# 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.)

• You found $AJ, MJ, NK, BK$. So you can calculate the areas of triangles $AJM, NBK$ and subtract them from $[\triangle ABC]$. – player3236 Nov 13 '20 at 15:43
• Oh @player3236 , completely forgot about that, I am so dumb. – Anonymous Nov 13 '20 at 15:58
• 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+ – dan_fulea Nov 13 '20 at 16:49

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 $$\boxed{18}$$.

$$[\triangle ABC]=30$$

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

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

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

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

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

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

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}

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