Isosceles trapezoid

In an isosceles trapezoid KLMN intersection of the diagonals is marked by the letter S. Calculate the area of trapezoid if /KS/: /SM/ = 2:1 and a triangle KSN is 14 cm².

Correct result:

S =  63 cm²

Solution:

$$\begin{gathered} S_1=14{\text{cm}}^2 \\ {S}_2=S_1=14=14{\text{cm}}^2 \\ {S}_1=2x{v}_1\mathrm{/}2 \\ {S}_3=x{v}_1\mathrm{/}2 \\ {S}_3=S_1\mathrm{/}2=14\mathrm{/}2=7{\text{cm}}^2 \\ {S}_1=1y{v}_2\mathrm{/}2 \\ {S}_4=2y{v}_2\mathrm{/}2 \\ {S}_4=y{v}_2=2\cdot S_1 \\ {S}_4=2\cdot S_1=2\cdot 14=28{\text{cm}}^2 \\ S=S_1+S_2+S_3+S_4=14+14+7+28=63{\text{cm}}^2 \end{gathered}$$

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