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Difficulty: MediumSurface Area and Volume of 3D Solids

A solid rectangular wooden block measures 10 cm10\text{ cm} by 14 cm14\text{ cm} by 12 cm12\text{ cm}. A cylindrical hole of radius 3.5 cm3.5\text{ cm} is drilled straight through the block along its height of 12 cm12\text{ cm}. Taking π=227\pi = \frac{22}{7}, what is the volume of the remaining solid in cm3\text{cm}^3?

Answer: 1218 cm³

Answer

The volume of the remaining solid is 1218 cm31218\text{ cm}^3.
The initial total volume of the rectangular block is 10×14×12=1680 cm310 \times 14 \times 12 = 1680\text{ cm}^3. The volume of the cylindrical hole drilled through it is 227×(3.5)2×12=462 cm3\frac{22}{7} \times (3.5)^2 \times 12 = 462\text{ cm}^3. Subtracting the removed cylindrical volume from the block gives 1680−462=1218 cm31680 - 462 = 1218\text{ cm}^3.

Step-by-Step Solution

1
Calculate the total volume of the rectangular block.
Volume of block = 1680 cm31680\text{ cm}^3
The initial volume of the cuboid before drilling is calculated by multiplying its length, width, and height: 10×14×12=1680 cm310 \times 14 \times 12 = 1680\text{ cm}^3.
2
Calculate the volume of the cylindrical hole removed from the block.
Volume of cylinder = 462 cm3462\text{ cm}^3
The cylindrical hole has radius r=3.5 cm=72 cmr = 3.5\text{ cm} = \frac{7}{2}\text{ cm} and height h=12 cmh = 12\text{ cm}. Using V=πr2hV = \pi r^2 h, we get 227×494×12=462 cm3\frac{22}{7} \times \frac{49}{4} \times 12 = 462\text{ cm}^3.
3
Subtract the volume of the cylindrical hole from the total volume of the block.
Remaining volume = 1218 cm31218\text{ cm}^3
Because material is removed by drilling, the remaining volume is 1680−462=1218 cm31680 - 462 = 1218\text{ cm}^3.

Key Concept

Volume of a composite solid (cuboid with a cylindrical cavity)
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Topics

Surface Area and Volume of 3D Solids

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