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
Step-by-Step Solution
Key Concept: 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.
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).
Correct Answer: Total surface area = $700 - 16\pi + 8\pi\sqrt{29}\ \text{cm}^2 \approx 785\ \text{cm}^2$