EXERCISE 12.1

1. Find the side of each cube\
Given volume \(V = 64\,\text{cm}^3\).\
\[\text{side} = a = \sqrt[3]{V} = \sqrt[3]{64} = 4\,\text{cm}\]
2. Dimensions of the cuboid after joining\
- Length (l) = two sides placed end to end = \(2a = 2\times4 = 8\,\text{cm}\)
- Breadth (w) = side of a cube = \(4\,\text{cm}\)
- Height (h) = side of a cube = \(4\,\text{cm}\)
3. Apply the surface area formula for a cuboid\
\[SA = 2(lw + lh + wh)\]
Substitute the values:\
\[SA = 2[(8\times4) + (8\times4) + (4\times4)]\]
\[= 2[32 + 32 + 16]\]
\[= 2\times80\]
\[= 160\,\text{cm}^2\]
4. Result\
The total surface area of the resulting cuboid is \(160\,\text{cm}^2\).

1. Find the radius of the hemisphere\
The diameter is 14 cm, therefore\
\[ r = \frac{\text{diameter}}{2}=\frac{14}{2}=7\ \text{cm}. \]
2. Determine the height of the cylindrical part\
The total height of the vessel = height of hemisphere + height of cylinder.\
Height of hemisphere = radius = 7 cm.\
Hence,\
\[ h_{\text{cyl}} = 13\ \text{cm} - 7\ \text{cm}=6\ \text{cm}. \]
3. Curved surface area of the hollow hemisphere\
\[ \text{CSA}_{\text{hemisphere}} = 2\pi r^{2}=2\pi (7)^{2}=2\pi \times 49 = 98\pi\ \text{cm}^{2}. \]
4. Curved surface area of the hollow cylinder\
\[ \text{CSA}_{\text{cylinder}} = 2\pi r h = 2\pi (7)(6)=84\pi\ \text{cm}^{2}. \]
5. Total inner surface area\
\[ \text{Total inner surface area}= \text{CSA}_{\text{hemisphere}}+\text{CSA}_{\text{cylinder}}\]
\[ = 98\pi + 84\pi = 182\pi\ \text{cm}^{2}. \]
6. Numerical value (optional)\
Using \(\pi \approx \frac{22}{7}\),\
\[ 182\pi \approx 182 \times \frac{22}{7}=572\ \text{cm}^{2}. \]
Or with \(\pi \approx 3.14\),\
\[ 182\pi \approx 182 \times 3.14 = 571.48\ \text{cm}^{2}. \]
Thus, the inner surface area of the vessel is \(182\pi\ \text{cm}^{2}\) (approximately \(572\ \text{cm}^{2}\)).

1. Given data
\[ r = 3.5\ \text{cm}, \quad \text{total height}=15.5\ \text{cm} \]
2. Height of the cone
The hemisphere contributes a height equal to its radius \(r\).
\[ h_{\text{cone}} = 15.5 - 3.5 = 12\ \text{cm} \]
3. Slant height of the cone
\[ l = \sqrt{r^{2}+h_{\text{cone}}^{2}} = \sqrt{(3.5)^{2} + (12)^{2}} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\ \text{cm} \]
4. Curved surface area of the cone
\[ \text{CSA}_{\text{cone}} = \pi r l = \pi \times 3.5 \times 12.5 = 43.75\pi \ \text{cm}^{2} \]
5. Curved surface area of the hemisphere
\[ \text{CSA}_{\text{hemisphere}} = 2\pi r^{2} = 2\pi \times (3.5)^{2} = 2\pi \times 12.25 = 24.5\pi \ \text{cm}^{2} \]
6. Total surface area of the toy (only the exposed curved surfaces)
\[ \text{Total SA} = \text{CSA}_{\text{cone}} + \text{CSA}_{\text{hemisphere}} \]
\[ = (43.75\pi + 24.5\pi) \ \text{cm}^{2} = 68.25\pi \ \text{cm}^{2} \]
7. Numerical value (using \(\pi = \frac{22}{7}\))
\[ 68.25\pi = 68.25 \times \frac{22}{7} = \frac{1501.5}{7} = 214.5 \ \text{cm}^{2} \]
Thus, the total surface area of the toy is \(68.25\pi\ \text{cm}^{2}\) or \(214.5\ \text{cm}^{2}\) (to 1 decimal place).

