If each side of a cube is reduced by 50%, the surface area will reduced by.
75%
Understanding the Cube and its Surface Area
The question asks about the reduction in the surface area of a cube when each side is reduced by a certain percentage. To solve this, we need to know how to calculate the surface area of a cube and how percentage reductions affect dimensions and areas.
A cube is a three-dimensional solid object bounded by six square faces, facets or sides, with three meeting at each vertex.
The surface area of a cube is the total area of all its six square faces. If the side length of the cube is denoted by 's', then the area of one face is \(s \times s = s^2\). Since there are six faces, the total surface area (SA) of the cube is:
\[SA = 6s^2\]
Calculating Original Surface Area
Let's assume the original side length of the cube is \(s_{original}\).
The original surface area (\(SA_{original}\)) is:
\[SA_{original} = 6 \times (s_{original})^2\]
Calculating New Side Length after Reduction
The problem states that each side of the cube is reduced by 50%. A 50% reduction means the new side length is 50% of the original side length.
Reduction amount = 50% of \(s_{original}\) = \(0.50 \times s_{original}\)
New side length (\(s_{new}\)) = \(s_{original}\) - Reduction amount
\[s_{new} = s_{original} - 0.50 \times s_{original}\]
\[s_{new} = (1 - 0.50) \times s_{original}\]
\[s_{new} = 0.50 \times s_{original}\]
So, the new side length is half of the original side length.
Calculating New Surface Area
Now, let's calculate the new surface area (\(SA_{new}\)) using the new side length, \(s_{new} = 0.50 \times s_{original}\).
\[SA_{new} = 6 \times (s_{new})^2\]
Substitute the value of \(s_{new}\):
\[SA_{new} = 6 \times (0.50 \times s_{original})^2\]
\[SA_{new} = 6 \times (0.50)^2 \times (s_{original})^2\]
\[SA_{new} = 6 \times 0.25 \times (s_{original})^2\]
We know that \(6 \times (s_{original})^2\) is the original surface area (\(SA_{original}\)). So, we can write:
\[SA_{new} = 0.25 \times SA_{original}\]
This means the new surface area is 0.25 times or 25% of the original surface area.
Calculating the Percentage Reduction in Surface Area
The question asks for the percentage reduction in surface area. To find this, we calculate the difference between the original surface area and the new surface area, and then express this difference as a percentage of the original surface area.
Reduction in surface area = \(SA_{original} - SA_{new}\)
Substitute \(SA_{new} = 0.25 \times SA_{original}\):
\[\text{Reduction} = SA_{original} - 0.25 \times SA_{original}\]
\[\text{Reduction} = (1 - 0.25) \times SA_{original}\]
\[\text{Reduction} = 0.75 \times SA_{original}\]
The reduction in surface area is 0.75 times the original surface area.
To express this as a percentage reduction:
\[\text{Percentage Reduction} = \left( \frac{\text{Reduction}}{SA_{original}} \right) \times 100\%\]
\[\text{Percentage Reduction} = \left( \frac{0.75 \times SA_{original}}{SA_{original}} \right) \times 100\%\]
\[\text{Percentage Reduction} = 0.75 \times 100\%\]
\[\text{Percentage Reduction} = 75\%\]
Therefore, when each side of a cube is reduced by 50%, the surface area is reduced by 75%.
Summary of Calculations
| Measure | Original (side = s) | New (side = 0.5s) |
|---|---|---|
| Side Length | \(s\) | \(0.5s\) |
| Surface Area | \(6s^2\) | \(6(0.5s)^2 = 6(0.25s^2) = 1.5s^2\) |
| SA Comparison | \(SA_{original}\) | \(0.25 \times SA_{original}\) |
| Reduction in SA | \(SA_{original} - SA_{new} = 6s^2 - 1.5s^2 = 4.5s^2\) | |
| Percentage Reduction | \(\left( \frac{4.5s^2}{6s^2} \right) \times 100\% = \left( \frac{4.5}{6} \right) \times 100\% = 0.75 \times 100\% = 75\%\) | |
Revision Table: Cube Geometry Concepts
| Concept | Formula | Notes |
|---|---|---|
| Side Length | \(s\) | Basic dimension of the cube |
| Area of One Face | \(s^2\) | Area of a square |
| Total Surface Area | \(6s^2\) | Sum of areas of 6 faces |
| Volume | \(s^3\) | Space occupied by the cube |
Additional Information: Scaling and Area/Volume
When the dimensions of a geometric shape are scaled by a factor, the area and volume scale by the square and cube of that factor, respectively.
- If the side length (a linear dimension) is scaled by a factor \(k\) (i.e., \(s_{new} = k \times s_{original}\)), then:
- The area (a 2D dimension) scales by \(k^2\). So, \(SA_{new} = k^2 \times SA_{original}\).
- The volume (a 3D dimension) scales by \(k^3\). So, \(V_{new} = k^3 \times V_{original}\).
In this problem, the side is reduced by 50%. This means the new side is 50% of the original, so the scaling factor \(k\) is 0.50.
- For the side length: \(s_{new} = 0.50 \times s_{original}\). The scaling factor is \(k=0.50\).
- For the surface area: The scaling factor for area is \(k^2 = (0.50)^2 = 0.25\). So, \(SA_{new} = 0.25 \times SA_{original}\). This confirms our calculation that the new surface area is 25% of the original.
- The reduction in surface area is \(SA_{original} - SA_{new} = SA_{original} - 0.25 \times SA_{original} = 0.75 \times SA_{original}\).
- The percentage reduction is \(\frac{0.75 \times SA_{original}}{SA_{original}} \times 100\% = 75\%\).
If the question were about the volume reduction, the new volume would be \(V_{new} = k^3 \times V_{original} = (0.50)^3 \times V_{original} = 0.125 \times V_{original}\). The volume would be reduced by \(V_{original} - 0.125 \times V_{original} = 0.875 \times V_{original}\), which is an 87.5% reduction in volume.
Important Questions from Solid Figures
The area of the floor of a cubical room is 192 m 2. The length of the longest rod that can be kept in that room is :
A solid metallic rectangular block of dimensions 112 cm × 44 cm × 25 cm is melted and recast into a cylinder of radius 35 cm. The curved surface area (in cm 2) of the cylinder is: (Take π = 22/7)
If the volume of a cube is 175616 cm 3, what is its side?
The volume of a right circular cone is 1232 cm 3. If the height of the cone is 24 cm, then what will be the radius of its base?
A right triangle contains the right angle between the sides 5 cm and 7 cm. A cone is generated by revolving about the side 5 cm. The volume of this cone is:
