Diagonal intersect

isosceles trapezoid ABCD with length bases | AB | = 6 cm, CD | = 4 cm is divided into 4 triangles by the diagonals intersecting at point S. How much of the area of the trapezoid are ABS and CDS triangles?

Correct result:

r1 =  0.36
r2 =  0.16


a=6 cm c=4 cm  h=h1+h2 h1:h2=a:c  h1=h aa+c  S=a+c2 h  S1=a h12 S2=c h22  r1=S1/S=a h12/(a+c) h2 r1=a h1(a+c) h  r1=a h aa+c(a+c) h r1=a aa+c(a+c)  r1=a2(a+c)2=62(6+4)2=925=0.36
r2=S2/S=c h22/(a+c) h2 r2=c h2(a+c) h  r2=c h ca+c(a+c) h r2=c ca+c(a+c)  r2=c2(a+c)2=42(6+4)2=425=0.16

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