# What's the measure of the gray area in the figure below??

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:

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 $$\triangle ADF$$,

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

...?

• just complete the upper triangle into a square Feb 14 at 16:48

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 $$\frac13 + 1 + 2 = \frac{10}{3}$$.

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

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

• @ACB Just curious, was there a problem with the text being next to the diagram, instead of below it?
– dxiv
Feb 15 at 17:42
• 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)
– ACB
Feb 15 at 18:32
• @ACB Thanks for the screenshots. I was wondering if it's about the phone layout.
– dxiv
Feb 15 at 18:45