Question 810599

the hypotenuse is {{{12}}}, and the base is {{{4}}} 

you did {{{a^2 + 4^2 = 12^2}}}....this is correct

 and you get {{{a^2 + 16=144}}}....also correct

now move {{{16}}} to the right side

{{{a^2 =144-16}}}

{{{a^2 =128}}}....find {{{a}}}

{{{a  =sqrt(128)}}}

{{{a  =sqrt(2*64)}}}

{{{a  =sqrt(2*8^2)}}}

{{{highlight(a  =8sqrt(2))}}}.....exact solution

you can find approximate value

{{{a  =8*1.4142135623730950488016887242097}}}

{{{a  =11.313708498984760390413509793678}}}..round it

{{{a  =11.314}}}


if the answer key says it's 16 {{{ sqrt ( 2 ) }}}, that is not correct; check and see

{{{(16sqrt ( 2 ))^2 + 16=144}}}

{{{16^2*2 + 16=144}}}

{{{256*2 + 16=144}}}

{{{512 + 16=144}}}

{{{528<>144}}}

let's check our solution


{{{(8sqrt ( 2 ))^2 + 16=144}}}

{{{8^2*2 + 16=144}}}

{{{64*2 + 16=144}}}

{{{128 + 16=144}}}

{{{144=144}}}