1. ## Trapezoid Area problem

ABCD Trapezoid

BC = 4

BD = 6

<DBC = 90

find the area of the darkened part of figure.

2. ## Re: Trapezoid Area problem

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,

$\displaystyle [ABC] = \frac{1}{2}(2)(8) = 8$

thanks!

4. ## Re: Trapezoid Area problem

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: .$\displaystyle \Delta AOB \:=\:\Delta BDA - \Delta ODA \:=\:24 - 16 \:=\:8$