A rectangle has a length L and a width W, where L > W. If the width, W, is increased
by 10%, which one of the following statements is correct for all values of L and W?
Options:
(A) Perimeter increases by 10%.
(B) Length of the diagonals increases by 10%.
(C) Area increases by 10%.
(D) The rectangle becomes a square.
Official Answer Key:
C
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.5)**
**1. Intuitive Physical Setup & Given Parameters**:
A rectangle has length $L$ and width $W$ with $L > W$.
The width $W$ is increased by $10\%$:
$W' = W(1 + 0.10) = 1.10 W$
The length $L$ remains unchanged ($L' = L$).
We need to determine which statement is **correct for all values of $L$ and $W$**.
**2. Step-by-Step Mathematical Derivation**:
1. **Area Comparison**:
- Initial area: $A = L \times W$
- New area:
$A' = L' \times W' = L \times (1.10 W) = 1.10 (L \times W) = 1.10 A$
- Percentage change in area:
$\Delta A\% = \frac{A' - A}{A} \times 100\% = \frac{1.10 A - A}{A} \times 100\% = 10\%$
- This holds strictly **for all values** of $L$ and $W$. Thus, Statement C is **CORRECT**.
2. **Evaluate Other Statements**:
- *Perimeter*:
$P = 2(L + W)$
$P' = 2(L + 1.10 W) = 2(L + W) + 0.20 W = P + 0.20 W$
$\frac{\Delta P}{P} = \frac{0.20 W}{2(L + W)} = \frac{0.10 W}{L + W} < 10\% \quad (\text{since } L > 0)$
Thus perimeter increases by less than $10\%$.
- *Diagonals*:
$D = \sqrt{L^2 + W^2}, \quad D' = \sqrt{L^2 + (1.1 W)^2} = \sqrt{L^2 + 1.21 W^2} < 1.10 \sqrt{L^2 + W^2}$
Diagonal increases by strictly less than $10\%$.
- *Square*: Since $L > W$, increasing $W$ by $10\%$ gives $1.1 W$, which only equals $L$ if $L = 1.1 W$, which is not true for arbitrary $L, W$.
**3. Comprehensive Options & Distractor Analysis**:
- **(A) Perimeter increases by 10%**: ❌ **INCORRECT**. Perimeter increase depends on $W/(L+W)$ and is strictly less than $10\%$.
- **(B) Length of the diagonals increases by 10%**: ❌ **INCORRECT**. $\sqrt{L^2 + 1.21W^2} < 1.1\sqrt{L^2+W^2}$.
- **(C) Area increases by 10%**: ✅ **CORRECT**. Area scales linearly with width: $A \propto W$, so $+10\%$ width produces exactly $+10\%$ area.
- **(D) The rectangle becomes a square**: ❌ **INCORRECT**. Only occurs if $L/W = 1.10$ exactly.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **C**
- **Exam Takeaway**: When one dimension of a multiplicative product $A = L \times W$ is scaled by factor $\lambda$, the product scales by exactly $\lambda$ regardless of the other dimension.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.