【高校数学Ⅰ】3-4-2 内接円と面積｜問題集

ヘロンの公式より、\(△ABC\)の面積\(S\)を求める。
\(\displaystyle s=\frac{4+3+2}{2}=\frac{9}{2}\)
\(\displaystyle S=\sqrt{\frac{9}{2}\times\left(\frac{9}{2}-4\right)\times\left(\frac{9}{2}-3\right)\times\left(\frac{9}{2}-2\right)}\)
\(\displaystyle \ \ =\sqrt{\frac{135}{16}}\)
\(\displaystyle \ \ =\frac{3\sqrt{15}}{4}\)
よって、内接円の半径\(r\)は
\(\displaystyle \frac{3\sqrt{15}}{4}=\frac{1}{2}r(4+3+2)\)
\(\displaystyle \frac{3\sqrt{15}}{4}=\frac{9}{2}r\)
\(\displaystyle \ \ r=\frac{\sqrt{15}}{6}\)

ヘロンの公式より、\(△ABC\)の面積\(S\)を求める。
\(\displaystyle s=\frac{7+9+10}{2}=13\)
\(S=\sqrt{13\times(13-7)\times(13-9)\times(13-10)}\)
\(\ \ =\sqrt{936}\)
\(\ \ =6\sqrt{26}\)
よって、内接円の半径\(r\)は
\(\displaystyle 6\sqrt{26}=\frac{1}{2}r(7+9+10)\)
\(6\sqrt{26}=13r\)
\(\displaystyle \ \ r=\frac{6\sqrt{26}}{13}\)

\(△ABC\)の面積\(S\)は、
\(\displaystyle S=\frac{1}{2}\times6\times8\sin60^{\circ}\)
\(\displaystyle \ \ =24\times\frac{\sqrt{3}}{2}\)
\(\displaystyle \ \ =12\sqrt{3}\)
余弦定理より、
\(c^2=6^2+8^2-2\times6\times8\cos60^{\circ}\)
\(\displaystyle \ \ \ =36+64-96\times\frac{1}{2}\)
\(\ \ \ =52\)
\(c>0\)より、
\(c=2\sqrt{13}\)
よって、内接円の半径\(r\)は
\(\displaystyle 12\sqrt{3}=\frac{1}{2}r(6+8+2\sqrt{13})\)
\(12\sqrt{3}=(7+\sqrt{13})r\)
\(\displaystyle r=\frac{12\sqrt{3}}{7+\sqrt{13}}\)
\(\displaystyle \ \ =\frac{12\sqrt{3}(7-\sqrt{13})}{(7+\sqrt{13})(7-\sqrt{13})}\)
\(\displaystyle \ \ =\frac{12\sqrt{3}(7-\sqrt{13})}{36}\)
\(\displaystyle \ \ =\frac{\sqrt{3}(7-\sqrt{13})}{3}\)

\(△ABC\)の面積\(S\)は、
\(\displaystyle S=\frac{1}{2}\times8\times\sqrt{2}\sin45^{\circ}\)
\(\displaystyle \ \ =4\sqrt{2}\times\frac{1}{\sqrt{2}}\)
\(\ \ =4\)
余弦定理より、
\(a^2=8^2+(\sqrt{2})^2-2\times8\times\sqrt{2}\cos45^{\circ}\)
\(\displaystyle \ \ \ =64+2-16\sqrt{2}\times\frac{1}{\sqrt{2}}\)
\(\ \ \ =50\)
\(a>0\)より、
\(a=5\sqrt{2}\)
よって、内接円の半径\(r\)は
\(\displaystyle 4=\frac{1}{2}r(8+\sqrt{2}+5\sqrt{2})\)
\(4=(4+3\sqrt{2})r\)
\(\displaystyle r=\frac{4}{4+3\sqrt{2}}\)
\(\displaystyle \ \ =\frac{4(4-3\sqrt{2})}{(4+3\sqrt{2})(4-3\sqrt{2})}\)
\(\displaystyle \ \ =\frac{16-12\sqrt{2}}{-2}\)
\(\ \ =6\sqrt{2}-8\)

\(△ABC\)において、余弦定理より、
\(AC^2=6^2+7^2-2\times6\times7\cos B\)
\(\ \ \ \ \ \ \ \ =85-84\cos B・・・(1)\)
\(△ADC\)において、余弦定理より、
\(AC^2=2^2+3^2-2\times2\times3\cos D\)
\(\ \ \ \ \ \ \ \ =13-12\cos(180°-B)\)
\(\ \ \ \ \ \ \ \ =13+12\cos B・・・(2)\)
\((1),(2)\)より、
\(85-84\cos B=13+12\cos B\)
\(\displaystyle \cos B=\frac{3}{4}\)
\((1)\)に代入すると、
\(\displaystyle AC^2=85-84\times\frac{3}{4}=22\)
\(AC>0\)より、
\(AC=\sqrt{22}\)

\(\displaystyle \sin B=\sqrt{1-\left(\frac{3}{4}\right)^2}=\frac{\sqrt{7}}{4}\)
よって、
\(\displaystyle S=\frac{1}{2}\times6\times7\sin B+\frac{1}{2}\times2\times3\sin D\)
\(\ \ =21\sin B+3\sin D\)
\(\ \ =21\sin B+3\sin(180°-B)\)
\(\ \ =21\sin B+3\sin B\)
\(\displaystyle \ \ =24\times\frac{\sqrt{7}}{4}\)
\(\ \ =6\sqrt{7}\)

余弦定理より、
\(AC^2=5^2+4^2-2\times5\times4\cos60^{\circ}\)
\(\displaystyle \ \ \ \ \ \ \ \ =25+16-40\times\frac{1}{2}\)
\(\ \ \ \ \ \ \ \ =21\)
\(AC>0\)より、
\(AC=\sqrt{21}\)

余弦定理より、
\((\sqrt{21})^2=4^2+AD^2-2\times4\times AD\cos120^{\circ}\)
\(\displaystyle 21=16+AD^2-8AD\times\left(-\frac{1}{2}\right)\)
\(AD^2+4AD-5=0\)
\((AD+5)(AD-1)=0\)
\(AD=1,-5\)
\(AD>0\)より、
\(AD=1\)

\(\displaystyle S=\frac{1}{2}\times5\times4\sin60^{\circ}+\frac{1}{2}\times4\times1\sin120^{\circ}\)
\(\displaystyle \ \ =10\times\frac{\sqrt{3}}{2}+2\times\frac{\sqrt{3}}{2}\)
\(\ \ =5\sqrt{3}+\sqrt{3}\)
\(\ \ =6\sqrt{3}\)