阅读材料:
已知,如图\(1\),在面积为\(S\)的\(\triangle ABC\)中,\(BC=a\),\(AC=b\),\(AB=c\),内切圆\(O\)的半径为\(r\)。连接\(OA\),\(OB\),\(OC\),\(\triangle ABC\)被划分为三个小三角形.
图\(1\) 图\(2\) 图\(3\)
\(∵S=S_{\triangle OBC}+S_{\triangle OAC}+S_{\triangle OAB}=\dfrac{1}{2}BC\cdot r+\dfrac{1}{2}AC\cdot r+\dfrac{1}{2}AB\cdot r=\dfrac{1}{2}(a+b+c)r\),
\(∴r=\dfrac{2S}{a+b+c}\).
\((\)Ⅰ\()\)类比推理:若面积为\(S\)的四边形\(ABCD\)存在内切圆\((\)与各边都相切的圆\()\),如图\(2\),各边长分别为\(AB=a\),\(BC=b\),\(CD=c\),\(AD=d\),求四边形的内切圆半径\(r\);
\((\)Ⅱ\()\)理解应用:如图\(3\),在四边形\(ABCD\)中,\(AB/\!/CD\),\(AD=BC=13\),\(CD=11\),\(AB=21\),\(⊙O_{1}\)与\(⊙O_{2}\)分别为\(\triangle ABD\)与\(\triangle BCD\)的内切圆,设它们的半径分别为\(r_{1}\)和\(r_{2}\),求\(\dfrac{{{r}_{1}}}{{{r}_{2}}}\)的值.