# 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:

$S_{ 1 } = 13.5 \ cm^2 \ \\ k_{ 2 } = (2/3)^{ 2 } = \dfrac{ 4 }{ 9 } \doteq 0.4444 \ \\ S_{ 2 } = k_{ 2 } \cdot \ S_{ 1 } = 0.4444 \cdot \ 13.5 = 6 \ cm^2 \ \\ k_{ 3 } = (2+3)/3 = \dfrac{ 5 }{ 3 } \doteq 1.6667 \ \\ S_{ 3 } = k_{ 3 } \cdot \ S_{ 1 } - S_{ 1 } = 1.6667 \cdot \ 13.5 - 13.5 = 9 \ cm^2 \ \\ k_{ 4 } = (2+3)/2 = \dfrac{ 5 }{ 2 } = 2.5 \ \\ S_{ 4 } = k_{ 4 } \cdot \ S_{ 2 } - S_{ 2 } = 2.5 \cdot \ 6 - 6 = 9 \ cm^2 \ \\ S = S_{ 1 }+S_{ 2 }+S_{ 3 }+S_{ 4 } = 13.5+6+9+9 = \dfrac{ 75 }{ 2 } = 37.5 = 37.5 \ cm^2$

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