My solution: Extend AD until it intersects point E on BC. We now have 2 triangles: △ABE with ∠BAE = 30°, ∠ABE = 90° → ∠AEB = 60° △CDE with ∠CED = 180° − 60° = 120° Since △ABE is a 30-60-90 triangle, with longer leg (opposite 60° angle) AB = 4, then BE = 4/√3 (shorter leg opposite 30° angle) AE = 8/√3 (hypotenuse) Let AD = BC = x In △CDE we have: CE = BC − BE = x − 4/√3 DE = AE − AD = 8/√3 − x CD = 2 ∠CED = 120° Using law of cosines, we get: CD² = CE² + DE² − 2(CE)(DE)cos(∠CED) 2² = (x − 4/√3)² + (8/√3 − x)² − 2(x−4/√3)(8/√3−x)(−1/2) 4 = (x² − 8x/√3 + 16/3) + (x² − 16x/√3 + 64/3) + (−x² + 8x/√3 + 4x/√3 − 32/3) 4 = x² − (4√3)x + 16 x² − (4√3)x + 12 = 0 (x − 2√3)² = 0 x = 2√3 Substituting, we get: CE = x − 4/√3 = 2√3 − 4/√3 = 6/√3 − 4/√3 = 2/√3 DE = AE − AD = 8/√3 − 2√3 = 8/√3 − 6/√3 = 2/√3 Area(ABCD) = Area(△ABE) + Area(△CDE) = 1/2 × AB × BE + 1/2 × CE × DE × sin(120°) = 1/2 × 4 × 4/√3 + 1/2 × 2/√3 × 2/√3 × √3/2 = 8/√3 + 1/√3 = 9/√3 = 3√3

Once you know that DB is 2, with DAB = 30o and AB = 4 = 2DB, you have a right triangle, so AD = 2Sqrt[3]. Area ADB = 2Sqrt[3]*2/2. Draw a horizontal through D, label the intersect with CB as E. Angle EDB = (360-60-90-90)/2 = 60o, so another 30-60-90 triangle, and the area of DBC = 1*Sqrt[3]. Add them together for area of 3Sqrt[3]. Avoids any trig.

At 12:55, we can note that quadrilateral CDEB is composed of 3 congruent triangles, ΔDBE, ΔDBF and ΔDCF. The area of each is (1/2)(1)(√3) = (√3)/2, total area (3√3)/2. Adding the area of ΔADE, (3√3)/2, the total area is 3√3.

Extend AD until it intersects BC in point E a=AB=4 c=AE= a/cos (30)= 2a/sqrt(3) b=BE=a tg 30=a/sqrt(3) u=CD=2 S1= ab/2= a²/(2sqrt(3)) area of ABE x=AD=BE S2= (c-x)(x-b) sin(120)/2= (c-x)(b-x)sqrt(3)/4 area of CED u²=(c-x)²+(x-b)²-2(c-x)(x-b)cos(120) u²=((c-x)+(x-b))²-2(c-x)(x-b)(1-1/2) (c-x)(x-b)=(c-b)²-u² substitute (c-x)(x-b)= 4 S2/sqrt(3) 4S2/sqrt(3)=(c-b)²-u² (a²-3u²)/3=4S2/sqrt(3) S2= (a²-3u²)/(4sqrt(3)) S= S1+S2 S=a²/(2sqrt(3))+(a²-3u²)/(4sqrt(3))=(a²-u²)sqrt(3)/4 =(16-4)sqrt(3)/4=3sqrt(3) is answer

for second solution EDC and ADB are right triangles because sin30 = 2/4, it's easier to calculate

In ADE, DE =a, AE =art3. In DEB, EB=DEtan30=a/rt3. AB=AE + EB, so art3+a/rt3 = 4, => a=rt3. This saves using cosine rule and solving a quadratic. Rest is no problem.

Io l'ho risolto così...posto a =AD...risulta arctg((4-√3/2a)/(a/2))+arcsin a/4=90...applico cos all'equazione,perciò svolgendo i calcoli,a^2-4√3a+2=0...a=2√3...adesso sommo l'area del trapezio più il triangolo..Atot=3√3..forse è un po' strano l'uso di arctg+arcsin +applicazione del coseno,ma altre forme mi davano equazioni di 4 grado in a

5.6 sq unit

there are 2 equations with 2 unknown dimensions: print sin(rad(45)) 10 dim x(3),y(3):sw=.1:w1=30:l2=4:l3=2:sw=.01:w2=sw:goto 50 20 dgu1=(l2-l3*sin(rad(w2)))/l3/cos(rad(w1)):dgu2=cos(rad(w2))/(1-sin(rad(w1))) 30 dg=dgu1-dgu2:return 50 gosub 20 60 dg1=dg:w21=w2:w2=w2+sw:if w2>90 then stop 70 w22=w2:gosub 20:if dg1*dg>0 then 60 80 l1=l3*sin(rad(w2))/(1-sin(rad(w1))):print w1,w2 90 print "l1=";l1,"der 2.winkel=";w2 100 x(0)=0:y(0)=0:x(1)=l2:y(1)=0:x(2)=x(1):y(2)=l1:x(3)=l1*cos(rad(w1)):y(3)=l1*sin(rad(w1)) 110 mass=800/l2:goto 130 120 xbu=x*mass:ybu=y*mass:return 130 xba=0:yba=0:for a=1 to 4:ia=a:if ia=4 then ia=0 140 x=x(ia):y=y(ia):gosub 120:xbn=xbu:ybn=ybu 150 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:next a 0.707106781 30 30 l1=2 der 2.winkel=30 > run in bbc basic sdl and hit ctrl tab to copy from the results windos

I see this as a coordinate geometry problem. Find expressions for the coordinates of C and D in terms of x… and use Pythagoras to make CD = 2.