A wooden pen stand is a cuboid \(10\times5\times4\,\text{cm}\) with four conical depressions (radius 0.5 cm, depth 2.1 cm) and a cubical depression of edge 3 cm. Find the volume of wood in the stand.
\(\displaystyle 173-0.7\pi\;\text{cm}^3\approx 170.8\,\text{cm}^3\)
Step 1: First, calculate the volume of the cuboid (the outer shape of the pen stand).
Formula: \( V = l \times b \times h \)
Here, \( l = 10\,\text{cm}, b = 5\,\text{cm}, h = 4\,\text{cm} \).
So, \( V = 10 \times 5 \times 4 = 200\,\text{cm}^3 \).
Step 2: Now calculate the volume of one conical depression (hole).
Formula for cone: \( V = \tfrac{1}{3}\pi r^2 h \)
Here, \( r = 0.5\,\text{cm}, h = 2.1\,\text{cm} \).
So, \( V = \tfrac{1}{3}\pi (0.5)^2 (2.1) \).
\( V = \tfrac{1}{3}\pi (0.25)(2.1) = \tfrac{1}{3}\pi (0.525) = 0.175\pi\,\text{cm}^3 \).
Step 3: Since there are 4 such conical depressions:
Total volume removed = \( 4 \times 0.175\pi = 0.7\pi\,\text{cm}^3 \).
Step 4: Calculate the volume of the cubical depression (hole).
Formula for cube: \( V = a^3 \)
Here, \( a = 3\,\text{cm} \).
So, \( V = 3^3 = 27\,\text{cm}^3 \).
Step 5: Subtract the volumes of all depressions (holes) from the cuboid volume.
Net volume = \( 200 - 27 - 0.7\pi \).
Step 6: Approximate value of \( \pi \approx 3.14 \).
So, \( 0.7\pi \approx 0.7 \times 3.14 = 2.198 \).
Net volume \( \approx 200 - 27 - 2.198 = 170.8\,\text{cm}^3 \).
Final Answer: The volume of wood in the pen stand is \( 173 - 0.7\pi\,\text{cm}^3 \approx 170.8\,\text{cm}^3 \).