Initial Room Dimensions and Area

Let the length, breadth, and height of the room be $L$, $B$, and $H$ respectively.

Given the ratio $L : B : H = 4 : 3 : 2$.

We can represent the dimensions as $L = 4x$, $B = 3x$, and $H = 2x$ for some constant $x$.

The area of the four walls is calculated as: Area $= 2 \times (\text{Length} + \text{Breadth}) \times \text{Height}$ $A_1 = 2 \times (L + B) \times H$

Substituting the initial dimensions:

$A_1 = 2 \times (4x + 3x) \times 2x$ $A_1 = 2 \times (7x) \times 2x$ $A_1 = 28x^2$

New Room Dimensions and Area

The dimensions are modified as follows:

• New Length ($L'$) = $\frac{L}{2} = \frac{4x}{2} = 2x$
• New Breadth ($B'$) = $B \times 2 = 3x \times 2 = 6x$
• New Height ($H'$) = $\frac{H}{2} = \frac{2x}{2} = x$

The new area of the four walls ($A_2$) is:

$A_2 = 2 \times (L' + B') \times H'$ $A_2 = 2 \times (2x + 6x) \times x$ $A_2 = 2 \times (8x) \times x$ $A_2 = 16x^2$

Percentage Change in Area

The change in the area of the four walls is:

$\Delta A = A_1 - A_2 = 28x^2 - 16x^2 = 12x^2$

The percentage decrease is calculated using the initial area ($A_1$) as the reference:

\text{Percentage Decrease} $= \frac{\Delta A}{A_1} \times 100$ \text{Percentage Decrease} $= \frac{12x^2}{28x^2} \times 100$ \text{Percentage Decrease} $= \frac{12}{28} \times 100$ \text{Percentage Decrease} $= \frac{3}{7} \times 100$

Calculating the value:

\frac{3}{7} \times 100 \approx 0.42857 \times 100 \approx 42.86\%

Therefore, the new area of the four walls decreases approximately by 42.86%.