Question 156797
We know {{{Area[T]=(1/2)bh}}} Sustituting when {{{h=b-2}}}, height is 2mm less than the base
{{{60=(1/2)(b)(b-2)}}}------------------------> working eqn
{{{60=(1/2)(b^2-2b)}}}, cross multiply
{{{120=b^2-2b}}} ---> {{{b^2-2b-120=0}}}
Solve by QUADRATIC, {{{a=1}}}, {{{b=-2}}}, & {{{c=-120}}}
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}
{{{x=(-(-2)+-sqrt(-2^2-4*1*-120))/(2*1)}}}
{{{x=(2+-sqrt(4+480))/2=(2+-sqrt(484))/2}}}
{{{x=(2+-22)/2}}}
2 values: {{{x=(2+22)/2}}}={{{24/2=12mm}}} BASE 
{{{x=(2-22)/2}}}={{{(-20/2)=(-10mm)}}}, DON'T USE
In doubt? Go back working eqn:
{{{60=(1/2)(12)(12-2)}}}
{{{60=(1/2)(12)(10)}}}
{{{60=(1/2)(120)}}}
{{{60=60}}}
Thank you,
Jojo