Lanjutan Luas Segitiga 2

D. 3 Aturan Sinus

Perhatikan ilustasi berikut

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

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

Sehingga luas segitiga dapat dituliskan sebagai berikut:

$\begin{array}{rl}1.L△ABC& ={\displaystyle \frac{1}{2}ab\sin \mathrm{\angle }C}\\ & ={\displaystyle \frac{1}{2}a(a.{\displaystyle \frac{\sin \mathrm{\angle }B}{\sin \mathrm{\angle }A}})\sin \mathrm{\angle }C}\\ & ={\displaystyle \frac{1}{2}{a}^2{\displaystyle \frac{\sin \mathrm{\angle }B\sin \mathrm{\angle }C}{\sin \mathrm{\angle }A}}}\\ 2.L△ABC& ={\displaystyle \frac{1}{2}bc\sin \mathrm{\angle }A}\\ & ={\displaystyle \frac{1}{2}b(b.{\displaystyle \frac{\sin \mathrm{\angle }C}{\sin \mathrm{\angle }B}})\sin \mathrm{\angle }A}\\ & ={\displaystyle \frac{1}{2}{b}^2{\displaystyle \frac{\sin \mathrm{\angle }B\sin \mathrm{\angle }A}{\sin \mathrm{\angle }B}}}\\ 3.L△ABC& ={\displaystyle \frac{1}{2}ac\sin \mathrm{\angle }B}\\ & ={\displaystyle \frac{1}{2}(c.{\displaystyle \frac{\sin \mathrm{\angle }A}{\sin \mathrm{\angle }C}})c\sin \mathrm{\angle }B}\\ & ={\displaystyle \frac{1}{2}{c}^2{\displaystyle \frac{\sin \mathrm{\angle }A\sin \mathrm{\angle }B}{\sin \mathrm{\angle }C}}}\end{array}$.

D. 4 Luas segitiga sama sisi

$\begin{array}{rl}{L}_{△}ABC& ={\displaystyle \frac{1}{2}ab\sin \mathrm{\angle }C,a=b=c}\\ & \text{dan}\mathrm{\angle }A=\mathrm{\angle }B\mathrm{\angle }C={60}^{\circ }\\ & ={\displaystyle \frac{1}{2}a.a\sin {60}^{\circ }}\\ & ={\displaystyle \frac{1}{2}{a}^2\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}\\ & ={\displaystyle \frac{1}{4}{a}^2\sqrt{3}}\end{array}$.

D. 5 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:

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

D. 6 Lingkaran dalam segitiga

Perhatikanlah gambar berikut

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

D. 7 Lingkaran singgung segitiga

Sebagai ilustrasinya adalah gambar berikut

$\begin{array}{rl}& \text{Diketahui}\\ & DO=EO=FO=r_a\\ & \text{maka}\\ & 1.L_{△}ABO={\displaystyle \frac{1}{2}(AB)(OD)={\displaystyle \frac{1}{2}c{r}_a}}\\ & 2.L_{△}ACO={\displaystyle \frac{1}{2}(AC)(OE)={\displaystyle \frac{1}{2}b{r}_a}}\\ & 3.L_{△}BCO={\displaystyle \frac{1}{2}(BC)(OF)={\displaystyle \frac{1}{2}a{r}_a}}\end{array}$.

$\begin{array}{rl}\text{Sehingga}& \\ L_{△}ABC& =\left[ABC\right]\\ & =\left[ACO\right]+\left[ABO\right]-\left[BCO\right]\\ & ={\displaystyle \frac{1}{2}b{r}_a+{\displaystyle \frac{1}{2}c{r}_a-{\displaystyle \frac{1}{2}a{r}_a}}}\\ & ={\displaystyle \frac{1}{2}{r}_a(b+c-a)}\\ & ={\displaystyle \frac{1}{2}{r}_a(a+b+c-2a)}\\ & ={\displaystyle \frac{1}{2}{r}_a(2s-2a)}\\ & =r_a(s-a)\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diberikan sembarang}△ABC.\text{Jika}r\\ & \text{merupakan jari-jari lingkaran singgung }\\ & \text{dalam pada}△ABC\text{dan}{r}_a,r_b,r_c\\ & \text{adalah jari-jari singgung luar pad}△ABC\\ & \text{tunjukkan bahwa}:{\displaystyle \frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}=\frac{1}{r}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\text{Diketahu}& \text{i}\\ L_{△}ABC& =r_a(s-a),\Rightarrow r_a={\displaystyle \frac{\left[ABC\right]}{s-a}}\\ L_{△}ABC& =r_b(s-b),\Rightarrow r_b={\displaystyle \frac{\left[ABC\right]}{s-b}}\\ L_{△}ABC& =r_c(s-c),\Rightarrow r_c={\displaystyle \frac{\left[ABC\right]}{s-c}}\\ \text{maka}& \\ {\displaystyle \frac{1}{r_a}+\frac{1}{r_b}}& +\frac{1}{r_c}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{\left[ABC\right]}{s-a}}}+{\displaystyle \frac{1}{{\displaystyle \frac{\left[ABC\right]}{s-b}}}+{\displaystyle \frac{1}{{\displaystyle \frac{\left[ABC\right]}{s-c}}}}}}\\ & ={\displaystyle \frac{s-a}{\left[ABC\right]}+{\displaystyle \frac{s-b}{\left[ABC\right]}+{\displaystyle \frac{s-c}{\left[ABC\right]}}}}\\ & ={\displaystyle \frac{s-a+s-b+s-c}{\left[ABC\right]}}\\ & ={\displaystyle \frac{3s-(a+b+c)}{\left[ABC\right]}}\\ & ={\displaystyle \frac{3s-2s}{\left[ABC\right]}}\\ & ={\displaystyle \frac{s}{\left[ABC\right]}}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{\left[ABC\right]}{s}}}}\\ & ={\displaystyle \frac{1}{r}◼}\end{array}\end{array}$.

