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
JHello, rcs!
Your answer is close, but . . .
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?
In right triangle $\displaystyle EAD$, the hypotenuse is: $\displaystyle DE \,=\,\sqrt{1^2+2^2} \,=\,\sqrt{5}$Code:1 E A o - - - - - o - - - - - o B | θ / | | / | | F / | | _ o | | √5 / * | 2 | / * | | / * | | / * | | / * | | / * | |/ θ * | D o - - - - - - - - - - - o C 2
Right triangles $\displaystyle EAD$ and $\displaystyle DFC$ have angle $\displaystyle \theta$.
. . Hence: .$\displaystyle \Delta EAD \sim \Delta DFC$
$\displaystyle \begin{array}{cccccccc}\text{In }\Delta DFC:& \sin\theta &=& \dfrac{CF}{2} \\ \\[-3mm] \text{In }\Delta EAD: & \sin\theta &=& \dfrac{2}{\sqrt{5}} \end{array}$
Therefore: .$\displaystyle \frac{CF}{2} \:=\:\frac{2}{\sqrt{5}} \quad\Rightarrow\quad CF \:=\:\frac{4}{\sqrt{5}} \;=\;\frac{4\sqrt{5}}{5}$