1. ## Circle

AB is a diameter of a circle of radius 1 unit CD is a chord perpendicular to AB that cuts Ab at E If the arc CAD is 2/3 of the circumference of the circle what is the length of the segment AE.?

2. Originally Posted by Dragon
AB is a diameter of a circle of radius 1 unit CD is a chord perpendicular to AB that cuts Ab at E If the arc CAD is 2/3 of the circumference of the circle what is the length of the segment AE.?
Hi,

the centre of the circle is M. So the triangle MEC is a right triangle.

The arc $\displaystyle BC = \frac{1}{6} \pi$

$\displaystyle \overline{ME}=\cos(\frac{1}{6} \pi)=\frac{1}{2} \cdot \sqrt{3}$

So the total length of

$\displaystyle \overline{AE}=1+\frac{1}{2} \cdot \sqrt{3}$

EB

3. Hello, Dragon!

Hmmm, I got a different answer . . .

$\displaystyle AB$ is a diameter of a circle of radius 1 unit.
$\displaystyle CD$ is a chord perpendicular to $\displaystyle AB$ that cuts $\displaystyle AB$ at $\displaystyle E$.
If the arc $\displaystyle CAD$ is 2/3 of the circumference of the circle,
what is the length of the segment $\displaystyle AE$?

Of course you made a sketch, right?
Code:
              * * *   C
*           *
*           / | *
*          1/  |  *
/   |
*    1     /60° |   *
A * - - - - * - - + - * B
*         O\    |E  *
\   |
*          1\  |  *
*           \ | *
*           *
* * *   D

Since major arc $\displaystyle CAD \:= \:\frac{2}{3}\cdot360^o \:=\:240^o$

. . then minor arc $\displaystyle CBD \,=\,120^o$ and central angle $\displaystyle COB \,=\,60^o.$

Right triangle $\displaystyle CEO$ has acute angle $\displaystyle 60^o$ and $\displaystyle hyp = 1.$

Hence: $\displaystyle OE \,= \,adj \,= \,\frac{1}{2}$

Therefore: .$\displaystyle AE\:=\:\frac{3}{2}$

4. Originally Posted by earboth
Hi,

the centre of the circle is M. So the triangle MEC is a right triangle.

The arc $\displaystyle BC = \frac{1}{6} \pi$

$\displaystyle \overline{ME}=\cos(\frac{1}{6} \pi)=\frac{1}{2} \cdot \sqrt{3}$

So the total length of

$\displaystyle \overline{AE}=1+\frac{1}{2} \cdot \sqrt{3}$

EB
The arc $\displaystyle BC = \frac{2}{3} \pi$ rad since the arc CAD is 2/3 the total circumference.

-Dan