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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.)

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    $\begingroup$ You found $AJ, MJ, NK, BK$. So you can calculate the areas of triangles $AJM, NBK$ and subtract them from $[\triangle ABC]$. $\endgroup$ – player3236 Nov 13 '20 at 15:43
  • $\begingroup$ Oh @player3236 , completely forgot about that, I am so dumb. $\endgroup$ – Anonymous Nov 13 '20 at 15:58
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    $\begingroup$ It is this kind of question that deserves more attention... It does not happen often to see an (almost solved) detailed question with a nice picture... 1+ $\endgroup$ – dan_fulea Nov 13 '20 at 16:49
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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 $\boxed{18}$.

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$$[\triangle ABC]=30$$

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

$$\frac{[\triangle ABC]}{[\triangle AMJ]}=(\frac{13}{1})^2=\frac{30}{[\triangle AMJ]}$$

$$[\triangle AMJ]=\frac{30}{169}$$

$$\frac{[\triangle ABC]}{[\triangle NBK]}=(\frac{13}{8})^2=\frac{30}{[\triangle NBK]}$$

$$[\triangle NBK]=\frac{1920}{169}$$

$$[CMJKN]=[\triangle ABC]-([\triangle AMJ]+[\triangle NBK])$$

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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: $$ \begin{aligned}{} [CMJKN] &=[ABC]-[AMJ]-[NBK]\\ &=S-\left(\frac 1{13}\right)^2S-\left(\frac 8{13}\right)^2S =\frac{13^2-1^2-8^2}{13^2}\cdot S \\ &=\frac{104}{169}\cdot S\approx 18.4615384615385\dots \end{aligned} $$

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  • $\begingroup$ oh... i am definitively too slow in typing, the issue is always accepted as i was submitting first. $\endgroup$ – dan_fulea Nov 13 '20 at 16:01
  • $\begingroup$ Nope, the answers are not that useful now, I already got my answer myself after the hint on the comment. $\endgroup$ – Anonymous Nov 13 '20 at 16:05

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