Question

Дано: ABCD-прямоугольник
Найти :CD AC Sabcd

Answer 1

<AOD=180-a
AD²=2AO²-2AO²cos<AOD=2AO²+2AO²cosa=2AO²(1+cosa)=
=2AO²*2cos²(a/2)=4AO²*cos²(a/2)
AO=AD/2cos(a/2)=a/2cos(a/2)
AC=2AO=a/cos(a/2)
CD=√(AC²-AD²)=√(a²/cos²(a/2)-a²)=√(a²(1-cos²(a/2))/cos²(a/2))=
=√(a²sin²(a/2)/cos²(a/2))=√(a²tg²(a/2))=atg(a/2)
S=AD*CD=a*atg(a/2)=a²tg(a/2)