LINGKARAN-DALAM, LUAR DAN SINGGUNG SEGITIGA-LANJUTAN

 I. Lingkaran Dalam, Luar, dan Singgung Segitiga

Perhatikan ilustasi berikut

$\Large\begin{array}{|c|}\hline \displaystyle \frac{a}{\sin A}= \frac{b}{\sin B}=\frac{c}{\sin C}=2R\\\hline \end{array}$.

$\begin{aligned}1.\quad a&=b.\displaystyle \frac{\sin \angle A}{\sin \angle B}=c.\displaystyle \frac{\sin \angle A}{\sin \angle C}=2R\sin \angle A\\ 2.\quad b&=c.\displaystyle \frac{\sin \angle B}{\sin \angle C}=a.\displaystyle \frac{\sin \angle B}{\sin \angle A}=2R\sin \angle B\\ 3.\quad c&=a.\displaystyle \frac{\sin \angle C}{\sin \angle A}=b.\displaystyle \frac{\sin \angle C}{\sin \angle B}=2R\sin \angle C \end{aligned}$.

Sehingga luas segitiga dapat dituliskan sebagai berikut:

$\begin{aligned} 1.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}ab\sin \angle C\\ &=\displaystyle \frac{1}{2}a\left ( a.\displaystyle \frac{\sin \angle B}{\sin \angle A} \right )\sin \angle C\\ &=\displaystyle \frac{1}{2}a^{2}\displaystyle \frac{\sin \angle B\sin \angle C}{\sin \angle A}\\ 2.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}bc\sin \angle A\\ &=\displaystyle \frac{1}{2}b\left ( b.\displaystyle \frac{\sin \angle C}{\sin \angle B} \right )\sin \angle A\\ &=\displaystyle \frac{1}{2}b^{2}\displaystyle \frac{\sin \angle B\sin \angle A}{\sin \angle B}\\ 3.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}ac\sin \angle B\\ &=\displaystyle \frac{1}{2}\left ( c.\displaystyle \frac{\sin \angle A}{\sin \angle C} \right )c\sin \angle B\\ &=\displaystyle \frac{1}{2}c^{2}\displaystyle \frac{\sin \angle A\sin \angle B}{\sin \angle C} \end{aligned}$.

I.1 Luas segitiga berdasar tiga sisinya (Heron's formula)

Bukti Luas Segitiga dengan sisi a, b, dan c

$\begin{aligned}& \textrm{Bagaimana membuktikan luas suatu}\\ &\textrm{segitiga jika diketahui sisinya}\: a,b\: \textrm{dan}\: c\\ &\textrm{berupa rumus}\\ &L_{\bigtriangleup }=\left [ ABC \right ]=\sqrt{s(s-a)(s-b)(s-c)}\\ & \textrm{dengan}\\ &\qquad s=\displaystyle \frac{1}{2}(a+b+c) \end{aligned}$.

Berikut akan dipaparkan buktinya

$\begin{aligned}\displaystyle \textrm{L}{\bigtriangleup }\textrm{ABC}&=\frac{1}{2}bc\sin\angle A\\ &=\displaystyle \frac{1}{2}bc\sqrt{\sin ^{2}\angle A}\\ &=\displaystyle \frac{1}{2}\sqrt{b^{2}c^{2}\left ( \sin ^{2}\angle A \right )}\\ &=\displaystyle \frac{1}{2}\sqrt{b^{2}c^{2}\left ( 1-\cos ^{2}\angle A \right )},\\ &\textrm{ingat bahwa};\: \cos \angle A=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ &=\displaystyle \frac{1}{2}\sqrt{b^{2}c^{2}\left ( 1-\left ( \frac{b^{2}+c^{2}-a^{2}}{2bc} \right )^{2} \right )}\\ &=\displaystyle \frac{1}{2}\sqrt{b^{2}c^{2}-\left ( \frac{b^{2}+c^{2}-a^{2}}{2} \right )^{2}}\\ &=\displaystyle \frac{1}{2}\sqrt{\frac{4b^{2}c^{2}-\left ( b^{2}+c^{2}-a^{2} \right )^{2}}{4}}\\ &=\displaystyle \frac{1}{2}.\frac{1}{2}\sqrt{\left ( 2bc \right )^{2}-\left ( b^{2}+c^{2}-a^{2} \right )^{2}}\\ &=\displaystyle \frac{1}{4}\sqrt{\left ( 2bc+b^{2}+c^{2}-a^{2} \right )\left (2bc-b^{2}-c^{2}+a^{2} \right )}\\ &=\frac{1}{4}\sqrt{\left \{ \left ( b+c \right )^{2}-a^{2} \right \}\left \{ a^{2}-\left ( b-c \right )^{2} \right \}}\\ &=\displaystyle \frac{1}{4}\sqrt{\left ( b+c+a \right )\left ( b+c-a \right )\left ( a+b-c \right )\left ( a-b+c \right )},\\ &\textrm{dengan mengingat bahwa}\: 2s=a+b+c\\ &=\displaystyle \frac{1}{4}\sqrt{(2s)(2s-2a)(2s-2b)(2s-2c)}\\ &=\displaystyle \frac{1}{4}.\sqrt{16.s(s-a)(s-b)(s-c)}\\ &=\displaystyle \frac{1}{4}.4\sqrt{s(s-a)(s-b)(s-c)}\\ \textrm{L}\bigtriangleup \textrm{ABC}&=\sqrt{s(s-a)(s-b)(s-c)}\quad \blacksquare \end{aligned}$.

