Question 1137216
 If the value of 
{{{a =18.0}}} 
{{{b =1.2 }}}
{{{c =5.0 }}}

what is the surface area of this shape? (Use {{{3.14}}} as the value of {{{pi}}}) 


 the surface area of this shape will be sum of:

{{{3}}} faces of the cube side length of {{{a}}}: {{{3a^2}}} (these faces have no cuts)
{{{2}}} faces (upper and one on the left side of big cube) have cut of {{{b^2}}} and area is: {{{2(a^2-b^2)}}}
{{{1 }}}face (front) has a  cut of {{{b^2}}} and circle, so area is: {{{a^2-(b^2+2c*pi)}}}
and, add surface area of three faces of small cube side length {{{b}}}:{{{3b^2}}}

{{{surface_ area=3a^2+2(a^2-b^2)+a^2-(b^2+c^2*pi)+3b^2}}}

{{{surface_ area=3*18^2+2(18^2-1.2^2)+18^2-(1.2^2+25*3.14)+3*1.44}}}

{{{surface_ area=972+2(324-1.44)+324-(1.44+78.5)+3*1.44}}}
{{{surface_ area=972+645.12+324-79.94+3*1.44}}}
{{{surface_ area=1865.5}}}


or, since we have first deducted {{{b^2}}} from three faces, then added same as faces of small cube, surface area is same as surface area of cube side length a minus area of the circle

{{{surface_ area=  6a^2-c^2*pi=6*18^2-25*3.14=1865.5}}}