$ABCD$ is a rectangle with $AD = 2$ , $AB = 4$. $P$ is on $AB$ such that $AP : PB = 2 : 1$. $CE \perp DP$ at $E$. Find $CE$.

Question

$ABCD$ is a rectangle with $AD = 2$, $AB = 4$. $P$ is on $AB$ such that $AP : PB = 2 : 1$. $CE \perp DP$ at $E$. Find $CE$.

What I Tried: Here is a picture :-

Since $AP : PB = 2 : 1$ and $AB = CD$, we have $AP = \frac{8}{3}$, $BP = \frac{4}{3}$.

Now :- $$[\Delta DPC] = \frac{[\square ABCD]}{2}$$ $$\Rightarrow [\Delta DPC] = 4$$

From Pythagoras Theorem :- $$(AP)^2 + (AD)^2 = (PD)^2$$ $$\Rightarrow PD = \frac{\sqrt{82}}{3}$$

So, $$[\Delta PDC] = \frac{1}{2} * PD * CE$$ $$\Rightarrow 4 = \frac{1}{2} * \frac{\sqrt{82}}{3} * CE$$ $$\Rightarrow CE = \frac{12 \sqrt{82}}{41}$$

But my answer was given as $\frac{12}{5}$ , so where did I go wrong?
Can anyone help me? Thank You.

Answer 1

$PD=\color{red}{\dfrac{\sqrt{82}}{3}}$ is false.

$$AD=\dfrac{\color{blue}{6}}{3}, AP=\dfrac{\color{blue}{8}}{3} \Rightarrow PD=\dfrac{\color{blue}{10}}{3}$$