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

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