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?

Result

r1 =  0.36
r2 =  0.16

Solution:

$a=6 \ \text{cm} \ \\ c=4 \ \text{cm} \ \\ \ \\ h=h_{1}+h_{2} \ \\ h_{1}:h_{2}=a:c \ \\ \ \\ h_{1}=h \cdot \ \dfrac{ a }{ a+c } \ \\ \ \\ S=\dfrac{ a+c }{ 2 } \cdot \ h \ \\ \ \\ S_{1}=\dfrac{ a \cdot \ h_{1} }{ 2 } \ \\ S_{2}=\dfrac{ c \cdot \ h_{2} }{ 2 } \ \\ \ \\ r_{1}=S_{1}/S=\dfrac{ a \cdot \ h_{1} }{ 2 } / \dfrac{ (a+c) \cdot \ h }{ 2 } \ \\ r_{1}=\dfrac{ a \cdot \ h_{1} }{ (a+c) \cdot \ h } \ \\ \ \\ r_{1}=\dfrac{ a \cdot \ h \cdot \ \dfrac{ a }{ a+c } }{ (a+c) \cdot \ h } \ \\ r_{1}=\dfrac{ a \cdot \ \dfrac{ a }{ a+c } }{ (a+c) } \ \\ \ \\ r_{1}=\dfrac{ a^2 }{ (a+c)^2 }=\dfrac{ 6^2 }{ (6+4)^2 }=\dfrac{ 9 }{ 25 }=0.36$
$r_{2}=S_{2}/S=\dfrac{ c \cdot \ h_{2} }{ 2 } / \dfrac{ (a+c) \cdot \ h }{ 2 } \ \\ r_{2}=\dfrac{ c \cdot \ h_{2} }{ (a+c) \cdot \ h } \ \\ \ \\ r_{2}=\dfrac{ c \cdot \ h \cdot \ \dfrac{ c }{ a+c } }{ (a+c) \cdot \ h } \ \\ r_{2}=\dfrac{ c \cdot \ \dfrac{ c }{ a+c } }{ (a+c) } \ \\ \ \\ r_{2}=\dfrac{ c^2 }{ (a+c)^2 }=\dfrac{ 4^2 }{ (6+4)^2 }=\dfrac{ 4 }{ 25 }=0.16$

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