What is the area of the quadrilateral?

Create square $EFGH$ . The sums of the pairs of opposite triangles' areas are the same. The four right triangles $\triangle EBF$ etc. have the same area. Thus $69+93=60+Area(DHMG)$

$Area(DHMG)=\boxed{102}$

Set up coordinates with $A$ at the origin, and let $B$ have coordinates $(X,0)$ , and let $M$ have coordinates $(a,b)$ . Suppose that $AE = x$ . Then $AEMH + CGMF \; = \; \left(\tfrac12xb + \tfrac12(X-x)a\right) + \left(\tfrac12(X-x)(X-a) + \tfrac12x(X-b)\right) \; = \; \tfrac12xX + \tfrac12(X-x)X = \tfrac12X^2$ and hence $X^2 = 324$ , and $X = 18$ . But then $DHMG \; = \; \tfrac12X^2 - BFME \; = \; 162 - 60 \; = \; \boxed{102}$

Let $HE = EF = FG = GH = c$ ; $\Delta AHE = \Delta BEF = \Delta CGF = \Delta DGH = x$ and $DHMG = y$

Note that $\Delta AHE = \dfrac{ch}{2} = 69 - x$ and $\Delta CFG = \dfrac{c(c-h)}{2} = 93 -x$

So we get $ch = 138 - 2x; c^2 - ch = 168 -2x$ where $h$ is the height of $\Delta AHE$ .

$\implies c^2 = 324 - 4x$

Similarly we get $c^2 = 120 + 2y - 4x$

Equating both equations

$\implies 324 - 4x = 120 +2y -4x$

$\implies y=\boxed{102}$ .

Empirically the area of square must be (side) squared. Known area of quadrilaterals is 222 and orange area largest, so must be larger than (16) squared. Logic shows (17) squared not large enough and (19) too large. Answer is (18) squared = 324. Orange = 324-222=102.

Join the vertices $M$ and each of $A, B, C$ and $D$ by a line segment.

Let $[XYZ]$ denotes the area of the polygon $XYZ$ .

Suppose $AE:EB=1:p$ . Then $BF:FC=CG:GD=DH:HA=1:p$ .

Suppose $[AEM]=r, [BFM]=s, [CGM]=t$ and $[DHM]=u$ . Then $[BEM]=pr, [CFM]=ps, [DGM]=pt$ and $[AHM]=pu$ .

We have the following:

$\begin{aligned} [BFME] &=& pr+s &=& 60\\ [CGMF] &=& ps+t &=& 93\\ [DHMG] &=& pt+u &=& ??\\ [AEMH] &=& pu+r &=& 69\\ && r+t &=& s+u\\ \end{aligned}$

Hence, $\begin{aligned} \left(pt+u \right) + \left(pr+s \right)&=& p\left(r+t \right) +\left(s+u\right)\\ &=& p\left( s+u\right) +\left(r+t\right)\\ &=& \left(ps+t \right) + \left(pu+r \right)\\ \end{aligned}$

This means that $\displaystyle [DHMG] =pt+u = \left(ps+t \right) + \left(pu+r \right) - \left(pr+s \right) = 93+69- 60=\boxed{102}$ .

Simply I solved by area square thought. Total area should be in the square format as it is sqaure. There are 225,256, 289 and 324 around that one( sum of remaining is 222 ). It is closed to 93 as it is opposite to that one (and it seems to be also). So 324 is the utopian one which gives the answer as 102 which is around 93. ( 324- 102)

S1 + S3 = S2 + S4

S5 + S7 = S6 + S8

Add up these two equations

(S1 + S5) + (S3 + S7) = (S2 + S6) + (S4 + S8)

S1 + S5 + 60 = 93 + 69

S1 + S5 = 102

Consider eight triangles: DGH, AHE, BEF, and GFC are all clearly identical. That leaves HME, GMF, EMF, and EMG.

The formula for an area of a triangle is (base*height)/2, where base and height are defined perpendicular to each other and in the plane of the triangle. The bases of all the unknown triangles are the lines HE, EF, FG, and GH, and must be identical due to the side lengths of ABCD and the AE=BF=CG=DH equivalences.

Now consider the square HEFG. Drawing lines through M perpendicular to the square will produce the height lines of the triangles, and these heights don't need to be directly calculated - the line passing from GH to EF through M must be the same length as the line passing from HE to FG passing through M. The formula A=bh/2 thus means that the areas of HME+GFC equals the areas of EMF+HMG. The identical triangles can be added to this, meaning that HMEA+GCFM=EMFB+HMCD.

Solving for HMCD produces HMCD=HMEA+GCFD-EMFB=69+93-60=102

This is not a very reliable method for future problems but someone will probably find it useful: First I tgought that the area of the square has to be the square of some number, so I added the known areas first, that is 222. Then I compared the area of DHMG to MFCG, and DHMG had to have an area equal or lager than MFCG, so I added 93. 315 was the minimum area of the square, so I rounded up to the closest square number, that was 324, or 18^2. After that I remained with 69+60+93+x=324. x=102.

Basically, I supposed the square was drawn to scale with each side being an integer. It seemed probably that the square was 18x18 = total area of 324.

The area can be presumed to be a square of a natural number. Since DGHM looks bigger than CGMF the nearest possible number that would lead to a square of 18×18=324 area would be 102: 102+60+69+93=18×18

I noticed all areas where divisible by three. And the blank one seemed bigger than the one with 93, so I first tried 99, that was false. Then I tried 102 and that was correct. A little educated guessing :p

I just guessed squares and subtracted the area from each result until I got a solution just bigger than 93, as that’s the way it looks to me. I landed on 18x18, which is 324, subtract 222 (60+69+93) from that and end up at 102.

69-60+93=102, that is wassup,

This is my solution: Let the small length be $a$ and the side of square be $L$ : Since $GCM+AEM=\dfrac12aL=DMH+BMF$

So we can swap the color as the right figure, so $?+60=69+93$ Obtain the answer $?=\boxed{102}$ .

Total area of the given parts = 222.

The square numbers great than 222 are 225, 256, 289, and 324.

324 - 222 = 102

This is the suitable solution.

Another approach using estimation: assuming the sides of the square are integers, and judging the orange area slightly larger than the green area, around 10%, say 100. Total Area is then an estimated 69+93+60+100=322. Then I try approaching the answer: Sides are 17 then 17x17=289 Sides are 18 then 18x18=324 (very close)

Therefore I assume the sides to be 18, resulting in an area for the orange area of 324-(60+69+93) = 102

Because there is only one solution, the lengths of $AE$ , $BF$ , $CG$ , and $DH$ should not matter. To simplify things, we assume that $AE=BF=CG=DH=0$ . Again, to find the answer, it does not matter what we set the lengths to, as long as the given conditions in the problem are satisfied. Now, we get four triangles which have areas of $69$ , $60$ , $93$ , and an unknown area. Let the distance that $M$ is from $AD$ be $m$ . Adding the areas of the red and green triangles together, we get $\frac{xm}{2}+\frac{x(x-m)}{2}=\frac{x(m+x-m)}{2}=\frac{x^2}{2}=69+93=162.$ We know that $x^2$ is the area of the square, so the area is $2\cdot162=324$ . We can find the area of the orange area by subtracting the area of the entire square by the area of the red, blue, and green. Doing so gives an answer of $324-69-93-60=\boxed{102}$ .