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Ex. 1 |
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Ans. |
 \begin{align*}
& \textbf {Given that -} \\
& \text {Diameter of Hemisphere, Let } d = 3.5 \text {cm}\\
& \text {Radius of Hemisphere, Let r = }\frac {d}{2}=\frac {3.5}{2}=1.75 \text { cm}\\
& \text {Height of Cone, Let } h \\
&\text{ = Height of top }-\text {Radius of hemisphere}\\
& 5-1.75 = 3.25 cm\\
& \textbf{Solution : }\\
& \text{We know that -} \\
& \text{TSA of the Lattu} = \text{CSA of hemisphere} + \text{CSA of cone}\\
& \textbf{So, CSA of hemisphere:}\\
& = \frac {1}{4}(4\pi r^{2}) = 2\pi r^{2}\\
& \text {Putting value -}\\
& = 2\pi r^{2} = 2\times \frac {22}{7}\times 1.75 \times 1.75\\
& = 2\pi r^{2} = 2\times \frac {22}{7}\times 3.0625\\
& = 19.25 \text { cm}^{2}\\
& \textbf{CSA of Cone:}\\
& = \pi rl\\
& \text {So, First we find } l \\
& \text {We know that -}\\
& l^{2} = h^{2}+r^{2}\\
& \text {Putting value -}\\
& l^{2} = (3.25)^{2}+(1.75)^{2}\\
& l^{2} = 10.56 + 3.0625 \\
& l^{2} = 13.6225 \\
& l = \sqrt {13.6225} \\
& l = 3.69 \text { cm}\\
& \textbf{So, CSA of Cone =} \frac {22}{7}\times 1.75\times 6.39\\
& = 19.8 \text { cm}^{2}\\
& \text {Hence TSA = } 19.25 + 19.8 = 39.05 \text { cm}^{2}
\end{align*}
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Ex. 2 | The decorative block shown in Fig. 12.7 is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Take \(\pi = \frac {22}{7}\))  |
Ans. | \begin{align*}
& \textbf {Given that -} \\
& \text {Side of square} = 5 \text { cm}\\
& \text {Diameter of Hemisphere} = 4.2 \text { cm}\\
& \text {So, Radius }= \frac {Diameter}{2}=\frac {4.2}{2}=2.1 \text { cm}\\
& \textbf {Solution :}\\
& \text {TSA of given figure =}\\
& \text {TSA of Cube } +\text {CSA of Hemisphere }\\
& - \text {Base area of hemisphere}\\
& \textbf {TSA of Cube } = 6 (side)^{2}\\
& \text {Putting value}\\
& = 6\times 5\times 5 = 150 \text { cm}^{2}\\
& \textbf {CSA of Hemisphere } = 2\pi r^{2}\\
& \text {Putting value}\\
& = 2\times \frac {22}{7} \times (2.1)^{2} = 27.72 \text { cm}^{2}\\
& \textbf {Base area of hemisphere :}\\
& \text {Base area of hemisphere is a circle with 2.1 cm radius }\\
& = \pi r^{2} = \frac {22}{7}\times 2.1\times 2.1\\
& = 22\times 0.3\times 2.1\\
& = 13.86 \text {cm}^{2}\\
& \text {So,}\\
& TSA = 150 + 27.72 - 13.86 = 163.86 \text {cm}^{2}\\
\end{align*}
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Ex. 3 | A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in Fig. 12.8. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take \(\pi = 3.14\))  |
Ans. | \begin{align*}
\text{Given that:}
& \text{Radius of Cone} = 2.5 \text{ cm} = \text{Let } r \\
& \text{Slant height of cone} = \text {Let, } l \\
& \text{Height of cone} = 6 \text{ cm} = \text{Let, } h \\
& \text{Radius of cylinder} = 1.5 \text{ cm} = \text{Let } r' \\
& \text{Height of cylinder} = 26 - 6 = 20 \text{ cm} = \text{Let } h' \\
& \text{First of all we find } l, \\
& l^{2} = r^{2} + h^{2} \\
& l = \sqrt{r^{2} + h^{2}} \\
& \text{Putting value} \\
& l = \sqrt{2.5^{2} + 6^{2}} \\
& l = \sqrt{6.25 + 36} \\
& l = \sqrt{42.25} \\
& l = 6.5 \text{ cm} \\
& \text{Area to be printed in Orange} =\\
