Area of Trapezoid

**jasonlewiz** · November 21st 2009, 11:51 PM

The plane area shown in the figure consists of an isosceles trapezoid(non-parallel sides equal) and a segment of a circle. If the non-parallel sides are tangent to the segment at points A and B, find the area of the composite figure.

Where should I start in this problem?

**ukorov** · November 22nd 2009, 12:10 AM

and what is that $60^o$ ?

**jasonlewiz** · November 22nd 2009, 12:34 AM

I think it is needed to compute for the r?

**ukorov** · November 22nd 2009, 12:40 AM

Originally Posted by jasonlewiz

I think it is needed to compute for the r?

you are not sure what is that $60^o$ for???

**ukorov** · November 22nd 2009, 02:04 AM

assume your $60^o$ is for $\angle DBF$ as shown in my attached figure.

I have worked out a pretty complicated, perhaps stupid, method...

produce all the lines and points as shown in figure, such that
AO = WO = BO = radius of a circle. O is centre of the circle.
AOZ, BOX and WOY are all straight lines.
AC//WY//BD and they are all perpendicular to both AB and EF
since AE and BF are tangents, $\angle EAO$ = $\angle FBO$ = $90^o$ hence $\angle CAZ$ = $\angle DBX$ = $30^o$ , and
$\angle BAZ$ = $\angle ABX$ = $60^o$
so we have a equilateral triangle ABO, with AO = BO = AB = 3
you can prove that WOY is bisector of angles AOB and XOZ, therefore
$\angle YOZ$ = $30^o$
Given that WOY = 5 and assumed that WO = 3, hence OY = 2

In triangle YOZ,
$cos30^o = \frac{OY}{OZ}$
$\frac{\sqrt{3}}{2} = \frac{2}{OZ}$
OZ = $\frac{4}{\sqrt{3}}$
therefore AZ = $3 + \frac{4}{\sqrt{3}}$

In triangle CAZ,
$cos30^o = \frac{AC}{AZ}$
$(cos30^o)(AZ) = AC$
AC = $(\frac{\sqrt{3}}{2})(3 + \frac{4}{\sqrt{3}})$
AC = $\frac{3\sqrt{3}}{2} + 2$

In triangle EAC,
$tan60^o = \frac{EC}{AC}$
( $tan60^o)(AC) = EC$
EC = $(\sqrt{3})(\frac{3\sqrt{3}}{2} + 2)$
EC = $\frac{9}{2} + 2\sqrt{3}$

The area of sector AWBO
$= (\frac{1}{6})(\pi)(3^2)$
$= \frac{3\pi}{2}$
and the area of triangle ABO
$= (\frac{1}{2})(3)(3)(sin60^o)$
$= (\frac{9}{2})(\frac{\sqrt{3}}{2})$
$= \frac{9\sqrt{3}}{4}$
Therefore the area of region AWB (green in figure)
= area of sector AWBO - area of triangle ABO
= 0.8153

area of trapezium AEFB
$= \frac{(2EC + CD + AB)(AC)}{2}$
$= \frac{(6 + 2(\frac{9}{2} + 2\sqrt{3}))(\frac{3\sqrt{3}}{2} + 2)}{2}$
$= (15 + 4\sqrt{3})(\frac{3\sqrt{3}}{4} + 1)$
.....in the end you will get this area = 50.4138

So the area of overall figure = 50.4138 + 0.8153 = 51.2291 ( $cm^2$ ??)

**denncy19** · December 14th 2010, 04:41 AM

how come u assume WO= 3 ?? is that reasonable??

**denncy19** · December 14th 2010, 04:43 AM

how come u assume WO=3?? and OY=2 is u need to find for their value like WO= x then OY= 5-x?

**denncy19** · December 14th 2010, 04:49 AM

is it there is a special triagle 30 by 60 so ur WO must be 1.5squareroot of 3 or 2.598

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