EXERCISE 12.2
Areas Related To Circles • 3 Questions
Question 1
Hint available
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.) Fig. 12.14 170
Key Idea
The volume of a composite solid is obtained by adding the volumes of its individual parts. Use the standard formulas: \(V_{cylinder}=\pi r^{2}h\) and \(V_{cone}=\frac{1}{3}\pi r^{2}h\).
Step-by-Step Solution
1. Identify the dimensions
- Diameter = 3 cm \(\Rightarrow\) radius \(r = \frac{3}{2}=1.5\) cm.
- Height of the cylindrical part = 12 cm.
- Height of each cone = 2 cm.
2. Volume of the cylindrical part
\[V_{cyl}=\pi r^{2}h = \pi (1.5)^{2}(12) = \pi \times 2.25 \times 12 = 27\pi\ \text{cm}^{3}.\]
3. Volume of one cone
\[V_{cone}=\frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi (1.5)^{2}(2) = \frac{1}{3}\pi \times 2.25 \times 2 = 1.5\pi\ \text{cm}^{3}.\]
4. Volume of two cones
\[V_{2\,cones}=2\times 1.5\pi = 3\pi\ \text{cm}^{3}.\]
5. Total volume of the model
\[V_{total}=V_{cyl}+V_{2\,cones}=27\pi+3\pi=30\pi\ \text{cm}^{3}.\]
6. Numerical value (optional)
Using \(\pi \approx 3.14\) or \(\frac{22}{7}\):
\[V_{total}\approx 30\times 3.14 = 94.2\ \text{cm}^{3}\] (or \(\frac{660}{7}\approx 94.3\ \text{cm}^{3}\)).
- Diameter = 3 cm \(\Rightarrow\) radius \(r = \frac{3}{2}=1.5\) cm.
- Height of the cylindrical part = 12 cm.
- Height of each cone = 2 cm.
2. Volume of the cylindrical part
\[V_{cyl}=\pi r^{2}h = \pi (1.5)^{2}(12) = \pi \times 2.25 \times 12 = 27\pi\ \text{cm}^{3}.\]
3. Volume of one cone
\[V_{cone}=\frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi (1.5)^{2}(2) = \frac{1}{3}\pi \times 2.25 \times 2 = 1.5\pi\ \text{cm}^{3}.\]
4. Volume of two cones
\[V_{2\,cones}=2\times 1.5\pi = 3\pi\ \text{cm}^{3}.\]
5. Total volume of the model
\[V_{total}=V_{cyl}+V_{2\,cones}=27\pi+3\pi=30\pi\ \text{cm}^{3}.\]
6. Numerical value (optional)
Using \(\pi \approx 3.14\) or \(\frac{22}{7}\):
\[V_{total}\approx 30\times 3.14 = 94.2\ \text{cm}^{3}\] (or \(\frac{660}{7}\approx 94.3\ \text{cm}^{3}\)).
Question 2
Hint available
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 12.15).
Key Idea
The jamun is a composite solid (cylinder + two hemispheres). Find its total volume using the formulae for volume of a cylinder $V_{cyl}=\pi r^{2}h$ and volume of a sphere $V_{sphere}=\frac{4}{3}\pi r^{3}$ (two hemispheres make a sphere). Then take 30% of this volume for syrup and multiply by 45.
Step-by-Step Solution
1. Identify dimensions\
- Diameter $d = 2.8\,\text{cm}$ \=> radius $r = \frac{d}{2}=1.4\,\text{cm}$.\
- Total length of the jamun $L = 5\,\text{cm}$.\
- Length of the cylindrical part $h = L - d = 5-2.8 = 2.2\,\text{cm}$ (because the two hemispherical ends together contribute a length equal to the diameter).\
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2. Volume of the cylindrical part\
$$V_{cyl}=\pi r^{2}h = \pi (1.4)^{2}(2.2) = \pi \times 1.96 \times 2.2 = \pi \times 4.312 \approx 3.1416 \times 4.312 \approx 13.55\,\text{cm}^{3}.$$\
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3. Volume of the spherical part (two hemispheres form a sphere)\
$$V_{sphere}=\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi (1.4)^{3}=\frac{4}{3}\pi \times 2.744 \approx \frac{10.976}{3}\pi \approx 3.6587\pi \approx 11.49\,\text{cm}^{3}.$$\
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4. Total volume of one gulab jamun\
$$V_{jamun}=V_{cyl}+V_{sphere}=13.55+11.49 \approx 25.04\,\text{cm}^{3}.$$\
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5. Volume of sugar syrup in one jamun (30% of total volume)\
$$V_{syrup\,per\,jamun}=0.30\times V_{jamun}=0.30\times 25.04 \approx 7.512\,\text{cm}^{3}.$$\
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6. Total syrup in 45 jamuns\
$$V_{syrup\,total}=45\times 7.512 \approx 338.04\,\text{cm}^{3}.$$\
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7. Result\
Since $1\,\text{cm}^{3}=1\,\text{mL}$, the amount of syrup is approximately $338\,\text{mL}$ (or $\approx 338\,\text{cm}^{3}$).
