Rectangle Geometry: Width Increase Analysis

This solution examines how a 10% increase in a rectangle's width affects its properties.

Given: Rectangle dimensions are length $L$ and width $W$, with $L > W$.

Modification: Width $W$ is increased by 10%.

New Width: $W' = W + 0.10W = 1.1W$. The length $L$ remains constant.

Analyzing Potential Changes

• Perimeter:

  Original Perimeter $P = 2(L + W)$.

  New Perimeter $P' = 2(L + W') = 2(L + 1.1W)$.

  The percentage change is calculated as $\frac{P' - P}{P} \times 100$. Substituting the values: $\frac{2(L + 1.1W) - 2(L + W)}{2(L+W)} \times 100 = \frac{0.2W}{2(L+W)} \times 100 = \frac{10W}{L+W}\%$. This value depends on the ratio $L:W$ and is not always 10%.

• Diagonal Length:

  Original Diagonal $D = \sqrt{L^2 + W^2}$.

  New Diagonal $D' = \sqrt{L^2 + (W')^2} = \sqrt{L^2 + (1.1W)^2} = \sqrt{L^2 + 1.21W^2}$.

  The percentage change is $\frac{D' - D}{D} \times 100 = \frac{\sqrt{L^2 + 1.21W^2} - \sqrt{L^2 + W^2}}{\sqrt{L^2 + W^2}} \times 100$. This value depends on the ratio $L:W$ and is not always 10%.

• Area:

  Original Area $A = L \times W$.

  New Area $A' = L \times W' = L \times (1.1W) = 1.1(LW)$.

  Therefore, $A' = 1.1A$.

  The percentage change in area is $\frac{A' - A}{A} \times 100 = \frac{1.1A - A}{A} \times 100 = \frac{0.1A}{A} \times 100 = 10\%$. This is true for all values of $L$ and $W$.

• Shape Transformation:

  A rectangle becomes a square when its length equals its width ($L = W'$).

  In this case, it would require $L = 1.1W$. This condition ($\frac{L}{W} = 1.1$) is specific and does not apply to all rectangles where $L > W$. Thus, the rectangle does not necessarily become a square.

Correct Statement Identification

Based on the analysis, the only statement that remains correct for all possible values of $L$ and $W$ (given $L > W$) is that the area increases by 10% when the width is increased by 10%.