A solid cube is cut into two cuboids of equal volume. What is the ratio of total surface area of the given cube to that of one of the cuboids ?
3 : 2
The question asks us to find the ratio of the total surface area of a solid cube to the total surface area of one of the two identical cuboids that are formed by cutting the cube exactly in half by volume. To solve this, we need to calculate the surface area of the original cube and the surface area of one of the resulting cuboids and then find their ratio.
Let's assume the original cube has a side length of \(a\) units.
A cube has 6 square faces, and each face has a side length of \(a\). The area of one face is \(a \times a = a^2\).
The total surface area of the cube is the sum of the areas of its 6 faces.
Total Surface Area of Cube \(= 6 \times (\text{Area of one face})\)
Total Surface Area of Cube \(= 6 \times a^2 = 6a^2\)
We can represent this using LaTeX:
\(\text{Surface Area}_{\text{cube}} = 6a^2\)
When the solid cube is cut into two cuboids of equal volume, the cut must be parallel to one of the faces and pass exactly through the midpoint of the dimension perpendicular to that face. This means one dimension of the cube is halved.
Let's say the original cube had dimensions \(a \times a \times a\). Cutting it in half along one dimension (for example, the length) results in two cuboids. Each cuboid will have dimensions \(a/2 \times a \times a\).
The volume of the original cube is \(a \times a \times a = a^3\).
The volume of one cuboid is \((a/2) \times a \times a = a^3/2\). Two such cuboids have a total volume of \(2 \times (a^3/2) = a^3\), which matches the original cube's volume, confirming the dimensions are correct.
So, the dimensions of one cuboid are: Length (\(l\)) \(= a/2\), Width (\(w\)) \(= a\), Height (\(h\)) \(= a\).
A cuboid has 6 rectangular faces. The formula for the total surface area of a cuboid with length \(l\), width \(w\), and height \(h\) is \(2(lw + wh + hl)\).
Using the dimensions \(l = a/2\), \(w = a\), \(h = a\) for one cuboid:
Surface Area of one Cuboid \(= 2 \times ((a/2) \times a + a \times a + a \times (a/2))\)
Let's calculate the areas of the pairs of faces:
Total Surface Area of one Cuboid \(= a^2 + 2a^2 + a^2 = 4a^2\)
Using the formula \(2(lw + wh + hl)\) with \(l=a/2, w=a, h=a\):
\(\text{Surface Area}_{\text{cuboid}} = 2 \left( \left(\frac{a}{2}\right)a + a \cdot a + a \left(\frac{a}{2}\right) \right)\)
\(\text{Surface Area}_{\text{cuboid}} = 2 \left( \frac{a^2}{2} + a^2 + \frac{a^2}{2} \right)\)
\(\text{Surface Area}_{\text{cuboid}} = 2 \left( a^2 + a^2 \right)\)
\(\text{Surface Area}_{\text{cuboid}} = 2 \left( 2a^2 \right)\)
\(\text{Surface Area}_{\text{cuboid}} = 4a^2\)
We need the ratio of the total surface area of the given cube to that of one of the cuboids.
Ratio \(= \frac{\text{Total Surface Area of Cube}}{\text{Total Surface Area of one Cuboid}}\)
Ratio \(= \frac{6a^2}{4a^2}\)
Cancel out the \(a^2\) terms and simplify the fraction:
Ratio \(= \frac{6}{4} = \frac{3}{2}\)
The ratio is \(3:2\).
| Shape | Side Length / Dimensions | Total Surface Area Formula |
|---|---|---|
| Cube | \(a\) | \(6a^2\) |
| Cuboid (cut from cube) | \(l=a/2, w=a, h=a\) | \(2(lw + wh + hl)\) |
When a solid object is cut, new surfaces are created, which increases the total surface area of the pieces compared to the original object. In this case, the original cube had a surface area of \(6a^2\). When cut into two cuboids, a new face of size \(a \times a\) is created on each cuboid at the site of the cut. So, for two cuboids, the total increase in surface area is \(2 \times a^2 = 2a^2\).
This sum (\(8a^2\)) is equal to the original cube's surface area (\(6a^2\)) plus the area of the two new cut faces (\(2a^2\)), i.e., \(6a^2 + 2a^2 = 8a^2\). This confirms the calculation of the cuboid's surface area is consistent with the physics of the cut.
The question specifically asks for the ratio of the cube's surface area to one cuboid's surface area, which is \(6a^2 : 4a^2 = 3:2\).
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