In $\Delta ABC, AB = 13, AC = 5, BC = 12.$ Points $M,N$ lie on $AC,BC$ respectively with $CM = CN = 4$. 
In $\Delta ABC, AB = 13, AC = 5, BC = 12.$ Points $M,N$ lie on $AC,BC$ respectively with $CM = CN = 4$. Points $J,K$ are on $AB$ such that  $MJ$ and $NK$ are perpendicular to $AB$. Find the area of the pentagon $(CMJKN)$ to the nearest whole number.

What I Tried: Here is a picture :-

Almost everything of what I did is shown in the picture, I am typing the rest. 
First of all, $\Delta ABC \sim \Delta AMJ \sim \Delta NBK$.
So we have $(AB = 13 , BC = 12 , AC = 5)$ as well as $(AM = 1 , BN = 8)$ . 
Hence :- $$AJ = \frac{5}{13} , MJ = \frac{12}{13}$$ and :-
$$NK = \frac{40}{13} , BK = \frac{96}{13}$$
Also $JK = 13 - \frac{101}{13} = \frac{68}{13}$ .
After joining $MN$ and $MK$ , I find $MN = 4\sqrt{2}$ and $MK = \frac{4\sqrt{298}}{13}$ .
With all these lengths known, I will be able to find $[\Delta MCN] , [\Delta JMK]$ by using normal area formula, and $[\Delta MNK]$ using Heron's Formula and then simply add them up to get my required solution.
The Problem is, finding $[\Delta MNK]$ using Heron's Formula will be a lot difficult for me, because the side lengths are all complicated.
Can anyone give me a solution to this problem which I am facing? Thank You. (Alternate Solutions are also welcome.)
 A: $$[\triangle ABC]=30$$
since by A.A.A similarity theorem $$\triangle ABC \sim \triangle AMJ \sim \triangle NBK$$
and the ratio between area of two similar triangles is equal to the square of the ratio of length of corresponding sides
$$\frac{[\triangle ABC]}{[\triangle AMJ]}=(\frac{13}{1})^2=\frac{30}{[\triangle AMJ]}$$
$$[\triangle AMJ]=\frac{30}{169}$$
$$\frac{[\triangle ABC]}{[\triangle NBK]}=(\frac{13}{8})^2=\frac{30}{[\triangle NBK]}$$
$$[\triangle NBK]=\frac{1920}{169}$$
$$[CMJKN]=[\triangle ABC]-([\triangle AMJ]+[\triangle NBK])$$
A: Because $\triangle AJM\sim\triangle ACB\sim\triangle KNB$, note that $[AJM] =
 \frac{[ACB]}{169}$ and $[KNB] = \frac{64[ACB]}{169}$. Thus, $[CMJKN] = [ACB] - 
[AJM] - [KNB] = \frac{104[ACB]}{169} = \frac{8[ACB]}{13}$. However, $[ACB] = 30$,
so $[CMJKN] = \frac{240}{13}\approx 18.46$, and your answer is $\boxed{18}$.
A: Let $S=\frac 12CA\cdot CB=30=[ABC]$ be the area of $\Delta ABC$. (We use brackets to denote the area of a convex polygon.)
Note the similitudes $\Delta ABC\sim \Delta AMJ\sim\Delta NBK$. We know the proportions for the three hypothenuse values $13, 1,8$. This implies:
$$
\begin{aligned}{}
[CMJKN]
&=[ABC]-[AMJ]-[NBK]\\
&=S-\left(\frac 1{13}\right)^2S-\left(\frac 8{13}\right)^2S
=\frac{13^2-1^2-8^2}{13^2}\cdot S
\\
&=\frac{104}{169}\cdot S\approx 18.4615384615385\dots 
\end{aligned}
$$