Rumus di atas lebih dikenal dengan istilah rumus Heron lihat Heron's formula di sini.

Sumber tulisan lagi di antara silahkan kunjungi di sini.

I.2 Luas segitiga sama sisi

$\begin{aligned}L_{\bigtriangleup }ABC&=\displaystyle \frac{1}{2}ab\sin \angle C,\quad a=b=c\\ &\qquad\quad\quad \textrm{dan}\: \: \angle A=\angle B\angle C=60^{\circ}\\ &=\displaystyle \frac{1}{2}a.a\sin 60^{\circ}\\ &=\displaystyle \frac{1}{2}a^{2}\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{1}{4}a^{2}\sqrt{3} \end{aligned}$.

I.3 Lingkaran Luar Segitiga

Perhatikan lagi lingkaran luar segitiga di atas, dari sana kita akan mendapatkan rumus luas segitiga yang dapat kita munculkan harga R nya, yaitu:

$\begin{aligned}1.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}ab\sin \angle C\\ &=\displaystyle \frac{1}{2}(2R\sin \angle A)(2R\sin \angle B)\sin \angle C\\ &=2R^{2}\sin \angle A\sin \angle B\sin \angle C\\ 2.\quad L\bigtriangleup ABC&=\displaystyle \frac{1}{2}ab\sin \angle C\\ &=\displaystyle \frac{1}{2}ab\left ( \displaystyle \frac{c}{2R} \right )\\ &=\displaystyle \frac{abc}{4R} \end{aligned}$.

I.4 Lingkaran dalam segitiga

Perhatikanlah gambar berikut

$\begin{aligned}\textrm{Diketahu}&\textrm{i}\\ L_{\bigtriangleup }AOB&=\displaystyle \frac{1}{2}(AB)(OD)=\displaystyle \frac{1}{2}cr\\ L_{\bigtriangleup }AOC&=\displaystyle \frac{1}{2}(AC)(OF)=\displaystyle \frac{1}{2}br\\ L_{\bigtriangleup }BOC&=\displaystyle \frac{1}{2}(BC)(OE)=\displaystyle \frac{1}{2}ar\\ \textrm{Sehingga}&\\ L_{\bigtriangleup }ABC&=\left [ ABC \right ]\\ &=\displaystyle \frac{1}{2}ar+\displaystyle \frac{1}{2}br+\displaystyle \frac{1}{2}cr\\ &=\displaystyle \frac{1}{2}r(a+b+c)\\ &=\displaystyle \frac{1}{2}r(2s)\\ &=rs \end{aligned}$.

I.5 Lingkaran singgung segitiga

Sebagai ilustrasinya adalah gambar berikut

$\begin{aligned}&\textrm{Diketahui}\\ &DO=EO=FO=r_{a}\\ &\textrm{maka}\\ &1.\quad L_{\bigtriangleup}ABO=\displaystyle \frac{1}{2}(AB)(OD)=\displaystyle \frac{1}{2}cr_{a}\\ &2.\quad L_{\bigtriangleup}ACO=\displaystyle \frac{1}{2}(AC)(OE)=\displaystyle \frac{1}{2}br_{a}\\ &3.\quad L_{\bigtriangleup}BCO=\displaystyle \frac{1}{2}(BC)(OF)=\displaystyle \frac{1}{2}ar_{a} \end{aligned}$.