& \text{CSA of the cone} + \text{Base area of cone} - \text{Base area of cylinder} \\
& = \pi rl + \pi r^{2} - \pi (r')^{2} \\
& = \pi [rl + r^{2} - (r')^{2}] \\
& \text{Putting value} \\
& = 3.14 [(2.5 \times 6.5) + (2.5 \times 2.5) - (1.5 \times 1.5)] \\
& = 3.14 (16.25 + 6.25 - 2.25) \\
& = 3.14 \times 20.25 \\
& = 63.585 \text{ cm}^{2} \\
& \text{Area to be printed in Yellow} =\\
& \text{CSA of the cylinder} + \text{Area of one base of cylinder} \\
& = 2 \pi r'h' + \pi (r')^{2} \\
& = \pi r' (2h' + r') \\
& \text{Putting value} \\
& = (3.14 \times 1.5)(2 \times 20 + 1.5) \\
& = 4.71 \times 41.5 \text{ cm}^{2} \\
& = 195.465 \text{ cm}^{2}
\end{align*}
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Ex. 4 | Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see Fig. 12.9). The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (Take \(\pi = \frac {22}{7}\))  |
Ans. | \begin{align*}
\text{Given that:} \\
& \text{Height of cylinder} = 1.45 \text{ m (Let } h) \\
& \text{Radius of cylinder} = 30 \text{ cm} = \frac{30}{100} = 0.30 \text{ m (Let } r) \\
& \text{Radius of cylinder} = \text{Radius of Hemisphere} \\
& \text{TSA of the bird-bath} = \text{CSA of cylinder} + \text{CSA of hemisphere} \\
& \Rightarrow 2\pi rh + 2\pi r^2 \\
& \Rightarrow 2\pi r(h + r) \\
& \text{Putting value} \\
& \Rightarrow 2 \times \frac{22}{7} \times 0.30 (1.45 + 0.30) \\
& \Rightarrow 2 \times \frac{22}{7} \times 0.30 \times 1.75 \\
& \Rightarrow 2 \times \frac{22}{7} \times 0.525 \\
& \Rightarrow \frac{44}{7} \times 0.525 = 3.3 \text{ m}^2 \\
\end{align*}
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Ex. 5 | Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see Fig. 12.12). If the base of the shed is of dimension 7 m × 15 m, and the height of the cuboidal portion is 8 m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of \(300 \text { m}^{3}\), and there are 20 workers, each of whom occupy about \(0.08 \text { m}^{3}\) space on an average. Then, how much air is in the shed? (Take \(\pi = \frac {22}{7}\))  |
Ans. | \begin{align*} \text{Given that:}
& \text{Length of cuboidal part} = 15 \text{ m} \\
& \text{Breadth of cuboidal part} = 7 \text{ m} \\
& \text{Height of cuboidal part} = 8 \text{ m} \\
& \text{Radius of half-cylinderical part} = \frac {7}{2} \text{ m} \\
& \text{Space occupy by each worker} = 0.08 \text{ m}^{3} \\
& \text {Number of worker } = 20\\
& \text {Total space occupy by machinery } = 300 \text { m}^{3}\\
& \text{Let, Height of half-cylinder} = h \text{ m} \\
\text {Solution :}
& \text {Volume of cuboidal part} = 15\times 7 \times 8 = 840 \text { m}^{3}\\
& \text {Volume of half cylinder }= \frac {1}{2}\pi r^{2} h\\
& \text {Putting value -}\\
& \Rightarrow \frac {1}{2}\times \frac {22}{7}\times \left(\frac {7}{2}\right)^{2}\times 15 \\
& \Rightarrow \frac {1}{2}\times \frac {22}{7}\times \frac {49}{4}\times 15\\
& \Rightarrow \frac {11\times 7\times 15}{4}\\
& \Rightarrow \frac {1155}{4}\\
& \Rightarrow 288.75 \text {m}^{3}\\
& \text {So, Volume of inside the shed }\\
& \text {When no worker or machine inside it} \\
& \Rightarrow (840 + 288.75)=1128.75 \text { m}^{3}\\
& \text {Total space occupy by 20 worker }= 20\times 0.08 = 1.6 \text { m}^{3}\\
