EXAMPLES
Areas Related To Circles • 7 Questions
Question 1
Hint available
Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere (see Fig 12.6). The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take = 22 7 )
Key Idea
The coloured area is the sum of the curved surface area of the cone and the curved surface area of the hemisphere. Use the formulas: \(\text{CSA of cone}=\pi r l\) where \(l=\sqrt{r^{2}+h^{2}}\), and \(\text{CSA of hemisphere}=2\pi r^{2}\). The cone height is obtained by subtracting the hemisphere radius from the total height.
Step-by-Step Solution
1. Given data
- Total height of the top, \(H = 5\,\text{cm}\)
- Diameter = 3.5 cm \(\Rightarrow\) radius \(r = \dfrac{3.5}{2}=1.75\,\text{cm}\)
- Height of a hemisphere = its radius \(= r = 1.75\,\text{cm}\)
2. Find the height of the cone
\[ h_{\text{cone}} = H - r = 5 - 1.75 = 3.25\,\text{cm} \]
3. Slant height of the cone
\[ l = \sqrt{r^{2}+h_{\text{cone}}^{2}} = \sqrt{(1.75)^{2}+(3.25)^{2}} \]
\[ (1.75)^{2}=3.0625,\;(3.25)^{2}=10.5625 \]
\[ l = \sqrt{13.625}\approx 3.69\,\text{cm} \]
4. Curved surface area of the cone
\[ \text{CSA}_{\text{cone}} = \pi r l = \frac{22}{7}\times 1.75 \times 3.69 \]
\[ = \frac{22}{7}\times \frac{7}{4}\times 3.69 = \frac{22}{4}\times 3.69 = 5.5\times 3.69 \approx 20.30\,\text{cm}^{2} \]
5. Curved surface area of the hemisphere
\[ \text{CSA}_{\text{hemisphere}} = 2\pi r^{2} = 2\times \frac{22}{7}\times (1.75)^{2} \]
\[ = \frac{44}{7}\times 3.0625 = \frac{134.75}{7} \approx 19.25\,\text{cm}^{2} \]
6. Total area to be coloured
\[ \text{Total area}= \text{CSA}_{\text{cone}}+\text{CSA}_{\text{hemisphere}} \]
\[ \approx 20.30 + 19.25 = 39.55\,\text{cm}^{2} \]
Rounding to one decimal place, \(\boxed{39.5\,\text{cm}^{2}}\).
- Total height of the top, \(H = 5\,\text{cm}\)
- Diameter = 3.5 cm \(\Rightarrow\) radius \(r = \dfrac{3.5}{2}=1.75\,\text{cm}\)
- Height of a hemisphere = its radius \(= r = 1.75\,\text{cm}\)
2. Find the height of the cone
\[ h_{\text{cone}} = H - r = 5 - 1.75 = 3.25\,\text{cm} \]
3. Slant height of the cone
\[ l = \sqrt{r^{2}+h_{\text{cone}}^{2}} = \sqrt{(1.75)^{2}+(3.25)^{2}} \]
\[ (1.75)^{2}=3.0625,\;(3.25)^{2}=10.5625 \]
\[ l = \sqrt{13.625}\approx 3.69\,\text{cm} \]
4. Curved surface area of the cone
\[ \text{CSA}_{\text{cone}} = \pi r l = \frac{22}{7}\times 1.75 \times 3.69 \]
\[ = \frac{22}{7}\times \frac{7}{4}\times 3.69 = \frac{22}{4}\times 3.69 = 5.5\times 3.69 \approx 20.30\,\text{cm}^{2} \]
5. Curved surface area of the hemisphere
\[ \text{CSA}_{\text{hemisphere}} = 2\pi r^{2} = 2\times \frac{22}{7}\times (1.75)^{2} \]
\[ = \frac{44}{7}\times 3.0625 = \frac{134.75}{7} \approx 19.25\,\text{cm}^{2} \]
6. Total area to be coloured
\[ \text{Total area}= \text{CSA}_{\text{cone}}+\text{CSA}_{\text{hemisphere}} \]
\[ \approx 20.30 + 19.25 = 39.55\,\text{cm}^{2} \]
Rounding to one decimal place, \(\boxed{39.5\,\text{cm}^{2}}\).
