# Thread: Area of Trapezoid

1. ## Area of Trapezoid

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?

2. and what is that $60^o$ ?

3. I think it is needed to compute for the r?

4. Originally Posted by jasonlewiz
I think it is needed to compute for the r?
you are not sure what is that $60^o$ for???

5. 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 ( $cm^2$??)

6. ## hey

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

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

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