It can easily be shown that triangles ADO and CBO are similar with a 2:1 ratio, using an angle-angle-angle argument. Therefore, $\displaystyle BO = \frac{1}{2}DO \Rightarrow BO = 2$.
Triangle AOB has a "base" of 2 and a height equal to AD, or 8 (since BDA is a right angle). Hence,
$\displaystyle [ABC] = \frac{1}{2}(2)(8) = 8$
Hello, Telo!
Trapezoid $\displaystyle ABCD\!:\;BC = 4,\;AD = 8,\;BD = 6,\;\angle ADB = 90^o,\;\angle DBC = 90^o$
Find the area of the shaded portion.
Code:B 4 C .* - - - * .*:| * * .*:::| * * .*::::*|O * .*:::* 6| * .*::* | * .*:* | * ** |* A * - - - - - - - * D 8
Since $\displaystyle \Delta OBC \,\sim\,\Delta ODA\!:\:OB = 2,\;OD = 4.$
Refering to the areas of the triangles:
. . $\displaystyle \Delta BDA \:=\:\tfrac{1}{2}(8)(6) \:=\:24$
. . $\displaystyle \Delta ODA \:=\:\tfrac{1}{2}(8)(4) \:=\:16$
Therefore: .$\displaystyle \Delta AOB \:=\:\Delta BDA - \Delta ODA \:=\:24 - 16 \:=\:8 $