Question 1011973
{{{drawing(370,330,-18.5,18.5,-1.5,31.5,
triangle(-17,0,0,0,0,29.445),
triangle(17,29.445,0,0,0,29.445),
rectangle(0,0,-1,1),
locate(-17.7,0,A),locate(-0.2,31,B),
locate(-0.2,0,D),locate(16.8,31,C),
red(arc(-17,0,6,6,-60,0)),
locate(-14,3,red(60^o)),
locate(-9,0,17),locate(-8.5,14.7,x)
)}}} In right triangle ABD, {{{cos(60^o)=1/2=17/x}}}--->{{{x=17*2=34}}}
So, {{{AD=BC=17}}} , {{{AB=CD+34}}} , and
{{{Perimeter=AB+BC+CD+AD=34+17+34+17=highlight(102)}}}