Question 2
Hint available
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 = 22 7 )
Key Idea
Find the exposed area of the cube (all faces except the part covered by the hemisphere) and add the curved surface area of the hemisphere. The circular base of the hemisphere is not exposed.
Step-by-Step Solution
1. Data given
- Edge of cube, \(a = 5\) cm ⇒ each face area = \(a^2 = 5^2 = 25\) cm².
- Diameter of hemisphere = 4.2 cm ⇒ radius \(r = \frac{4.2}{2}=2.1\) cm = \(\frac{21}{10}\) cm.
- \(π = \frac{22}{7}\).
2. Area of the cube that remains exposed
- The cube has 6 faces. The bottom and the four side faces are completely exposed:
\[5\text{ faces} \times 25\text{ cm}^2 = 125\text{ cm}^2.\]
- The top face is partially covered by the circular base of the hemisphere. The area of the circular base is
\[\text{Area}_{\text{circle}} = πr^2 = \frac{22}{7}\times\left(\frac{21}{10}\right)^2 = \frac{22}{7}\times\frac{441}{100}=\frac{9702}{700}=13.86\text{ cm}^2.\]
- Exposed part of the top face = square area – circular area
\[25 - 13.86 = 11.14\text{ cm}^2.\]
- Total exposed area of the cube = \(125 + 11.14 = 136.14\) cm².
3. Curved surface area of the hemisphere
- Curved surface area of a hemisphere = \(2πr^2\).
- \[2πr^2 = 2\times\frac{22}{7}\times\frac{441}{100}=\frac{44\times441}{700}=\frac{19404}{700}=27.72\text{ cm}^2.\]
4. Total surface area of the block
\[\text{Total SA}= \text{exposed area of cube} + \text{curved area of hemisphere}
= 136.14 + 27.72 = 163.86\text{ cm}^2.\]
- Rounding to the nearest whole number, the total surface area ≈ 164 cm².
5. Answer
The total surface area of the decorative block is \(\boxed{163.86\text{ cm}^2 \;(\approx 164\text{ cm}^2)}\).
- Edge of cube, \(a = 5\) cm ⇒ each face area = \(a^2 = 5^2 = 25\) cm².
- Diameter of hemisphere = 4.2 cm ⇒ radius \(r = \frac{4.2}{2}=2.1\) cm = \(\frac{21}{10}\) cm.
- \(π = \frac{22}{7}\).
2. Area of the cube that remains exposed
- The cube has 6 faces. The bottom and the four side faces are completely exposed:
\[5\text{ faces} \times 25\text{ cm}^2 = 125\text{ cm}^2.\]
- The top face is partially covered by the circular base of the hemisphere. The area of the circular base is
\[\text{Area}_{\text{circle}} = πr^2 = \frac{22}{7}\times\left(\frac{21}{10}\right)^2 = \frac{22}{7}\times\frac{441}{100}=\frac{9702}{700}=13.86\text{ cm}^2.\]
- Exposed part of the top face = square area – circular area
\[25 - 13.86 = 11.14\text{ cm}^2.\]
- Total exposed area of the cube = \(125 + 11.14 = 136.14\) cm².
3. Curved surface area of the hemisphere
- Curved surface area of a hemisphere = \(2πr^2\).
- \[2πr^2 = 2\times\frac{22}{7}\times\frac{441}{100}=\frac{44\times441}{700}=\frac{19404}{700}=27.72\text{ cm}^2.\]
4. Total surface area of the block
\[\text{Total SA}= \text{exposed area of cube} + \text{curved area of hemisphere}
= 136.14 + 27.72 = 163.86\text{ cm}^2.\]
- Rounding to the nearest whole number, the total surface area ≈ 164 cm².
