# perpendicular segment measure

• Dec 22nd 2012, 06:23 AM
rcs
perpendicular segment measure
Let ABCD be a square with each side of length 2cm. Let E be the midpoint of AB. How long is the perpendicular segment C to DE?

can anyone help me on this please
• Dec 22nd 2012, 06:29 AM
skeeter
Re: perpendicular segment measure
make a sketch ... use the fact that the sides of similar triangles are proportional.
• Dec 22nd 2012, 07:07 AM
rcs
Re: perpendicular segment measure
is this correct 2* sqrt(5) / 5
• Dec 22nd 2012, 07:46 PM
Soroban
Re: perpendicular segment measure
JHello, rcs!

Quote:

Let ABCD be a square with each side of length 2cm.
Let E be the midpoint of AB.
How long is the perpendicular segment from C to DE?

Code:

            1    E     A o - - - - - o - - - - - o B       |        θ /            |       |        /            |       |      F /              |       |    _  o              |       |  √5 /  *            |     2 |    /    *          |       |    /        *        |       |  /          *      |       |  /              *    |       | /                *  |       |/ θ                  * |     D o - - - - - - - - - - - o C                   2
In right triangle $EAD$, the hypotenuse is: $DE \,=\,\sqrt{1^2+2^2} \,=\,\sqrt{5}$

Right triangles $EAD$ and $DFC$ have angle $\theta$.
. . Hence: . $\Delta EAD \sim \Delta DFC$

$\begin{array}{cccccccc}\text{In }\Delta DFC:& \sin\theta &=& \dfrac{CF}{2} \\ \\[-3mm] \text{In }\Delta EAD: & \sin\theta &=& \dfrac{2}{\sqrt{5}} \end{array}$

Therefore: . $\frac{CF}{2} \:=\:\frac{2}{\sqrt{5}} \quad\Rightarrow\quad CF \:=\:\frac{4}{\sqrt{5}} \;=\;\frac{4\sqrt{5}}{5}$