$\begin{aligned} \textrm{Sehingga}&\\ L_{\bigtriangleup }ABC&=\left [ ABC \right ]\\ &=\left [ ACO \right ]+\left [ ABO \right ]-\left [ BCO \right ]\\ &=\displaystyle \frac{1}{2}br_{a}+\displaystyle \frac{1}{2}cr_{a}-\displaystyle \frac{1}{2}ar_{a}\\ &=\displaystyle \frac{1}{2}r_{a}(b+c-a)\\ &=\displaystyle \frac{1}{2}r_{a}(a+b+c-2a)\\ &=\displaystyle \frac{1}{2}r_{a}(2s-2a)\\ &=r_{a}(s-a) \end{aligned}$.

$\LARGE\fbox{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diberikan sembarang}\: \: \bigtriangleup ABC\: .\: \textrm{Jika}\: \: r\\ & \textrm{merupakan jari-jari lingkaran singgung }\\ &\textrm{dalam pada}\: \: \bigtriangleup ABC\: \: \textrm{dan}\: \: r_{a},\: r_{b},\: r_{c}\\ &\textrm{adalah jari-jari singgung luar pad}\: \: \bigtriangleup ABC\\ &\textrm{tunjukkan bahwa}:\: \displaystyle \frac{1}{r_{a}}+\frac{1}{r_{b}}+\frac{1}{r_{c}}=\frac{1}{r}\\\\ &\textbf{Bukti}:\\ &\begin{aligned} \textrm{Diketahu}&\textrm{i}\\ L_{\bigtriangleup }ABC&=r_{a}(s-a),\: \Rightarrow r_{a}=\displaystyle \frac{\left [ ABC \right ]}{s-a}\\ L_{\bigtriangleup }ABC&=r_{b}(s-b),\: \Rightarrow r_{b}=\displaystyle \frac{\left [ ABC \right ]}{s-b}\\ L_{\bigtriangleup }ABC&=r_{c}(s-c),\: \Rightarrow r_{c}=\displaystyle \frac{\left [ ABC \right ]}{s-c}\\ \textrm{maka}\: \quad&\\ \displaystyle \frac{1}{r_{a}}+\frac{1}{r_{b}}&+\frac{1}{r_{c}}\\ &=\displaystyle \frac{1}{\displaystyle \frac{\left [ ABC \right ]}{s-a}}+\displaystyle \frac{1}{\displaystyle \frac{\left [ ABC \right ]}{s-b}}+\displaystyle \frac{1}{\displaystyle \frac{\left [ ABC \right ]}{s-c}}\\ &=\displaystyle \frac{s-a}{\left [ ABC \right ]}+\displaystyle \frac{s-b}{\left [ ABC \right ]}+\displaystyle \frac{s-c}{\left [ ABC \right ]}\\ &=\displaystyle \frac{s-a+s-b+s-c}{\left [ ABC \right ]}\\ &=\displaystyle \frac{3s-(a+b+c)}{\left [ ABC \right ]}\\ &=\displaystyle \frac{3s-2s}{\left [ ABC \right ]}\\ &=\displaystyle \frac{s}{\left [ ABC \right ]}\\ &=\displaystyle \frac{1}{\displaystyle \frac{\left [ ABC \right ]}{s}}\\ &=\displaystyle \frac{1}{r}\qquad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Pada}\: \: \bigtriangleup ABC\: ,\: \textrm{Jika}\: \: AB=20\: \textrm{cm},\: BC=12\: \textrm{cm}\\ &AC=16\:\textrm{cm}.\quad \textrm{Tentukan jari-jari lingkaran}\\ & \textrm{dalam}\quad \bigtriangleup ABC\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{array}$.

$\begin{aligned}\qquad&\textrm{Karena}\:\: BC^{\displaystyle 2}+AC^{\displaystyle 2}=AB^{\displaystyle 2},\:\: \textrm{maka}\:\: \bigtriangleup ABC\\ &\textrm{adalah segitiga siku-siku di}\:\: C\\ & \textrm{Dan karena}\:\: r\:\: \textrm{jari-jari lingkaran dalam}\:\: \bigtriangleup ABC,\\ &\textrm{maka}\\ &BC=12-r+16-r\\ &\Leftrightarrow 20=28-2r\\ &\Leftrightarrow r=4 \end{aligned}$.

DAFTRA PUSTAKA

1. Isnaini, H.F., Santoso, N.E. 2023. Matematika untuk SMA/SMK/MAK Kelas 11A Kurikulum Merdeka. Yogyakarta: PENERBIT INTAN PARIWARA.
2. Maulan, S.F. 2010. Juara Olimpiade Matematika SMA. Jakarta: WAHYUMEDIA.
3. Sembiring, S., Sukino. 2020. Super Master KSN Matematika SMA/MA. Bandung: YRAMA WIDYA.