Lanjutan Luas Segitiga 1

C. Luas Segitiga dengan Integral Tentu

Misalkan suatu garis linear di sekitar  sumbu-X yang tidak sejajar dan sumbu-X itu sendiri yang membatasi suatu daerah di antara keduanya serta dibatasi pula oleh 2 garis yang sejajar dengan sumbu Y yang keduanya tidak berimpit, maka hasil dari proses integral tentu ini akan menghasilkan luas segitiga.

Secara rumus integral tentu untuk model di atas adalah:

$\begin{array}{rl}& \text{Integral Tentu}\\ & ={\displaystyle {\int }_p^{q}f(x)dx={\displaystyle {[F(x)]}_p^q}}\\ & =F(x){|}_p^q=F(q)-F(p)\end{array}$.

$\begin{array}{rl}& \text{Diketahui}\\ & y=mx,\text{dengan}m={\displaystyle \frac{b}{a},}\\ & \text{sehingga}\\ & y={\displaystyle \frac{b}{a}x\text{atau}f(x)={\displaystyle \frac{b}{a}x}}\\ & \text{maka}\\ & L_{area}={\displaystyle {\int }_a^{b}f(x)dx}\\ & ={\displaystyle {\int }_a^b{\displaystyle \frac{b}{a}xdx}}\\ & ={\displaystyle \frac{b}{a}{\displaystyle {\int }_a^{b}xdx}}\\ & ={\displaystyle \frac{b}{a}\left({\displaystyle \frac{1}{2}{x}^2}\right)\begin{array}{c}a\\ |\\ 0\end{array}}\\ & ={\displaystyle \frac{b}{a}\left({\displaystyle \frac{1}{2}(a{)}^2}\right)-{\displaystyle \frac{b}{a}\left({\displaystyle \frac{1}{2}(0{)}^2}\right)}}\\ & ={\displaystyle \frac{b}{a}\left({\displaystyle \frac{1}{2}(a{)}^2}\right)-0}\\ & ={\displaystyle \frac{b\times a^2}{2a}}\\ & ={\displaystyle \frac{a\times b}{2}}\\ & ={\displaystyle \frac{1}{2}\times ab}\end{array}$.

D. Luas segitiga dengan Trigonometri

D. 1 Segitiga siku-siku

Perhatikan ilustrasi segitiga siku-siku berikut

$\begin{array}{rl}& 1.\sin \mathrm{\angle }B={\displaystyle \frac{b}{a}\text{dan}\cos \mathrm{\angle }C={\displaystyle \frac{b}{a}}}\\ & \text{sehingga}b=a\sin \mathrm{\angle }B=a\cos \mathrm{\angle }C\\ & 2.\sin \mathrm{\angle }C={\displaystyle \frac{c}{a}\text{dan}\cos \mathrm{\angle }B={\displaystyle \frac{c}{a}}}\\ & \text{sehingga}c=a\sin \mathrm{\angle }C=a\cos \mathrm{\angle }B\\ & 3.\tan \mathrm{\angle }B={\displaystyle \frac{b}{c}\text{dan}\cot \mathrm{\angle }C={\displaystyle \frac{b}{c}}}\\ & \text{sehingga}b=c\tan \mathrm{\angle }B=c\cot \mathrm{\angle }C\\ & 4.\tan \mathrm{\angle }C={\displaystyle \frac{c}{b}\text{dan}\cot \mathrm{\angle }B={\displaystyle \frac{c}{b}}}\\ & \text{sehingga}c=b\tan \mathrm{\angle }C=b\cot \mathrm{\angle }B\\ & 5.\sin \mathrm{\angle }B={\displaystyle \frac{h}{c}\text{dan}\sin \mathrm{\angle }C={\displaystyle \frac{h}{b}}}\\ & \text{sehingga}h=c\sin \mathrm{\angle }B=b\sin \mathrm{\angle }C\end{array}$.

