# Trapezoid Area problem

• Jun 14th 2012, 12:11 PM
Telo
Trapezoid Area problem
Attachment 24080

ABCD Trapezoid

BC = 4

BD = 6

<DBC = 90

find the area of the darkened part of figure.

• Jun 14th 2012, 12:16 PM
richard1234
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,

$\displaystyle [ABC] = \frac{1}{2}(2)(8) = 8$
• Jun 14th 2012, 01:49 PM
Telo
Re: Trapezoid Area problem
thanks!
• Jun 14th 2012, 03:07 PM
Soroban
Re: Trapezoid Area problem
Hello, Telo!

Quote:

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