Surface Area of a Cuboid: Increasing a specific element by addition of.....or multiplication by.......

To solve this problem, let's start with the basic formula for the surface area of a rectangular prism:

The original surface area $SA_{\text{original}}$ is given by: $SA_{\text{original}} = 2lw + 2lh + 2wh$

When we double the dimensions, each dimension is multiplied by 2, so the new dimensions are $2l$, $2w$, and $2h$.

The new surface area $SA_{\text{new}}$ is calculated as follows: $SA_{\text{new}} = 2(2l)(2w) + 2(2l)(2h) + 2(2w)(2h)$

This simplifies to: $\begin{aligned} SA_{\text{new}} & = 2(4lw) + 2(4lh) + 2(4wh) \\ & = 8lw + 8lh + 8wh \end{aligned}$

To find the increase in surface area, subtract the original surface area from the new surface area:

$\Delta SA = SA_{\text{new}} - SA_{\text{original}}$

Therefore, $\begin{aligned} \Delta SA & = (8lw + 8lh + 8wh) - (2lw + 2lh + 2wh) \\ & = 6lw + 6lh + 6wh \end{aligned}$

The difference or increase in the surface area is expressed as: $6(lw + lh + wh)$

After multiplying by 2, each pair of dimensions $(width + height) \times length$ gives the entire side areas that change. We find that the surface area increases by $(width + height)\cdot length \cdot2$.

Therefore, this matches with choice 3.

It will increase by $(width + height)\cdot length \cdot2$.

To solve this problem, let's follow these steps:

• Step 1: Calculate the original surface area of the prism.
• Step 2: Calculate the surface area after tripling the width.
• Step 3: Determine the change in surface area.

First, the original surface area of the rectangular prism is given by the formula:
$SA = 2(lw + lh + wh)$.

Step 1: Substitute the original dimensions:
$SA_{\text{original}} = 2(lw + lh + wh)$.

Step 2: Now, when we triple the width, the new width is $3w$. Substitute $3w$ into the surface area formula:
$SA_{\text{new}} = 2(l(3w) + lh + (3w)h)$.
This simplifies to:
$SA_{\text{new}} = 2(3lw + lh + 3wh) = 6lw + 2lh + 6wh$.

Step 3: Subtract the original surface area from the new one to find the change:
$\Delta SA = SA_{\text{new}} - SA_{\text{original}} = (6lw + 2lh + 6wh) - (2lw + 2lh + 2wh)$.
Thus, $\Delta SA = 4lw + 4wh$.

This change can be factorized further as:
$\Delta SA = 4(l(w + h))$.

Therefore, the surface area will increase by $4(w + h)l$.

Thus, the correct answer is: It will increase by $(w + h) \cdot l \cdot 4$. This is choice 3 and 4.