- Diameter $d = 2.8\,\text{cm}$ \=> radius $r = \frac{d}{2}=1.4\,\text{cm}$.\
- Total length of the jamun $L = 5\,\text{cm}$.\
- Length of the cylindrical part $h = L - d = 5-2.8 = 2.2\,\text{cm}$ (because the two hemispherical ends together contribute a length equal to the diameter).\
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2. Volume of the cylindrical part\
$$V_{cyl}=\pi r^{2}h = \pi (1.4)^{2}(2.2) = \pi \times 1.96 \times 2.2 = \pi \times 4.312 \approx 3.1416 \times 4.312 \approx 13.55\,\text{cm}^{3}.$$\
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3. Volume of the spherical part (two hemispheres form a sphere)\
$$V_{sphere}=\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi (1.4)^{3}=\frac{4}{3}\pi \times 2.744 \approx \frac{10.976}{3}\pi \approx 3.6587\pi \approx 11.49\,\text{cm}^{3}.$$\
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4. Total volume of one gulab jamun\
$$V_{jamun}=V_{cyl}+V_{sphere}=13.55+11.49 \approx 25.04\,\text{cm}^{3}.$$\
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5. Volume of sugar syrup in one jamun (30% of total volume)\
$$V_{syrup\,per\,jamun}=0.30\times V_{jamun}=0.30\times 25.04 \approx 7.512\,\text{cm}^{3}.$$\
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6. Total syrup in 45 jamuns\
$$V_{syrup\,total}=45\times 7.512 \approx 338.04\,\text{cm}^{3}.$$\
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7. Result\
Since $1\,\text{cm}^{3}=1\,\text{mL}$, the amount of syrup is approximately $338\,\text{mL}$ (or $\approx 338\,\text{cm}^{3}$).
Question 3
Hint available
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by
Key Idea
Total surface area = (surface area of the cuboid) – (area of the four circular openings) + (lateral surface area of the four cones). Use the formulae: Surface area of cuboid = 2(lb + bh + hl), area of a circle = πr², slant height of a cone l = √(r² + h²), lateral surface area of a cone = πrl.
Step-by-Step Solution
1. Surface area of the cuboid\
\[\text{SA}_{\text{cuboid}} = 2(lb + bh + hl)\]\
where \(l = 15\,\text{cm},\; b = 10\,\text{cm},\; h = 8\,\text{cm}.\
\[\text{SA}_{\text{cuboid}} = 2(15\times10 + 10\times8 + 8\times15)\]
\[= 2(150 + 80 + 120) = 2\times350 = 700\ \text{cm}^2.\
2. Area removed due to the four circular openings\
Radius of each circle \(r = 2\,\text{cm}.\
\[\text{Area of one circle} = \pi r^2 = \pi \times 2^2 = 4\pi\]\
For four circles: \[\text{Area}_{\text{4 circles}} = 4 \times 4\pi = 16\pi\ \text{cm}^2.\
3. Lateral surface area of one conical depression\
Depth (height) of cone \(h_c = 5\,\text{cm}.\
Slant height \(l_c = \sqrt{r^2 + h_c^2} = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}\,\text{cm}.\
Lateral area of one cone \[\text{LA}_{\text{cone}} = \pi r l_c = \pi \times 2 \times \sqrt{29} = 2\pi\sqrt{29}\ \text{cm}^2.\
4. Lateral surface area of the four cones\
\[\text{LA}_{\text{4 cones}} = 4 \times 2\pi\sqrt{29} = 8\pi\sqrt{29}\ \text{cm}^2.\
5. Total surface area of the pen stand\
\[\text{Total SA} = \text{SA}_{\text{cuboid}} - \text{Area}_{\text{4 circles}} + \text{LA}_{\text{4 cones}}\]
\[= 700 - 16\pi + 8\pi\sqrt{29}\ \text{cm}^2.\
For a numerical value (using \(\pi \approx 3.14\)):\
\[16\pi \approx 50.24, \quad 8\pi\sqrt{29} \approx 135.23\]\
\[\text{Total SA} \approx 700 - 50.24 + 135.23 \approx 785\ \text{cm}^2.\
Hence, the total surface area is \(700 - 16\pi + 8\pi\sqrt{29}\ \text{cm}^2\) (≈ 785 cm²).
\[\text{SA}_{\text{cuboid}} = 2(lb + bh + hl)\]\
where \(l = 15\,\text{cm},\; b = 10\,\text{cm},\; h = 8\,\text{cm}.\
\[\text{SA}_{\text{cuboid}} = 2(15\times10 + 10\times8 + 8\times15)\]
\[= 2(150 + 80 + 120) = 2\times350 = 700\ \text{cm}^2.\
2. Area removed due to the four circular openings\
Radius of each circle \(r = 2\,\text{cm}.\
\[\text{Area of one circle} = \pi r^2 = \pi \times 2^2 = 4\pi\]\
For four circles: \[\text{Area}_{\text{4 circles}} = 4 \times 4\pi = 16\pi\ \text{cm}^2.\
3. Lateral surface area of one conical depression\
Depth (height) of cone \(h_c = 5\,\text{cm}.\
Slant height \(l_c = \sqrt{r^2 + h_c^2} = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}\,\text{cm}.\
Lateral area of one cone \[\text{LA}_{\text{cone}} = \pi r l_c = \pi \times 2 \times \sqrt{29} = 2\pi\sqrt{29}\ \text{cm}^2.\
4. Lateral surface area of the four cones\
\[\text{LA}_{\text{4 cones}} = 4 \times 2\pi\sqrt{29} = 8\pi\sqrt{29}\ \text{cm}^2.\
5. Total surface area of the pen stand\
\[\text{Total SA} = \text{SA}_{\text{cuboid}} - \text{Area}_{\text{4 circles}} + \text{LA}_{\text{4 cones}}\]
\[= 700 - 16\pi + 8\pi\sqrt{29}\ \text{cm}^2.\
For a numerical value (using \(\pi \approx 3.14\)):\
\[16\pi \approx 50.24, \quad 8\pi\sqrt{29} \approx 135.23\]\
\[\text{Total SA} \approx 700 - 50.24 + 135.23 \approx 785\ \text{cm}^2.\
Hence, the total surface area is \(700 - 16\pi + 8\pi\sqrt{29}\ \text{cm}^2\) (≈ 785 cm²).