1. Greatest possible diameter
- Let the side of the cube be $a = 7\,\text{cm}$. The hemisphere rests on the top face of the cube. For the hemisphere to fit without overhanging, its diameter $d$ must be at most the side length of the cube.
- Hence the greatest diameter is $$d_{\max}=a = 7\,\text{cm}.$$
- The radius of the hemisphere is therefore $$r = \frac{d_{\max}}{2}=\frac{7}{2}=3.5\,\text{cm}.$$

2. Surface area of the solid
- Curved surface area of the hemisphere: For a hemisphere, the curved surface area (excluding the base circle) is $$\text{CSA}_{\text{hemisphere}} = 2\pi r^{2}.$$
Substituting $r = 3.5\,\text{cm}$,
$$\text{CSA}_{\text{hemisphere}} = 2\pi (3.5)^{2}=2\pi \times 12.25 = 24.5\pi\,\text{cm}^{2}.$$

- Exposed faces of the cube: The cube has six faces, each of area $a^{2}=7^{2}=49\,\text{cm}^{2}$. The top face is completely covered by the hemisphere, so it does not contribute to the external surface. The remaining five faces (four vertical faces + bottom face) are exposed.
$$\text{Area}_{\text{cube}} = 5 \times 49 = 245\,\text{cm}^{2}.$$

- Total surface area:
$$\text{Total SA} = \text{CSA}_{\text{hemisphere}} + \text{Area}_{\text{cube}}
= 24.5\pi + 245\,\text{cm}^{2}$$
or, writing the fractional form,
$$\text{Total SA}=\frac{49\pi}{2}+245\,\text{cm}^{2}.$$

3. Numerical value (optional)
- Using $\pi \approx 3.1416$,
$$\text{Total SA} \approx 245 + 24.5 \times 3.1416 \approx 245 + 76.97 \approx 321.97\,\text{cm}^{2}.$$

1. Identify the given data
- Edge of the cube = \(l\)
- Diameter of the hemisphere = \(l\) \(\Rightarrow\) radius \(r = \frac{l}{2}\)

2. Surface area of the original cube
The cube has 6 equal faces, each of area \(l^{2}\).
$$\text{SA}_{\text{cube}} = 6l^{2}$$

3. Effect of cutting the hemispherical depression
- One face (area \(l^{2}\)) is removed.
- The curved surface of the hemisphere becomes exposed.
- Curved surface area of a hemisphere = \(2\pi r^{2}\).

4. Compute the curved surface area of the hemisphere
$$2\pi r^{2}=2\pi\left(\frac{l}{2}\right)^{2}=2\pi\cdot\frac{l^{2}}{4}=\frac{\pi l^{2}}{2}$$

5. Total surface area of the remaining solid
- Keep the 5 untouched faces of the cube: \(5l^{2}\).
- Add the curved surface of the hemisphere: \(\frac{\pi l^{2}}{2}\).
$$\text{SA}_{\text{remaining}} = 5l^{2}+\frac{\pi l^{2}}{2}=l^{2}\left(5+\frac{\pi}{2}\right)$$

6. Final answer
$$\boxed{\text{Surface area}=l^{2}\left(5+\frac{\pi}{2}\right)}$$

1. Identify the dimensions\
- Diameter $=5\,\text{mm}\;\Rightarrow\;$ radius $r=\dfrac{5}{2}=2.5\,\text{mm}$.\
- Total length of capsule $=14\,\text{mm}$.\
- The two hemispherical ends together contribute a length of $2r = 5\,\text{mm}$.\
- Hence the length of the cylindrical part (height $h$) is\
$$h = 14\,\text{mm} - 2r = 14 - 5 = 9\,\text{mm}.$$\
2. Surface area of the cylindrical part\
$$\text{Lateral area of cylinder}=2\pi r h = 2\pi (2.5)(9) = 45\pi\,\text{mm}^2.$$\
3. Surface area of the two hemispheres\
Two hemispheres make a complete sphere, whose surface area is\
$$\text{Surface area of sphere}=4\pi r^{2}=4\pi (2.5)^{2}=4\pi (6.25)=25\pi\,\text{mm}^2.$$\
4. Total surface area of the capsule\
$$\text{Total SA}=\text{Lateral area of cylinder}+\text{Surface area of sphere}
=45\pi+25\pi = 70\pi\,\text{mm}^2.$$\
5. Numerical value (optional)\
$$70\pi \approx 70 \times 3.14 = 219.8\,\text{mm}^2 \;\text{(≈ 220 mm}^2\text{)}.$$

