# 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 trapezoid area.

Result

S =  36 cm2

#### Solution:

$a=8 \ \text{cm} \ \\ h=6 \ \text{cm} \ \\ h_{1}=4 \ \text{cm} \ \\ \ \\ h_{2}=h - h_{1}=6 - 4=2 \ \text{cm} \ \\ a:c=h_{1}:h_{2} \ \\ c=a \cdot \ \dfrac{ h_{2} }{ h_{1} }=8 \cdot \ \dfrac{ 2 }{ 4 }=4 \ \text{cm} \ \\ \ \\ S=\dfrac{ a+c }{ 2 } \cdot \ h=\dfrac{ 8+4 }{ 2 } \cdot \ 6=36 \ \text{cm}^2$

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