Question

Find the dimensions of the cuboidal box.

Answer 1

Given:
- A hemispherical bowl is packed in a cuboidal box.
- Inner radius of bowl = 10 cm.
- Outer radius of bowl = 10.5 cm.
- The bowl fits snugly in the box.

Step 1: Determine cuboidal box dimensions
- Hemisphere diameter = \(2 \times 10 = 20\) cm.
- Box base: square or rectangle with length and breadth = hemisphere diameter = 20 cm.
- Box height = bowl outer radius = 10.5 cm (accommodating thickness).

Step 2: Box dimensions
\[\text{Length} = 20 \, \text{cm}\] \[\text{Breadth} = 20 \, \text{cm}\] \[\text{Height} = 10.5 \, \text{cm}\]

Final Answer:
Cuboidal box dimensions:
\[\boxed{20\, \text{cm} \times 20\, \text{cm} \times 10.5\, \text{cm}}\]

Answer 2

Given:
Cuboidal box dimensions:
Length \(l = 20 \, \text{cm}\), Breadth \(b = 20 \, \text{cm}\), Height \(h = 10.5 \, \text{cm}\).

Step 1: Cuboid surface area formula
\[\text{Surface Area} = 2(lb + bh + hl)\]

Step 2: Calculation
\[= 2 \left(20 \times 20 + 20 \times 10.5 + 10.5 \times 20\right)\] \[= 2 \left(400 + 210 + 210\right) = 2 \times 820 = 1640 \, \text{cm}^2\]

Answer:
Total surface area:
\[\boxed{1640 \, \text{cm}^2}\]

Answer 3

Given:
- Inner radius of hemispherical bowl \(r = 10\, \text{cm}\)
- Dimensions of cuboidal box: \(20 \times 20 \times 10.5\, \text{cm}\)
- Use \(\pi = 3.14\)

Step 1: Hemispherical bowl volume calculation
Volume of hemisphere:
\[V_{\text{bowl}} = \frac{2}{3} \pi r^3 = \frac{2}{3} \times 3.14 \times (10)^3 = \frac{2}{3} \times 3.14 \times 1000 = \frac{2}{3} \times 3140 = 2093.33\, \text{cm}^3\]

Step 2: Cuboidal box volume calculation
\[V_{\text{box}} = l \times b \times h = 20 \times 20 \times 10.5 = 4200\, \text{cm}^3\]

Step 3: Volume difference calculation
\[\text{Difference} = V_{\text{box}} - V_{\text{bowl}} = 4200 - 2093.33 = 2106.67\, \text{cm}^3\]

Final Answer:
The difference in volumes is:
\[\boxed{2106.67\, \text{cm}^3}\]

Answer 4

Given:
- Inner radius of the bowl \(r = 10\, \text{cm}\)
- Outer radius of the bowl \(R = 10.5\, \text{cm}\)
- Use \(\pi = 3.14\)

Step 1: Calculate inner curved surface area (CSA) of the bowl
\[\text{Inner CSA} = 2 \pi r^2 = 2 \times 3.14 \times (10)^2 = 2 \times 3.14 \times 100 = 628\, \text{cm}^2\]

Step 2: Calculate outer curved surface area (CSA) of the bowl
\[\text{Outer CSA} = 2 \pi R^2 = 2 \times 3.14 \times (10.5)^2 = 2 \times 3.14 \times 110.25 = 692.67\, \text{cm}^2\]

Step 3: Calculate the area of the bowl's thickness
Thickness area is the difference between outer and inner surface areas:
\[\text{Thickness area} = 692.67 - 628 = 64.67\, \text{cm}^2\]

Step 4: Calculate the total area to be painted
\[\text{Total area} = \text{Inner CSA} + \text{Thickness area} = 628 + 64.67 = 692.67\, \text{cm}^2\]

Final Answer:
Area to be painted:
\[\boxed{692.67\, \text{cm}^2}\]