Quiz 7:內切圓半徑
\(\vartriangle ABC\) 之邊長各為5,10,13,則 \(\vartriangle ABC\) 之內切圓半徑為
\(\sqrt 2 \)
\(\frac{{5\sqrt 7 }}{4}\)
\(\frac{{3\sqrt {14} }}{7}\)
\(\frac{{7\sqrt {13} }}{{12}}\)