Question 627172
{{{A=(b*h)/2}}} and {{{h=4+3b}}} following the instructions from the problem... then we get by substitution that {{{32=b(4+3b)/2}}} then {{{64=4b+3b^2}}} then {{{0=3b^2+4b-64=0}}} which becomes {{{(3b+16)(b-4)=0}}} which leaves the only possible answer for {{{b=4}}} and then by substitution we obtain that {{{h=16}}}