Trapezoid - intersection of diagonals

In the ABCD trapezoid is AB = 8 cm long, trapezium height 6 cm, and distance of diagonals intersection from AB is 4 cm. Calculate trapezoid area.

Correct result:

S =  36 cm²

Solution:

$$\begin{gathered} a=8\text{ cm } \\ h=6\text{ cm } \\ {h}_1=4\text{ cm } \\ {h}_2=h-h_1=6-4=2\text{ cm } \\ a:c=h_1:h_2 \\ c=a\cdot {\displaystyle \frac{h_2}{h_1}}=8\cdot {\displaystyle \frac{2}{4}}=4\text{ cm } \\ S={\displaystyle \frac{a+c}{2}}\cdot h={\displaystyle \frac{8+4}{2}}\cdot 6=36{\text{cm}}^2 \end{gathered}$$

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