【高校数学Ⅰ】3-4-1 三角形の面積｜要点まとめ

\(\displaystyle S=\frac{1}{2}\times3\times4\sin30^{\circ}\)
\(\displaystyle \ \ =6\times\left(\frac{1}{2}\right)\)
\(\ \ =3\)

\(\displaystyle s=\frac{4+5+7}{2}=8\)
よって、
\(S=\sqrt{8(8-4)(8-5)(8-7)}\)
\(\ \ =\sqrt{96}\)
\(\ \ =4\sqrt{6}\)

\(\displaystyle S=\frac{1}{2}\times3\times4\sin60^{\circ}\)
\(\displaystyle \ \ =6\times\left(\frac{\sqrt{3}}{2}\right)\)
\(\ \ =3\sqrt{3}\)

\(AD=x\)とおく。
\(△ABD\)の面積を\(S_{1}\)とすると、
\(\displaystyle S_{1}=\frac{1}{2}\times4\times x\sin30^{\circ}\)
\(\displaystyle \ \ \ \ =2x\times\left(\frac{1}{2}\right)\)
\(\ \ \ \ =x\)
\(△ACD\)の面積を\(S_{2}\)とすると、
\(\displaystyle S_{2}=\frac{1}{2}\times3\times x\sin30^{\circ}\)
\(\displaystyle \ \ \ \ =\frac{3}{2}x\times\left(\frac{1}{2}\right)\)
\(\displaystyle \ \ \ \ =\frac{3}{4}x\)
\(S=S_{1}+S_{2}\)より、
\(\displaystyle 3\sqrt{3}=x+\frac{3}{4}x\)
\(\displaystyle x=\frac{12\sqrt{3}}{7}\)
よって、
\(\displaystyle AD=\frac{12\sqrt{3}}{7}\)