5. Answer
The total surface area of the decorative block is \(\boxed{163.86\text{ cm}^2 \;(\approx 164\text{ cm}^2)}\).
Question 3
Hint available
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 = 3.14)
Key Idea
Use the curved surface area formulas: \(\text{CSA of cone}=\pi r l\) where \(l=\sqrt{r^{2}+h^{2}}\), and \(\text{CSA of cylinder}=2\pi r h\). For the cylinder also add the area of its lower base (\(\pi r^{2}\)) because the upper base is attached to the cone and is not painted.
Step-by-Step Solution
1. Given data
- Height of whole rocket = 26 cm
- Height of cone (h_c) = 6 cm
- Height of cylinder (h_{cy}) = 26 – 6 = 20 cm
- Diameter of cone = 5 cm \(\Rightarrow\) radius of cone \(r_c = \frac{5}{2}=2.5\) cm
- Diameter of cylinder = 3 cm \(\Rightarrow\) radius of cylinder \(r_{cy}=\frac{3}{2}=1.5\) cm
2. Slant height of the cone
\[ l = \sqrt{r_c^{2}+h_c^{2}} = \sqrt{(2.5)^{2}+6^{2}} = \sqrt{6.25+36}=\sqrt{42.25}=6.5\text{ cm}\]
3. Area to be painted orange (cone)
Only the curved surface of the cone is exposed.
\[ \text{CSA}_{cone}=\pi r_c l = 3.14 \times 2.5 \times 6.5 = 3.14 \times 16.25 = 51.055\text{ cm}^{2} \]
Rounded to one decimal place: 51.1 cm².
4. Area to be painted yellow (cylinder)
- Curved surface area of cylinder:
\[ \text{CSA}_{cyl}=2\pi r_{cy} h_{cy}=2 \times 3.14 \times 1.5 \times 20 = 6.28 \times 30 = 188.4\text{ cm}^{2} \]
- Lower base of cylinder (the upper base is attached to the cone and is not painted):
\[ \text{Base area}=\pi r_{cy}^{2}=3.14 \times (1.5)^{2}=3.14 \times 2.25 = 7.065\text{ cm}^{2} \]
- Total yellow area:
\[ \text{Yellow area}=188.4 + 7.065 = 195.465\text{ cm}^{2} \]
Rounded to one decimal place: 195.5 cm².
5. Result
- Orange (cone) = 51.1 cm²
- Yellow (cylinder) = 195.5 cm²
- Height of whole rocket = 26 cm
- Height of cone (h_c) = 6 cm
- Height of cylinder (h_{cy}) = 26 – 6 = 20 cm
- Diameter of cone = 5 cm \(\Rightarrow\) radius of cone \(r_c = \frac{5}{2}=2.5\) cm
- Diameter of cylinder = 3 cm \(\Rightarrow\) radius of cylinder \(r_{cy}=\frac{3}{2}=1.5\) cm
2. Slant height of the cone
\[ l = \sqrt{r_c^{2}+h_c^{2}} = \sqrt{(2.5)^{2}+6^{2}} = \sqrt{6.25+36}=\sqrt{42.25}=6.5\text{ cm}\]
3. Area to be painted orange (cone)
Only the curved surface of the cone is exposed.
\[ \text{CSA}_{cone}=\pi r_c l = 3.14 \times 2.5 \times 6.5 = 3.14 \times 16.25 = 51.055\text{ cm}^{2} \]
Rounded to one decimal place: 51.1 cm².
4. Area to be painted yellow (cylinder)
- Curved surface area of cylinder:
\[ \text{CSA}_{cyl}=2\pi r_{cy} h_{cy}=2 \times 3.14 \times 1.5 \times 20 = 6.28 \times 30 = 188.4\text{ cm}^{2} \]
- Lower base of cylinder (the upper base is attached to the cone and is not painted):
\[ \text{Base area}=\pi r_{cy}^{2}=3.14 \times (1.5)^{2}=3.14 \times 2.25 = 7.065\text{ cm}^{2} \]
- Total yellow area:
\[ \text{Yellow area}=188.4 + 7.065 = 195.465\text{ cm}^{2} \]
Rounded to one decimal place: 195.5 cm².
