Diskusi 3. Menghitung Luas Segitiga Def

Diketahui luas segitiga ABC pada gambar yaitu 50 cm$^2$. AD : DB = BE : EC = CF : FA = 2. Luas segitiga DEF yaitu ....

Pembahasan:

$\begin{align} Luas \ ABC &= \frac{1}{2}.AC.AB.\sin \ A \\ &= \frac{1}{2}.5b.5c.\sin A \\ Luas \ ABC &= \frac{1}{2}.25bc.\sin A \end{align}$

$\begin{align} Luas \ ABC &= \frac{1}{2}.BC.AB.\sin \ B \\ &= \frac{1}{2}.5a.5c.\sin A \\ Luas \ ABC &= \frac{1}{2}.25ac.\sin B \end{align}$

$\begin{align} Luas \ ABC &= \frac{1}{2}.AC.BC.\sin \ C \\ &= \frac{1}{2}.5b.5a.\sin C \\ Luas \ ABC &= \frac{1}{2}.25ab.\sin C \end{align}$ 

Jumlahkan ketiga persamaan di atas, maka:
$3.Luas \ ABC = \frac{25}{2}(bc.\sin A + ac.\sin B + ab.\sin C)$
$3 \times 50 = \frac{25}{2}.(bc.\sin A + ac.\sin B + ab.\sin C)$
$12 = bc.\sin A + ac.\sin B + ab.\sin C$

$Luas \ DAF = \frac{1}{2}.3b.2c.\sin A = 3bc.\sin A$
$Luas \ DBE = \frac{1}{2}.3c.2a.\sin B = 3ac.\sin B$
$Luas \ FCE = \frac{1}{2}.2b.3a.\sin C = 3ab.\sin C$

Berdasarkan gambar
$[DAF] + [DBE] + [FCE] + [DEF] = [ABC]$
$3bc.\sin A + 3ac.\sin B + 3ab.\sin C + [DEF] = 50$
$3(bc.\sin A + ac.\sin B + ab.\sin C) + [DEF] = 50$
$3.12 + [DEF] = 50$
$[DEF] = 50 - 36$
$[DEF] = 14$

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