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 the trapezoid area.

Final Answer:

S =  36 cm²

Step-by-step explanation:

$$\begin{gathered} a=8\text{cm} \\ h=6\text{cm} \\ {h}_1=4\text{cm} \\ {h}_2=h-h_1=6-4=2\text{cm} \\ a:c=h_1:h_2 \\ c=a\cdot \frac{h_2}{h_1}=8\cdot \frac{2}{4}=\frac{8\cdot 2}{4}=\frac{16}{4}=4\text{cm} \\ S=\frac{a+c}{2}\cdot h=\frac{8+4}{2}\cdot 6=36{\text{cm}}^2 \end{gathered}$$

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