5. Result
- Orange (cone) = 51.1 cm²
- Yellow (cylinder) = 195.5 cm²
Question 4
Hint available
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 = 22 7 )
Key Idea
Total surface area = curved surface area of the cylinder + curved surface area of the hemispherical depression + area of the circular base. Use 2πrh for the curved surface of a cylinder, 2πr² for a hemisphere (half of the sphere’s surface area 4πr²), and πr² for the base.
Step-by-Step Solution
1. Convert all dimensions to the same unit\
Radius \(r = 30\,\text{cm}=0.30\,\text{m}\)\
Height of the cylindrical part \(h = 1.45\,\text{m}\)\
2. Curved surface area of the cylinder\
$$\text{CSA}_{\text{cyl}} = 2\pi r h = 2 \times \frac{22}{7} \times 0.30 \times 1.45$$\
$$= 2 \times \frac{22}{7} \times 0.435 = \frac{44}{7} \times 0.435 = 2.734\,\text{m}^2$$\
3. Curved surface area of the hemispherical depression\
The surface area of a full sphere is \(4\pi r^2\); half of it (hemisphere) is \(2\pi r^2\).\
$$\text{CSA}_{\text{hem}} = 2\pi r^2 = 2 \times \frac{22}{7} \times (0.30)^2$$\
$$= 2 \times \frac{22}{7} \times 0.09 = \frac{44}{7} \times 0.09 = 0.566\,\text{m}^2$$\
4. Area of the circular base\
$$\text{Base area} = \pi r^2 = \frac{22}{7} \times (0.30)^2 = \frac{22}{7} \times 0.09 = 0.283\,\text{m}^2$$\
5. Total surface area\
$$\text{Total SA} = \text{CSA}_{\text{cyl}} + \text{CSA}_{\text{hem}} + \text{Base area}$$\
$$= 2.734 + 0.566 + 0.283 = 3.583\,\text{m}^2$$\
(Rounded to two decimal places, \(\approx 3.58\,\text{m}^2\)).
Radius \(r = 30\,\text{cm}=0.30\,\text{m}\)\
Height of the cylindrical part \(h = 1.45\,\text{m}\)\
2. Curved surface area of the cylinder\
$$\text{CSA}_{\text{cyl}} = 2\pi r h = 2 \times \frac{22}{7} \times 0.30 \times 1.45$$\
$$= 2 \times \frac{22}{7} \times 0.435 = \frac{44}{7} \times 0.435 = 2.734\,\text{m}^2$$\
3. Curved surface area of the hemispherical depression\
The surface area of a full sphere is \(4\pi r^2\); half of it (hemisphere) is \(2\pi r^2\).\
$$\text{CSA}_{\text{hem}} = 2\pi r^2 = 2 \times \frac{22}{7} \times (0.30)^2$$\
$$= 2 \times \frac{22}{7} \times 0.09 = \frac{44}{7} \times 0.09 = 0.566\,\text{m}^2$$\
4. Area of the circular base\
$$\text{Base area} = \pi r^2 = \frac{22}{7} \times (0.30)^2 = \frac{22}{7} \times 0.09 = 0.283\,\text{m}^2$$\
5. Total surface area\
$$\text{Total SA} = \text{CSA}_{\text{cyl}} + \text{CSA}_{\text{hem}} + \text{Base area}$$\
$$= 2.734 + 0.566 + 0.283 = 3.583\,\text{m}^2$$\
(Rounded to two decimal places, \(\approx 3.58\,\text{m}^2\)).