1. Identify the dimensions\
- Diameter of cylindrical part $=4\,\text{m}$ \=> radius $r = \dfrac{4}{2}=2\,\text{m}$\
- Height of cylindrical part $h = 2.1\,\text{m}$\
- Slant height of conical top $l = 2.8\,\text{m}$\
2. Lateral surface area of the cylinder\
$$\text{Area}_{\text{cyl}} = 2\pi r h = 2\pi (2)(2.1) = 8.4\pi \;\text{m}^2$$\
3. Curved surface area of the cone\
$$\text{Area}_{\text{cone}} = \pi r l = \pi (2)(2.8) = 5.6\pi \;\text{m}^2$$\
4. Total canvas area (base is not covered)\
$$\text{Total area}=\text{Area}_{\text{cyl}}+\text{Area}_{\text{cone}} = 8.4\pi +5.6\pi = 14\pi \;\text{m}^2$$\
Using $\pi = \dfrac{22}{7}$ (as used in NCERT),\
$$\text{Total area}=14\times\frac{22}{7}=44\;\text{m}^2$$\
5. Cost of the canvas\
Rate = ` 500 per m$^2$\
$$\text{Cost}=44\times 500 = 22,000$$\
Hence, the canvas required is $44\,\text{m}^2$ and the cost is ` 22,000.

1. Given data
- Height of cylinder (and cone) \(h = 2.4\,\text{cm}\)
- Diameter = 1.4 cm \(\Rightarrow\) radius \(r = \frac{1.4}{2}=0.7\,\text{cm}\)

2. Slant height of the cone
\[ l = \sqrt{r^{2}+h^{2}} = \sqrt{0.7^{2}+2.4^{2}} = \sqrt{0.49+5.76}=\sqrt{6.25}=2.5\,\text{cm} \]

3. Curved surface area of the cylinder
\[ \text{CSA}_{\text{cyl}} = 2\pi r h = 2\pi(0.7)(2.4)=3.36\pi\ \text{cm}^{2} \]

4. Curved surface area of the cone (interior surface)
\[ \text{CSA}_{\text{cone}} = \pi r l = \pi(0.7)(2.5)=1.75\pi\ \text{cm}^{2} \]

5. Area of the remaining circular base of the cylinder
\[ \text{Base area}=\pi r^{2}=\pi(0.7)^{2}=0.49\pi\ \text{cm}^{2} \]

6. Total surface area of the remaining solid
\[ \text{TSA}=\text{CSA}_{\text{cyl}}+\text{CSA}_{\text{cone}}+\text{Base area}
=(3.36\pi+1.75\pi+0.49\pi)\ \text{cm}^{2}
=5.60\pi\ \text{cm}^{2} \]

7. Numerical value
\[ 5.60\pi \approx 5.60 \times 3.1416 = 17.59\ \text{cm}^{2} \]
Rounded to the nearest square centimetre, \(\boxed{18\ \text{cm}^{2}}\).

Hence, the total surface area of the solid after the conical cavity is removed is 18 cm² (nearest integer).

1. Identify the given data
- Radius of cylinder (and also radius of each hemisphere) $r = 3.5\,\text{cm}$
- Height of the cylinder $h = 10\,\text{cm}$

2. Curved surface area of the cylinder
$$\text{CSA}_{\text{cyl}} = 2\pi r h = 2\pi (3.5)(10) = 70\pi\ \text{cm}^2$$

3. Surface area contributed by the two hemispheres
- Two hemispheres together make a complete sphere of radius $r$.
- Surface area of a sphere: $\text{SA}_{\text{sphere}} = 4\pi r^2$
$$\text{SA}_{\text{hemis}} = 4\pi (3.5)^2 = 4\pi \times 12.25 = 49\pi\ \text{cm}^2$$

4. Total surface area of the wooden article
$$\text{Total SA} = \text{CSA}_{\text{cyl}} + \text{SA}_{\text{hemis}} = 70\pi + 49\pi = 119\pi\ \text{cm}^2$$

5. Numerical value (using $\pi \approx \frac{22}{7}$ or $3.14$)
- Using $\pi = \frac{22}{7}$:
$$119\pi = 119 \times \frac{22}{7} = 374\ \text{cm}^2$$
- Using $\pi \approx 3.14$:
$$119\pi \approx 119 \times 3.14 = 373.66\ \text{cm}^2$$

Hence, the total surface area of the article is $119\pi\,\text{cm}^2$ (approximately $374\,\text{cm}^2$).