【数学I】例題4.3.2：多角形の面積（One More）★★

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% 例題I4.3.2：多角形の面積（One More）★★
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(1) 半径$a$の円に内接する正八角形の面積$S$を求めよ．

(2) 1辺の長さが1の正十二角形の面積$S$を求めよ．

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% 例題I4.3.2の解答（One More）★★
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(1) 右の図のように，正八角形を対角線によって8個の合同な三角形に分け，3点O，A，Bをとると，
\[
\angle\mathrm{AOB}=360^{\circ}\div 8=45^{\circ}
\]

よって，求める面積$S$は，

\[
S=8\times\triangle\mathrm{OAB}=8\times\frac{1}{2}a^2\sin 45^{\circ}=8\times\frac{\sqrt{2}}{4}a^2=2\sqrt{2}a^2
\]

(2) 右の図のように，1辺の長さが1の正十二角形を対角線によって12個の合同な三角形に分け，3点O，A，Bをとると，

\[
\angle\mathrm{AOB}=360^{\circ}\div 12=30^{\circ}
\]

$\mathrm{OA}=\mathrm{OB}=a$とすると，

$\triangle\mathrm{OAB}$において，余弦定理より，

\[
1^2=a^2+a^2-2a\cdot a\cdot\cos 30^{\circ}
\]

したがって，$1=(2-\sqrt{3})a^2$

ゆえに，$a^2=\frac{1}{2-\sqrt{3}}=\frac{2+\sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})}=2+\sqrt{3}$

よって，求める面積$S$は，$S=12\times\triangle\mathrm{OAB}=12\times\frac{1}{2}a^2\sin 30^{\circ}=6+3\sqrt{3}$

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% 問題I4.3.2：多角形の面積（One More）★★
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(1) 1辺の長さが1の正八角形の面積$S$を求めよ．

(2) 半径1の円に内接する正十二角形の面積$S$を求めよ．

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% 問題I4.3.2の解答（One More）★★
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(1) 右の図のように，1辺の長さが1の正八角形を対角線によって8個の合同な三角形に分け，3点O，A，Bをとると，
\[
\angle\mathrm{AOB}=360^{\circ}\div 8=45^{\circ}
\]

余弦定理より，

\[
1^2=a^2+a^2-2a\cdot a\cdot\cos 45^{\circ}
\]

\[
1=2a^2(1-\frac{\sqrt{2}}{2})=2a^2\cdot\frac{2-\sqrt{2}}{2}=a^2(2-\sqrt{2})
\]

したがって，

\[
a^2=\frac{1}{2-\sqrt{2}}=\frac{2+\sqrt{2}}{(2-\sqrt{2})(2+\sqrt{2})}=\frac{2+\sqrt{2}}{2}
\]

よって，求める面積$S$は，

\[
S=8\times\triangle\mathrm{OAB}=8\times\frac{1}{2}a^2\sin 45^{\circ}=8\times\frac{\sqrt{2}}{4}a^2=\sqrt{2}(2+\sqrt{2})=2+2\sqrt{2}
\]

(2) 右の図のように，正十二角形を対角線によって12個の合同な三角形に分け，3点O，A，Bをとると，

\[
\angle\mathrm{AOB}=360^{\circ}\div 12=30^{\circ}
\]

よって，求める面積$S$は，

\[
S=12\times\triangle\mathrm{OAB}=12\times\frac{1}{2}\cdot 1^2\cdot\sin 30^{\circ}=3
\]

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% 例題I4.3.2の側注：多角形の面積（One More）★★
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% 側注1
一般に，半径$a$の円に内接する正$n$角形の面積$S$は，
\[
S=\frac{1}{2}na^2\sin\frac{360^{\circ}}{n}
\]

% 側注2
\begin{align*}
&\mathrm{AB}^2=\mathrm{OA}^2+\mathrm{OB}^2\\
&-2\mathrm{OA}\cdot\mathrm{OB}\cdot\cos\angle\mathrm{AOB}
\end{align*}

% 側注3
$a$の値まで求めなくてよい（$a^2$のまま扱うとよい）．

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% 問題I4.3.2の側注：多角形の面積（One More）★★
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% 側注1
\begin{align*}
&\mathrm{AB}^2=\mathrm{OA}^2+\mathrm{OB}^2\\
&-2\mathrm{OA}\cdot\mathrm{OB}\cdot\cos\angle\mathrm{AOB}
\end{align*}

% 側注2
$a$の値まで求めなくてよい（$a^2$のまま扱うとよい）．

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% 例題I4.3.2の考え方・補足など：多角形の面積（One More）★★
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% 考え方
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正多角形をいくつかの三角形に分割することを考える．例題の場合，対角線によって合同な三角形に分けるとよい．

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正多角形は，合同な三角形に分割して考える．