Question 5
Hint available
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 m3, and there are 20 workers, each of whom occupy about 0.08 m3 space on an average. Then, how much air is in the shed? (Take = 22 7 ) Fig. 12.12 Fig. 12.11
Key Idea
The volume of the shed is the sum of the volume of the cuboidal part and the volume of the half‑cylinder placed on its top. Use V(cuboid)=l·b·h and V(half‑cylinder)=½·π·r²·L, where r is the radius of the cylinder (half of the shorter side of the base) and L is its length (the longer side of the base). After obtaining the total volume, subtract the space occupied by machinery and workers to get the volume of air remaining.
Step-by-Step Solution
1. Identify dimensions\
- Base of cuboid: length \(l = 15\) m, breadth \(b = 7\) m.\
- Height of cuboidal part: \(h_c = 8\) m.\
- The half‑cylinder sits on the top of the cuboid. Its length equals the longer side of the base, i.e. \(L = 15\) m.\
- Diameter of the cylinder = breadth of the base = 7 m, therefore radius \(r = \dfrac{7}{2}=3.5\) m.
2. Volume of the cuboidal portion\
$$V_{cuboid}=l\times b\times h_c = 15\times 7\times 8 = 840\ \text{m}^3$$
3. Volume of the half‑cylinder\
- Full cylinder volume: $$V_{cyl}=\pi r^{2}L$$\
- Using \(\pi = \dfrac{22}{7}\):\
$$\pi r^{2}=\frac{22}{7}\times (3.5)^{2}=\frac{22}{7}\times 12.25=\frac{269.5}{7}=38.5$$\
- Hence, \(V_{cyl}=38.5\times 15=577.5\) m³.\
- Half‑cylinder volume: $$V_{half\,cyl}=\frac{1}{2}V_{cyl}=\frac{1}{2}\times 577.5=288.75\ \text{m}^3$$
4. Total volume of the shed (air space when empty)\
$$V_{total}=V_{cuboid}+V_{half\,cyl}=840+288.75=1128.75\ \text{m}^3$$
5. Space occupied by machinery\
$$V_{mach}=300\ \text{m}^3$$
6. Space occupied by workers\
- Each worker occupies \(0.08\) m³.\
- For 20 workers: $$V_{workers}=20\times 0.08=1.6\ \text{m}^3$$
7. Volume of air actually present\
$$V_{air}=V_{total}-V_{mach}-V_{workers}
=1128.75-300-1.6=827.15\ \text{m}^3$$
8. Answer (rounded to one decimal place)\
- Volume of shed = \(1128.8\) m³ (≈ \(1.13\times10^{3}\) m³).\
- Volume of air after accounting for machinery and workers = \(827.2\) m³.
- Base of cuboid: length \(l = 15\) m, breadth \(b = 7\) m.\
- Height of cuboidal part: \(h_c = 8\) m.\
- The half‑cylinder sits on the top of the cuboid. Its length equals the longer side of the base, i.e. \(L = 15\) m.\
- Diameter of the cylinder = breadth of the base = 7 m, therefore radius \(r = \dfrac{7}{2}=3.5\) m.
2. Volume of the cuboidal portion\
$$V_{cuboid}=l\times b\times h_c = 15\times 7\times 8 = 840\ \text{m}^3$$
3. Volume of the half‑cylinder\
- Full cylinder volume: $$V_{cyl}=\pi r^{2}L$$\
- Using \(\pi = \dfrac{22}{7}\):\
$$\pi r^{2}=\frac{22}{7}\times (3.5)^{2}=\frac{22}{7}\times 12.25=\frac{269.5}{7}=38.5$$\
- Hence, \(V_{cyl}=38.5\times 15=577.5\) m³.\
- Half‑cylinder volume: $$V_{half\,cyl}=\frac{1}{2}V_{cyl}=\frac{1}{2}\times 577.5=288.75\ \text{m}^3$$
4. Total volume of the shed (air space when empty)\
$$V_{total}=V_{cuboid}+V_{half\,cyl}=840+288.75=1128.75\ \text{m}^3$$
5. Space occupied by machinery\
$$V_{mach}=300\ \text{m}^3$$
6. Space occupied by workers\
- Each worker occupies \(0.08\) m³.\
- For 20 workers: $$V_{workers}=20\times 0.08=1.6\ \text{m}^3$$
7. Volume of air actually present\
$$V_{air}=V_{total}-V_{mach}-V_{workers}
=1128.75-300-1.6=827.15\ \text{m}^3$$
8. Answer (rounded to one decimal place)\
- Volume of shed = \(1128.8\) m³ (≈ \(1.13\times10^{3}\) m³).\
- Volume of air after accounting for machinery and workers = \(827.2\) m³.
