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In the figure the triangle $DCF$ is joined to the square $ABCD$ of side $2$, $M$ is middle of $CD$. So the gray area is:

image description here

My progress:

$S_{AMB}=\frac{4}2=2\implies S_{MCB}=S_{ADM}=1$

$\triangle CMF = \cong \triangle BMC\\ \therefore S_{CMF}=1$

$\frac{S_{MDG}}{S_{DFC}}=\frac{DG\cdot DM}{DF\cdot CF}=\frac{DG}{\sqrt2}$

$S_{DGM} = 1-S_{FGM}$

From Stweart's theorem; $FM =\sqrt5$

Using law of cosines in $\triangle ADF$,

$\angle ADG = 135^\circ\implies DF = 2\sqrt2$

...?

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    $\begingroup$ just complete the upper triangle into a square $\endgroup$
    – G Cab
    Feb 14 at 16:48

2 Answers 2

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We have $S_{MAB} = \frac12 S_{ABCD} = 2$ and $S_{MCF} = \frac14 S_{ABCD} = 1$, so it's only $S_{MGD}$ that needs tricky calculation.

We have \begin{align} S_{MGD} : S_{MFG} &= GD : FG & \text{(triangles with same height, different base)} \\ &= S_{BGD} : S_{BFG} & \text{(triangles with same height, different base)} \\ &= S_{BMD} : S_{BFM} & \text{(subtract equal ratios)} \\ &= 1 : 2. \end{align} (To explain the penultimate step: since $S_{MGD} : S_{MFG} = S_{BGD} : S_{BFG}$, it is also equal to the difference $(S_{BGD} - S_{MGD}) : (S_{BFG} - S_{MFG})$, which simplifies to $S_{BMD} : S_{BFM}$.)

Since $S_{MGD} + S_{MFG} = S_{MFD} = 1$, we conclude that $S_{MGD} = \frac13$, so the total area is $\frac13 + 1 + 2 = \frac{10}{3}$.

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  • $\begingroup$ Can you explain $SMGD:SMFG = SBMD:SBFM$? $\endgroup$ Feb 14 at 17:55
  • $\begingroup$ I have added an explanation. Is it more clear now? $\endgroup$ Feb 14 at 18:08
  • $\begingroup$ I already understood this part..what I didn't see was the passage of $SMGD:SMFG=SBGD:SBFG$ it is also equal $(SBGD−SMGD):(SBFG−SMFG)$... $\endgroup$ Feb 14 at 18:59
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    $\begingroup$ That's just algebra: if $\frac ab = \frac cd$, then both are also equal to $\frac{a+c}{b+d}$ and $\frac{a-c}{b-d}$... $\endgroup$ Feb 14 at 19:00
  • $\begingroup$ Now I understand..you used the ratios of proportions,,thanks again $\endgroup$ Feb 14 at 19:44
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enter image description here

Hint: $\,G\,$ is the centroid of $\,\triangle DCE\,$, so $\,S_{DMG}=\dots\,$

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  • $\begingroup$ @ACB Just curious, was there a problem with the text being next to the diagram, instead of below it? $\endgroup$
    – dxiv
    Feb 15 at 17:42
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    $\begingroup$ I don't know whether it was intentional (sorry for that). But I find this way (text below the picture) better than text stick to the side of image. (Btw, I am using a mobile phone. Here's a screenshot: i.stack.imgur.com/QS8yJ.jpg) $\endgroup$
    – ACB
    Feb 15 at 18:32
  • $\begingroup$ @ACB Thanks for the screenshots. I was wondering if it's about the phone layout. $\endgroup$
    – dxiv
    Feb 15 at 18:45

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