& \text {Volume of the air inside the shed}\\
& \text {when there are machine and wrokders inside it}\\
& \Rightarrow 1128.75-1.6-300 = 827.15 \text { m}^{3}
\end{align*} |
Ex. 6 | A juice seller was serving his customers using glasses as shown in Fig. 12.13. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent capacity of the glass and its actual capacity. \(Use \pi = 3.14\)  |
Ans. | \begin{align*} \text{Given that:}
& \text {Diameter of cylinderical glass }=5 \text { cm}\\
& \text {Radius of cylinderical glass }=\frac {D}{2}=\frac {5}{2} \text { cm}\\
& \text {Height of cylinderical glass }=10 \text { cm}\\
\text {Solution :}
& \text{As per given image:}\\
& \text {Apparent capacity of the glass} = \text {Volume of cylinder}=\pi r^{2}h\\
& \text {Putting value -}\\
& \Rightarrow 3.14\times \left(\frac {5}{2}\right)^{2}\times 10\\
& \Rightarrow 3.14\times \frac {25}{4}\times 10\\
& \Rightarrow 3.14\times 6.25\times 10\\
& \Rightarrow 196.25 \text { cm}^{3}\\
& \text {As per given figure}\\
& \text {Acual capacity of glass} = \text {Volume of cylinder}-\text {Volume of hemisphere}\\
& \text {We know that volume of hemisphere }= \frac {2}{3}\pi r^{3}\\
& \text {Putting value -}\\\
& \Rightarrow \frac {2}{3}\times 3.14 \left(\frac {5}{2}\right)^{3}\\
& \Rightarrow \frac {2}{3}\times 3.14 \frac {125}{8}\\
& \Rightarrow \frac {2}{3}\times 3.14 \times 15.625\\
& \Rightarrow 32.7 \text { cm}^{3}\\
& \therefore \text {Actual capacity } = 196.25-32.7 = 163.55 \text { cm}^{3}
\end{align*} |
Ex. 7. | A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. \(Use \pi = 3.14\)  |
Ans. | \begin{align*} \text{Given that:}
& \text {Height of Cone }=OA = 2 \text { cm}, Let h\\
& \text {Diameter of Cone }=BC = 4 \text { cm}\\
& \text {So, Radus }=\frac {D}{2}=\frac {4}{2}=2 \text { cm}\\
\text {Solution :}
& \text {Volume of toy }= \text {Volume of cone}+\text {Volume of Hemisphere}\\
& \Rightarrow \text {Volume of toy }=\frac {1}{3}\pi r^{2}h + \frac {2}{3}\pi r^{3}\\
& \text {putting value}\\
& \Rightarrow \left( \frac {1}{3}\times 3.14 \times (2)^{2} \times (2)\right) + \left( \frac {2}{3}\times 3.14\times (2)^{3}\right) \\
& \Rightarrow \left(\frac {1\times 3.14\times 4 \times 2}{3}\right)+\left(\frac {2\times3.14\times8}{3}\right)\\
& \Rightarrow \frac {25.12}{3}+\frac {50.24}{3}\\
& \Rightarrow \frac {25.12+50.24}{3}\\
& \Rightarrow \frac {75.36}{3}\\
& \Rightarrow 25.12 \text { cm}^{3}\\
& \text {Now, We need to find the difference}\\
& \text {of volumes between cylinder and toy}\\
& \text {As per given : cylinder circumscribes the toy, so}\\
& \text {Diameter of cylinder } = GH = CB = 4\text { cm}\\
& \text {So, Radius } = \frac {4}{2}=2\text{ cm}\\
& \text {And, Height of cylinder} = AO + PO \\
& = \text {Height of cone + Radius of Hemisphere}\\
& \Rightarrow 2+2 = 4 \text { cm}\\
& \text {Now, Volume of cylinder }= \pi r^{2}h\\
& \text {Putting value -}\\
& = 3.14 \times (2)^{2}\times (4)\\
& = 3.14 \times 4\times 4\\
& = 3.14 \times 16\\
& = 50.24 \text { cm}^{2}\\
& \text {Difference in volume }= \text {volume of cylinder}-\text {volume of toy}\\
& \text {Difference in volume }=50.24 - 25.12 = 25.12 \text { cm}^{2}
\end{align*} |
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