Contoh Soal Menghitung Panjang Garis Berat Segitiga
Anda harus paham materi dan silakan baca jika belum paham teori dan rumus dasarnya:
Soal:
Terdapat segitiga ABC dengan garis berat AD = $ \sqrt{10}, \, BE = \sqrt{31}, \, $ CF dan panjang AB = 4 cm. Tentukan panjang sisi-sisi segitiga lainnya dan panjang garis berat CF!
Penyelesaian:
Persamaan dari panjang garis berat.
Garis berat $ AD = \sqrt{10} $
$ \begin{align} AD^2 & = \frac{1}{2} AB^2 + \frac{1}{2}AC^2 - \frac{1}{4} BC^2 \\ (\sqrt{10})^2 & = \frac{1}{2} .4^2 + \frac{1}{2} b^2 - \frac{1}{4} a^2 \\ 10 & = 8 + \frac{1}{2} b^2 - \frac{1}{4} a^2 \, \, \, \, \, \text{(kali 4)} \\ 40 & = 32 + 2b^2 - a^2 \\ -a^2 + 2b^2 & = 8 \, \, \, \, \, \text{....pers(i)} \end{align} $
Garis berat $ BE = \sqrt{31} $
$ \begin{align} BE^2 & = \frac{1}{2} AB^2 + \frac{1}{2}BC^2 - \frac{1}{4} AC^2 \\ (\sqrt{31})^2 & = \frac{1}{2} .4^2 + \frac{1}{2} a^2 - \frac{1}{4} b^2 \\ 31 & = 8 + \frac{1}{2} a^2 - \frac{1}{4} b^2 \, \, \, \, \, \text{(kali 4)} \\ 124 & = 32 + 2a^2 - b^2 \\ 2a^2 - b^2 & = 92 \, \, \, \, \, \text{....pers(ii)} \end{align} $
Eliminasi pers(i) dan pers(ii) :
$ \begin{array}{c|c|cc} -a^2 + 2b^2 = 8 & \text{kali 2} & -2a^2 + 4b^2 = 16 & \\ 2a^2 - b^2 = 92 & \text{kali 2} & 2a^2 - b^2 = 92 & + \\ \hline & & 3b^2 = 108 & \\ & & b^2 = 36 & \\ & & b = 6 & \end{array} $
Pers(i) : $ -a^2 + 2b^2 = 8 \rightarrow -a^2 + 2. 6^2 = 8 \rightarrow a^2 = 64 \rightarrow a = 8 $.
Kita peroleh panjang sisi-sisi segitiganya : AB = 4 cm, BC = 8 cm, dan AC = 6 cm.
Panjang garis berat CF,
$ \begin{align} CF^2 & = \frac{1}{2} BC^2 + \frac{1}{2}AC^2 - \frac{1}{4} AB^2 \\ & = \frac{1}{2} .8^2 + \frac{1}{2}.6^2 - \frac{1}{4} .4^2 \\ & = 32 + 18 - 4 \\ CF^2 & = 46 \\ CF & = \sqrt{46} \end{align} $
Jadi, panjang garis berat $ CF = \sqrt{46} \, $ cm
Garis berat $ AD = \sqrt{10} $
$ \begin{align} AD^2 & = \frac{1}{2} AB^2 + \frac{1}{2}AC^2 - \frac{1}{4} BC^2 \\ (\sqrt{10})^2 & = \frac{1}{2} .4^2 + \frac{1}{2} b^2 - \frac{1}{4} a^2 \\ 10 & = 8 + \frac{1}{2} b^2 - \frac{1}{4} a^2 \, \, \, \, \, \text{(kali 4)} \\ 40 & = 32 + 2b^2 - a^2 \\ -a^2 + 2b^2 & = 8 \, \, \, \, \, \text{....pers(i)} \end{align} $
Garis berat $ BE = \sqrt{31} $
$ \begin{align} BE^2 & = \frac{1}{2} AB^2 + \frac{1}{2}BC^2 - \frac{1}{4} AC^2 \\ (\sqrt{31})^2 & = \frac{1}{2} .4^2 + \frac{1}{2} a^2 - \frac{1}{4} b^2 \\ 31 & = 8 + \frac{1}{2} a^2 - \frac{1}{4} b^2 \, \, \, \, \, \text{(kali 4)} \\ 124 & = 32 + 2a^2 - b^2 \\ 2a^2 - b^2 & = 92 \, \, \, \, \, \text{....pers(ii)} \end{align} $
Eliminasi pers(i) dan pers(ii) :
$ \begin{array}{c|c|cc} -a^2 + 2b^2 = 8 & \text{kali 2} & -2a^2 + 4b^2 = 16 & \\ 2a^2 - b^2 = 92 & \text{kali 2} & 2a^2 - b^2 = 92 & + \\ \hline & & 3b^2 = 108 & \\ & & b^2 = 36 & \\ & & b = 6 & \end{array} $
Pers(i) : $ -a^2 + 2b^2 = 8 \rightarrow -a^2 + 2. 6^2 = 8 \rightarrow a^2 = 64 \rightarrow a = 8 $.
Kita peroleh panjang sisi-sisi segitiganya : AB = 4 cm, BC = 8 cm, dan AC = 6 cm.
Panjang garis berat CF,
$ \begin{align} CF^2 & = \frac{1}{2} BC^2 + \frac{1}{2}AC^2 - \frac{1}{4} AB^2 \\ & = \frac{1}{2} .8^2 + \frac{1}{2}.6^2 - \frac{1}{4} .4^2 \\ & = 32 + 18 - 4 \\ CF^2 & = 46 \\ CF & = \sqrt{46} \end{align} $
Jadi, panjang garis berat $ CF = \sqrt{46} \, $ cm
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