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.

Result

S =  37.5 cm2

Solution:

S1=13.5 cm2 k2=(2/3)2=49≐0.4444 S2=k2⋅ S1=0.4444⋅ 13.5=6 cm2 k3=(2+3)/3=53≐1.6667 S3=k3⋅ S1−S1=1.6667⋅ 13.5−13.5=9 cm2 k4=(2+3)/2=52=2.5 S4=k4⋅ S2−S2=2.5⋅ 6−6=9 cm2 S=S1+S2+S3+S4=13.5+6+9+9=752=37.5 cm2S_{1}=13.5 \ \text{cm}^2 \ \\ k_{2}=(2/3)^{ 2 }=\dfrac{ 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)/3=\dfrac{ 5 }{ 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)/2=\dfrac{ 5 }{ 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=\dfrac{ 75 }{ 2 }=37.5 \ \text{cm}^2

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