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 $\displaystyle 60^o$ ?

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 $\displaystyle 60^o$ for???

5. assume your $\displaystyle 60^o$ is for $\displaystyle \angle DBF$ 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, $\displaystyle \angle EAO$ = $\displaystyle \angle FBO$ = $\displaystyle 90^o$hence $\displaystyle \angle CAZ$ = $\displaystyle \angle DBX$ = $\displaystyle 30^o$, and
$\displaystyle \angle BAZ$ = $\displaystyle \angle ABX$ = $\displaystyle 60^o$
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
$\displaystyle \angle YOZ$ = $\displaystyle 30^o$
Given that WOY = 5 and assumed that WO = 3, hence OY = 2

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

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

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

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

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

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

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

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the plane area shown in the figure consist of isosceles trapezoid (non-parallel sides are 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 pf the composite figure

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