# Question a6d51

Oct 25, 2015

Here's what I got.

#### Explanation:

The idea here is that you need to determine the volume of the wood needed to make the box. To do that, you will have to consider two boxes

• an "outside" box with the dimensions given to you in the problem
• an "inside" box with the dimensions that take into account the wood's thickness

The volume of the wood used to make the box will be the diference between these two volumes.

${V}_{\text{wood" = V_"outside" - V_"inside}}$

I assume that "ecteenal" is actualy external. The volume of the outside box will be

${V}_{\text{outside" = "20 cm" xx "15 cm" xx "10 cm" = "3000 cm}}^{3}$

Now, since the wood is $\text{1 cm}$ thick, it follows that the inside of the box will have the following dimensions - keep in mind that you need to take thickness into account on both sides of the box!

${V}_{\text{inside" = [20-(1+1)]"cm" xx [15 - (1+1)]"cm" xx [10-(1+1])"cm}}$

${V}_{\text{inside" = "18 cm" xx "13 cm" xx "8 cm" = "1872 cm}}^{3}$

The volume of the wood will thus be

${V}_{\text{wood" = "3000 cm"^3 - "1872 cm"^3 = "1128 cm}}^{3}$

Convert this value to cubic decimeters to make it match the units you have for the density of wood

1128color(red)(cancel(color(black)("cm"^3))) * "1 dm"^3/(10^3color(red)(cancel(color(black)("cm"^3)))) = "1.128 dm"^3

Now use the given density to find the mass of this volume of wood

1.128color(red)(cancel(color(black)("dm"^3))) * "1 kg"/(1color(red)(cancel(color(black)("dm"^3)))) = "1.128 kg"

I'll leave the answer rounded to two sig figs

m_"box" = color(green)("1.1 kg")#