Question 6
Hint available
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 = 3.14.)
Key Idea
Use the formula for volume of a cylinder \(V_{cyl}=\pi r^{2}h\) to obtain the apparent capacity. The bottom contains a hemispherical protrusion, whose volume is half of a sphere: \(V_{hem}=\frac{2}{3}\pi r^{3}\). The actual capacity is the cylindrical volume minus the hemispherical volume.
Step-by-Step Solution
1. Given data
- Inner diameter of glass = 5 cm \(\Rightarrow\) radius \(r = \frac{5}{2}=2.5\) cm.
- Height of glass (cylindrical part) = 10 cm.
- \(\pi = 3.14\).
2. Apparent capacity (volume of cylinder)
\[V_{\text{cyl}} = \pi r^{2} h = 3.14 \times (2.5)^{2} \times 10\]
\[(2.5)^{2}=6.25\]
\[V_{\text{cyl}} = 3.14 \times 6.25 \times 10 = 3.14 \times 62.5 = 196.25 \text{ cm}^{3}\]
Hence, apparent capacity = 196.25 cm³.
3. Volume of the hemispherical raised portion
Volume of a sphere = \(\frac{4}{3}\pi r^{3}\).
Therefore, volume of a hemisphere = \(\frac{1}{2}\times \frac{4}{3}\pi r^{3}=\frac{2}{3}\pi r^{3}\).
\[V_{\text{hem}} = \frac{2}{3} \times 3.14 \times (2.5)^{3}\]
\[(2.5)^{3}=15.625\]
\[V_{\text{hem}} = \frac{2}{3} \times 3.14 \times 15.625 = \frac{2 \times 49.0625}{3}= \frac{98.125}{3}= 32.7083 \text{ cm}^{3}\]
Approximate to two decimal places: 32.71 cm³.
4. Actual capacity
\[V_{\text{actual}} = V_{\text{cyl}} - V_{\text{hem}} = 196.25 - 32.71 = 163.54 \text{ cm}^{3}\]
Hence, actual capacity = 163.54 cm³ (approximately).
5. Result
- Apparent capacity = 196.25 cm³
- Actual capacity = 163.54 cm³
- Inner diameter of glass = 5 cm \(\Rightarrow\) radius \(r = \frac{5}{2}=2.5\) cm.
- Height of glass (cylindrical part) = 10 cm.
- \(\pi = 3.14\).
2. Apparent capacity (volume of cylinder)
\[V_{\text{cyl}} = \pi r^{2} h = 3.14 \times (2.5)^{2} \times 10\]
\[(2.5)^{2}=6.25\]
\[V_{\text{cyl}} = 3.14 \times 6.25 \times 10 = 3.14 \times 62.5 = 196.25 \text{ cm}^{3}\]
Hence, apparent capacity = 196.25 cm³.
3. Volume of the hemispherical raised portion
Volume of a sphere = \(\frac{4}{3}\pi r^{3}\).
Therefore, volume of a hemisphere = \(\frac{1}{2}\times \frac{4}{3}\pi r^{3}=\frac{2}{3}\pi r^{3}\).
\[V_{\text{hem}} = \frac{2}{3} \times 3.14 \times (2.5)^{3}\]
\[(2.5)^{3}=15.625\]
\[V_{\text{hem}} = \frac{2}{3} \times 3.14 \times 15.625 = \frac{2 \times 49.0625}{3}= \frac{98.125}{3}= 32.7083 \text{ cm}^{3}\]
Approximate to two decimal places: 32.71 cm³.
