Question 933448


the length of the diagonal of a rectangular box with a: 

length of  {{{L=5cm}}}, width of {{{W=4cm}}}, and height of {{{h=2cm}}} 

The base of the box is a rectangle of sides {{{L=5cm}}} and {{{W=4cm}}}. Thus its diagonal , {{{d}}} (of the base only), satisfies {{{d^2=L^2+W^2}}} by the Pythagorean theorem. 


 {{{d^2=L^2+W^2}}}...if  {{{L=5cm}}}, and {{{W=4cm}}} we will have

 {{{d^2=(5cm)^2+(4cm)^2}}}

 {{{d^2=25cm^2+16cm^2}}}

 {{{d^2=41cm^2}}}

 {{{d=sqrt(41cm^2)}}}

 {{{d=6.4cm}}}



Next consider the rectangle formed by this diagonal, the diagonal above it (parallel to it), and the two vertical sides connecting the two diagonals. This rectangle has the diagonal {{{D}}} connecting opposite sides of the box. 
Since the dimensions of this rectangle are {{{d}}} and {{{h}}}, again by the Pythagorean theorem {{{D^2=d^2+h^2}}}, then we have

{{{D^2=41cm^2+4cm^2}}}

{{{D^2=45cm^2}}}

{{{D=sqrt(45cm^2)}}}

{{{D=6.7cm}}}