# Question #a6d51

##### 1 Answer

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 probleman "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_"wood" = V_"outside" - V_"inside"#

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

#V_"outside" = "20 cm" xx "15 cm" xx "10 cm" = "3000 cm"^3#

Now, since the wood is **on both sides of the box**!

#V_"inside" = [20-(1+1)]"cm" xx [15 - (1+1)]"cm" xx [10-(1+1])"cm"#

#V_"inside" = "18 cm" xx "13 cm" xx "8 cm" = "1872 cm"^3#

The volume of the wood will thus be

#V_"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")#