Aeronautics Technological Institute

The question says that $x$ , $y$ and $z$ are in Arithmetic Progression. Now, that’s a tricky question if we don’t realise at first the right way to put those letters in AP. We have to name a new letter, like “ $w$ ” for example, and represent the AP from ( $x$ , $y$ , $z$ ) to ( $w - d$ , $w$ , $w + d$ ), where " $d$ " is the common difference . Then, it should be easy to find $w$ . Since their sum is equal to $33$ , we can now use the sum formula for AP:

${ S }_{ n }\quad =\quad \frac { n }{ 2 } \left( { a }_{ 1 }\quad+\quad { a }_{ n } \right)$

${ S }_{ n }\quad =\quad 33$

$33\quad =\quad \frac { 3 }{ 2 } \left( w-d\quad +\quad w+d \right)$

$33\quad =\quad \frac { 3\cdot 2w }{ 2 }$

$33\quad =\quad 3w$

Then we get:

$w=11$

Replacing the result in our AP we get: ( $11 - d$ , $11$ , $11 + d$ )

Where $x = 11 - d$ , $y = 11$ and $z = 11 + d$

“...the total area of the parallelepiped is equal to $694cm^{2}$ ”. We have to apply the total area formula using the new expressions showed above:

${ S }_{ t }\quad =\quad 2\left( ab\quad +\quad ac\quad +\quad bc \right)$

${ S }_{ t }\quad =\quad 694$

$694\quad =\quad 2\left[ \left( 11-d \right) \cdot 11\quad +\quad \left( 11-d \right) \cdot \left( 11+d \right) \quad +\quad 11\cdot \left( 11+d \right) \right]$

$694\quad =\quad 2\left[ 121-11d\quad +\quad 121-{ d }^{ 2 }\quad +\quad 121+11d \right]$ --->divide both sides by 2

$347\quad =\quad 121-11d\quad +\quad 121-{ d }^{ 2 }\quad +\quad 121+11d$

$347\quad =\quad 363-{ d }^{ 2 }$

${ d }^{ 2 }\quad =\quad 16$

Then we find that $d=4$

Replacing on the expressions we get:

$x = 11-4$ $\rightarrow$ $x = 7$ ;

$y = 11$ ;

$z = 11+4$ $\rightarrow$ $z = 15$

And so, we just have to use the volume formula for the parallelepiped rectangle:

$V\quad =\quad a\cdot b\cdot c$

$V\quad =\quad 7\cdot 11\cdot 15$

$\boxed{V = 1155cm^{3}}$