Dari fakta-fakta di atas dapat ditunjukkan beberapa rumus segitiga, yaitu:

$\begin{array}{|ccl|}\hline 1.& L_{△}& {\displaystyle \frac{1}{2}bc}\\ & & \begin{array}{rl}& {\displaystyle \frac{1}{2}(a\sin \mathrm{\angle }B)(a\sin \mathrm{\angle }C)}\\ & ={\displaystyle \frac{1}{2}{a}^2\sin \mathrm{\angle }B\sin \mathrm{\angle }C}\end{array}\\ & & \begin{array}{rl}& {\displaystyle \frac{1}{2}(a\cos \mathrm{\angle }C)(a\cos \mathrm{\angle }B)}\\ & ={\displaystyle \frac{1}{2}{a}^2\cos \mathrm{\angle }B\cos \mathrm{\angle }C}\end{array}\\ & & \begin{array}{rl}& {\displaystyle \frac{1}{2}(b)(b\tan \mathrm{\angle }C)}\\ & ={\displaystyle \frac{1}{2}{b}^2\tan \mathrm{\angle }C}\end{array}\\ & & \begin{array}{rl}& {\displaystyle \frac{1}{2}(c\tan \mathrm{\angle }B)(c)}\\ & ={\displaystyle \frac{1}{2}{c}^2\tan \mathrm{\angle }B}\end{array}\\ & & \begin{array}{rl}& {\displaystyle \frac{1}{2}(b)(b\cot \mathrm{\angle }B)}\\ & ={\displaystyle \frac{1}{2}{b}^2\cot \mathrm{\angle }B}\end{array}\\ & & \begin{array}{rl}& {\displaystyle \frac{1}{2}(c\cot \mathrm{\angle }C)(c)}\\ & ={\displaystyle \frac{1}{2}{c}^2\cot \mathrm{\angle }C}\end{array}\\ 2.& L_{△}& {\displaystyle \frac{1}{2}ha}\\ & & \begin{array}{rl}& {\displaystyle \frac{1}{2}(c\sin \mathrm{\angle }B)(a)}\\ & ={\displaystyle \frac{1}{2}ac\sin \mathrm{\angle }B}\end{array}\\ & & \begin{array}{rl}& {\displaystyle \frac{1}{2}(b\sin \mathrm{\angle }C)(a)}\\ & ={\displaystyle \frac{1}{2}ab\sin \mathrm{\angle }C}\end{array}\\ \hline\end{array}$.

D. 2 Segitiga tidak siku-siku

Perhatikan ilustrasi segitiga tidak siku-siku berikut

Jika dari tiap titik sudut ditarik garis tinggi sampai memotong sisi di depannya, misal titik A, maka garis tingginya sebagaimana gambar berikut:

$\begin{array}{rl}1.L_{△}& ={\displaystyle \frac{1}{2}{h}_A.a}\\ & ={\displaystyle \frac{1}{2}c\sin \mathrm{\angle }B.(a)={\displaystyle \frac{1}{2}ac\sin \mathrm{\angle }B}}\\ L_{△}& ={\displaystyle \frac{1}{2}{h}_A.a}\\ & ={\displaystyle \frac{1}{2}b\sin \mathrm{\angle }C.(a)={\displaystyle \frac{1}{2}ab\sin \mathrm{\angle }C}}\\ 2.L_{△}& ={\displaystyle \frac{1}{2}{h}_B.b}\\ & ={\displaystyle \frac{1}{2}c\sin \mathrm{\angle }A.(b)={\displaystyle \frac{1}{2}bc\sin \mathrm{\angle }A}}\\ L_{△}& ={\displaystyle \frac{1}{2}{h}_B.b}\\ & ={\displaystyle \frac{1}{2}a\sin \mathrm{\angle }C.(b)={\displaystyle \frac{1}{2}ab\sin \mathrm{\angle }C}}\\ 3.L_{△}& ={\displaystyle \frac{1}{2}{h}_C.c}\\ & ={\displaystyle \frac{1}{2}a\sin \mathrm{\angle }B.(c)={\displaystyle \frac{1}{2}ac\sin \mathrm{\angle }B}}\\ L_{△}& ={\displaystyle \frac{1}{2}{h}_C.c}\\ & ={\displaystyle \frac{1}{2}b\sin \mathrm{\angle }A.(c)={\displaystyle \frac{1}{2}bc\sin \mathrm{\angle }A}}\end{array}$.

Jika diringkas menjadi

$\begin{array}{rl}\mathbf{\text{Luas}}\mathrm{△}ABC& =\frac{1}{2}bc.\sin \mathrm{\angle }A\\ & =\frac{1}{2}ac.\sin \mathrm{\angle }B\\ & =\frac{1}{2}ab.\sin \mathrm{\angle }C\end{array}$.