MO Z9–I–2 - 2017

In the VODY trapezoid, VO is a longer base and the diagonal intersection K divides the VD line in a 3:2 ratio. The area of the KOV triangle is 13.5 cm². Find the area of the entire trapezoid.

Correct result:

S =  37.5 cm²

Solution:

$$\begin{gathered} S_1=13.5{\text{cm}}^2 \\ {k}_2=(2\mathrm{/}3{)}^2={\displaystyle \frac{4}{9}}\doteq 0.4444 \\ {S}_2=k_2\cdot S_1=0.4444\cdot 13.5=6{\text{cm}}^2 \\ {k}_3=(2+3)\mathrm{/}3={\displaystyle \frac{5}{3}}=1\frac{2}{3}\doteq 1.6667 \\ {S}_3=k_3\cdot S_1-S_1=1.6667\cdot 13.5-13.5=9{\text{cm}}^2 \\ {k}_4=(2+3)\mathrm{/}2={\displaystyle \frac{5}{2}}=2\frac{1}{2}=2.5 \\ {S}_4=k_4\cdot S_2-S_2=2.5\cdot 6-6=9{\text{cm}}^2 \\ S=S_1+S_2+S_3+S_4=13.5+6+9+9={\displaystyle \frac{75}{2}}=37\frac{1}{2}={\displaystyle \frac{75}{2}}{\text{cm}}^2=37.5{\text{cm}}^2 \end{gathered}$$

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