 1885.714\text{ m}^2$

$\displaystyle \text{Therefore the length of the canvas needed } = \frac{1885.714}{1.5} = 1257.143\text{ m }$

However, $\displaystyle 10\%$ of the canvas is consumed in folds and stitches. Hence the length needed $\displaystyle = 1257.143 \times 1.1 = 1382.85\text{ m }$

The total cost of the canvas $\displaystyle = 24 \times 1257.143 = 33188.57 \text{ Rs. }$

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Question 10: A solid cone of $\displaystyle \text{Height } 8 \text{ cm }$ and base radius $\displaystyle 6 \text{ cm }$ is melted and recast into identical cones, each of $\displaystyle \text{Height } 2 \text{ cm }$ and $\displaystyle \text{Diameter } 1 \text{ cm } .$ Find the number of cones formed.

Answer:

Large Cone: $\displaystyle \text{ Height } (h_1) = 8 cm , \text{ Radius } (r_1) = 6 \text{ cm }$

Small Cones: $\displaystyle \text{ Height } (h_2) = 2 cm , \text{ Radius } (r_2) = 0.5 \text{ cm }$

$\displaystyle \text{Therefore the number of cones made } = \frac{\frac{1}{3} \pi r_1^2 h_1}{\frac{1}{3} \pi r_2^2 h_2}$

$\displaystyle = \frac{\frac{1}{3} \times \pi \times \times (6)^2 \times (8)}{\frac{1}{3} \times \pi \times (0.5)^2 \times (2)} = \frac{36 \times 4}{0.25} = 576$

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Question 11: The total surface area of a right circular cone of slant $\displaystyle \text{Height } 13 \text{ cm }$ is $\displaystyle 90 \pi \text{ cm}^2 .$ calculate: (i) its radius in $\displaystyle \text{ cm } .$ (ii) its volume in $\displaystyle \text{ cm}^3 .$ [Take $\displaystyle \pi = 3.14$ ]

Answer:

Slant length $\displaystyle (l) = 13 \text{ cm }$

Let the Radius $\displaystyle = r \text{ cm }$

Total Surface Area $\displaystyle = 90 \pi \text{ cm}^2$

$\displaystyle \text{Therefore } 90 \pi = \pi (r)2 + \pi r . (13)$

$\displaystyle \Rightarrow 90 = r^2+13 r$

$\displaystyle \Rightarrow r^2 + 18 r - 5 r -90= 0$

$\displaystyle \Rightarrow r(r+18) -5(r+18) = 0$

$\displaystyle \Rightarrow (r-5)(r+18) = 0$

$\displaystyle \Rightarrow r = 5\text{ as } r = -18$ is not possible.

$\displaystyle \text{Hence Height } (h) = \sqrt{13^2-5^2} = 12 \text{ cm }$

$\displaystyle \text{Volume } = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times (5)^2 \times (12) = 314 \text{ cm}^3$

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Question 12: The area of the base of a conical solid is $\displaystyle 38.5 \text{ cm}^2$ and its volume is $\displaystyle 154 \text{ cm}^3 .$ Find the curved surface area of the solid.

Answer:

$\displaystyle \text{Area of the base } = 38.5 \text{ cm}^2$

Volume of the cone $\displaystyle = 154 \text{ cm}^2$

$\displaystyle \text{Therefore } \pi r^2 = 38.5 \Rightarrow r = \sqrt{\frac{38.5 \times 7}{22}} = 3.5 \text{ cm }$

$\displaystyle \frac{1}{3} \pi r^2 h = 154$

$\displaystyle \Rightarrow h = \frac{154 \times 3 \times 7}{22 \times 3.5^2} = 12 \text{ cm }$

$\displaystyle \text{Therefore Slant Length } (l) = \sqrt{12^2+3.5^2} = 12.5 \text{ cm }$

$\displaystyle \text{Curved Surface Area } = \pi rl = \frac{22}{7} \times 3.5 \times 12.5 = 137.5 \text{ cm}^2$

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Question 13: A vessel in the form of an inverted cone, is filled with water to the brim. Its height is $\displaystyle 32 \text{ cm }$ and diameter of the base is $\displaystyle 25.2 \text{ cm } .$ Six equal solid cones are dropped in it, so that they are fully submerged. As a result. one-fourth of water in the original cone overflows. what is the volume of each of the solid cones submerged?

Answer:

$\displaystyle \text{Height } (h) = 32 \text{ cm }$

$\displaystyle \text{Diameter } (d) = 25.2 \text{ cm }$

$\displaystyle \text{Therefore Radius } (r) = 12.6 \text{ cm }$

Let the volume of each of the solid cone be $\displaystyle V_1$

$\displaystyle \text{Therefore } 6 \times V_1 = \frac{1}{4} \{ \frac{1}{3} \times \pi \times (12.6)^2 \times (32) \}$

$\displaystyle \Rightarrow V_1 = 221.76 \text{ cm}^3$

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Question 14: The volume of a conical tent is $\displaystyle 1232\text{ m}^3$ and the area of the base floor is $\displaystyle 154\text{ m}^2$ . Calculate the: (i) radius of the floor (ii) height of the tent (iii) length of the canvas required to cover this conical tent if its width is $\displaystyle 2\text{ m }$ . [2008]