Harap bantuannya tentang luas segitiga BCD soal tertera digambar

<A = 45
AD = 4cm
AB  = 4√2
<DBC = 60
anggap <C = <A = 45° (tandanya sama dgn yg  45°)
maka <BDC = 180 - 45 - 60 = 75

BD²= AB² + AD² - 2 (AB(AD) cos 45
BD² =32 + 16 - 2(4√2)(4) (1/2 √2)
BD² = 48 - 32 = 16
BD = 4

BD/sin C = DC/sin  60
DC = sin 60/sin 45 . 4
DC = (1/2 √3)/(1/2 √2) (4)= 2√6

luas  BCD = 1/2 (BD)(DC) sin BDC
L = 1/2 (4)(2√6)(sin 75)
L = (4√6) sin (45+30)
L= 4√6 { sin 45 cos 30 + cos 45 sin 30)
L  = 4√6 ( 1/2√2 . 1/2√3  + 1/2√2 . 1/2)
L = 4√6 ( 1/4 √6 + 1/4 √2)
L = 4√6 (1/4)(√6  +√2)
L = √6 (√6+√2)
L = 6 + √12
L = 6 + 2√3