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

Re: perpendicular segment measure

make a sketch ... use the fact that the sides of similar triangles are proportional.

Re: perpendicular segment measure

is this correct 2* sqrt(5) / 5

Re: perpendicular segment measure

JHello, rcs!

Your answer is close, but . . .

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