Trapezoid Area problem

**Telo** · June 14th 2012, 12:11 PM

ABCD Trapezoid

BC = 4

AD = 8

BD = 6

<ADB = 90
<DBC = 90

find the area of the darkened part of figure.

please help

**richard1234** · June 14th 2012, 12:16 PM

It can easily be shown that triangles ADO and CBO are similar with a 2:1 ratio, using an angle-angle-angle argument. Therefore, $BO = \frac{1}{2}DO \Rightarrow BO = 2$ .

Triangle AOB has a "base" of 2 and a height equal to AD, or 8 (since BDA is a right angle). Hence,

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

**Telo** · June 14th 2012, 01:49 PM

thanks!

**Soroban** · June 14th 2012, 03:07 PM

Hello, Telo!

Trapezoid $ABCD\!:\;BC = 4,\;AD = 8,\;BD = 6,\;\angle ADB = 90^o,\;\angle DBC = 90^o$

Find the area of the shaded portion.

Code:

                      B   4   C
                     .* - - - * 
                   .*:|    * *
                 .*:::| *   *
               .*::::*|O   *
             .*:::*  6|   *
           .*::*      |  *
         .*:*         | *
        **            |*
    A * - - - - - - - * D
              8

Since $\Delta OBC \,\sim\,\Delta ODA\!:\:OB = 2,\;OD = 4.$

Refering to the areas of the triangles:

. . $\Delta BDA \:=\:\tfrac{1}{2}(8)(6) \:=\:24$

. . $\Delta ODA \:=\:\tfrac{1}{2}(8)(4) \:=\:16$

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

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