Question

If each edge of a cube is increased by 50%, find the percentage increase in its surface area.

• A. 125
• B. 150
• C. 175
• D. 110

Hint:

For any area-related percentage increase problem where dimensions increase symmetrically by \(x\%\), the net percentage increase in area is always \(2x + \frac{x^2}{100}\).
For \(x = 50\): \(2(50) + \frac{2500}{100} = 100 + 25 = 125\%\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The total surface area of a three-dimensional geometric figure scales quadratically with changes in its linear dimensions.
Understanding this relationship allows us to find percentage changes without knowing the actual dimensions.
Key Formula or Approach:
The total surface area (\(A\)) of a cube with side length \(a\) is given by:
\[ A = 6a^2 \]
If the side length increases by \(x\%\), the new side length is \(a \cdot (1 + \frac{x}{100})\).
Alternatively, we can use the successive percentage change formula for two-dimensional area scaling:
\[ \text{Net Percentage Change} = x + y + \frac{x \cdot y}{100} \]

Step 2: Detailed Explanation:
Let us calculate the change using both methods to ensure clarity:
Method 1: Direct Substitution
- Let the initial side of the cube be \(a\).
- Original Surface Area, \(A_1 = 6a^2\).
- Since each edge is increased by \(50\%\), the new side length \(a_2\) is:
\[ a_2 = a + 0.50a = 1.5a \]
- New Surface Area, \(A_2 = 6a_2^2\):
\[ A_2 = 6(1.5a)^2 = 6 \cdot 2.25a^2 = 13.5a^2 \]
- Percentage increase in surface area:
\[ \text{Percentage Increase} = \frac{A_2 - A_1}{A_1} \times 100 \]
\[ \text{Percentage Increase} = \frac{13.5a^2 - 6a^2}{6a^2} \times 100 = \frac{7.5}{6} \times 100 = 125\% \]
Method 2: Successive Percentage Change
Since surface area depends on the product of two linear dimensions (\(a \times a\)), we can apply the formula with \(x = 50\) and \(y = 50\):
\[ \text{Net Increase} = 50 + 50 + \frac{50 \times 50}{100} = 100 + 25 = 125\% \]

Step 3: Final Answer:
The percentage increase in the surface area of the cube is 125% (Option A).