4. Actual capacity
\[V_{\text{actual}} = V_{\text{cyl}} - V_{\text{hem}} = 196.25 - 32.71 = 163.54 \text{ cm}^{3}\]
Hence, actual capacity = 163.54 cm³ (approximately).
5. Result
- Apparent capacity = 196.25 cm³
- Actual capacity = 163.54 cm³
Question 7
Hint available
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. (Take = 3.14)
Key Idea
Use the standard volume formulas from NCERT: \(V_{cone}=\frac13\pi r^{2}h\), \(V_{hemisphere}=\frac23\pi r^{3}\), and \(V_{cylinder}=\pi r^{2}h\). The radius of the base is half the given diameter. The height of the cylinder equals the total height of the toy (height of cone + radius of hemisphere).
Step-by-Step Solution
1. Find the radius
The diameter of the base is 4 cm, so \(r = \frac{4}{2} = 2\) cm.
2. Volume of the cone
Height of cone, \(h_c = 2\) cm.
\[V_{cone}=\frac13\pi r^{2}h_c = \frac13\times \pi \times 2^{2}\times 2 = \frac{8}{3}\pi\]
3. Volume of the hemisphere
Height of a hemisphere = its radius = 2 cm.
\[V_{hemisphere}=\frac23\pi r^{3}=\frac23\times \pi \times 2^{3}=\frac{16}{3}\pi\]
4. Total volume of the toy
\[V_{toy}=V_{cone}+V_{hemisphere}=\frac{8}{3}\pi+\frac{16}{3}\pi=\frac{24}{3}\pi=8\pi\]
Substituting \(\pi = 3.14\):
\[V_{toy}=8\times 3.14 = 25.12\ \text{cm}^{3}\]
5. Dimensions of the circumscribing cylinder
- Radius = same as toy = 2 cm.
- Height = height of cone + height of hemisphere = 2 cm + 2 cm = 4 cm.
6. Volume of the cylinder
\[V_{cyl}=\pi r^{2}h = \pi \times 2^{2}\times 4 = 16\pi\]
With \(\pi = 3.14\):
\[V_{cyl}=16\times 3.14 = 50.24\ \text{cm}^{3}\]
7. Difference of volumes
\[\Delta V = V_{cyl}-V_{toy}=16\pi-8\pi = 8\pi\]
Numerically,
\[\Delta V = 8\times 3.14 = 25.12\ \text{cm}^{3}\]
The diameter of the base is 4 cm, so \(r = \frac{4}{2} = 2\) cm.
2. Volume of the cone
Height of cone, \(h_c = 2\) cm.
\[V_{cone}=\frac13\pi r^{2}h_c = \frac13\times \pi \times 2^{2}\times 2 = \frac{8}{3}\pi\]
3. Volume of the hemisphere
Height of a hemisphere = its radius = 2 cm.
\[V_{hemisphere}=\frac23\pi r^{3}=\frac23\times \pi \times 2^{3}=\frac{16}{3}\pi\]
4. Total volume of the toy
\[V_{toy}=V_{cone}+V_{hemisphere}=\frac{8}{3}\pi+\frac{16}{3}\pi=\frac{24}{3}\pi=8\pi\]
Substituting \(\pi = 3.14\):
\[V_{toy}=8\times 3.14 = 25.12\ \text{cm}^{3}\]
5. Dimensions of the circumscribing cylinder
- Radius = same as toy = 2 cm.
- Height = height of cone + height of hemisphere = 2 cm + 2 cm = 4 cm.
6. Volume of the cylinder
\[V_{cyl}=\pi r^{2}h = \pi \times 2^{2}\times 4 = 16\pi\]
With \(\pi = 3.14\):
\[V_{cyl}=16\times 3.14 = 50.24\ \text{cm}^{3}\]
7. Difference of volumes
\[\Delta V = V_{cyl}-V_{toy}=16\pi-8\pi = 8\pi\]
Numerically,
\[\Delta V = 8\times 3.14 = 25.12\ \text{cm}^{3}\]