Diagonal intersect

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

Correct answer:

r₁ =  0.36
r₂ =  0.16

Step-by-step explanation:

$$\begin{gathered} a=6\text{cm} \\ c=4\text{cm} \\ h=h_1+h_2 \\ {h}_1:h_2=a:c \\ {h}_1=h\cdot \frac{a}{a+c} \\ S=\frac{a+c}{2}\cdot h \\ {S}_1=\frac{a\cdot h_1}{2} \\ {S}_2=\frac{c\cdot h_2}{2} \\ {r}_1=S_1/S=\frac{a\cdot h_1}{2}/\frac{(a+c)\cdot h}{2} \\ {r}_1=\frac{a\cdot h_1}{(a+c)\cdot h} \\ {r}_1=\frac{a\cdot h\cdot \frac{a}{a+c}}{(a+c)\cdot h} \\ {r}_1=\frac{a\cdot \frac{a}{a+c}}{(a+c)} \\ {r}_1=\frac{a^2}{(a+c{)}^2}=\frac{6^2}{(6+4{)}^2}=\frac{9}{25}=0.36 \end{gathered}$$

$$\begin{gathered} r_2=S_2/S=\frac{c\cdot h_2}{2}/\frac{(a+c)\cdot h}{2} \\ {r}_2=\frac{c\cdot h_2}{(a+c)\cdot h} \\ {r}_2=\frac{c\cdot h\cdot \frac{c}{a+c}}{(a+c)\cdot h} \\ {r}_2=\frac{c\cdot \frac{c}{a+c}}{(a+c)} \\ {r}_2=\frac{c^2}{(a+c{)}^2}=\frac{4^2}{(6+4{)}^2}=\frac{4}{25}=0.16 \end{gathered}$$

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