If this was an 8th grade question they were probably just supposed to realize that the total area had to be a perfect square. The 3 known areas combine to make 106. So the eligible perfect squares are 121, 144, and 169. If 121 the red area is 15, if 144 it's 38, and if 169 it's 63. Only 38 makes sense, hence the answer is 38.

I tried this in Sketchup with the junction of the internal lines placed anywhere inside the square. The unknown quadrant is always the smallest known quadrant plus the difference of the two largest known quadrants.

I wouldn't have been able to solve this problem.
Thank you for the solution.

The constraint that the base shape be a square can be relaxed so that any quadrillateral would suffice so long as the sides are all bisected.  It can be generalised further by using any polygon with an even number of sides.  The polygon  doesn't  even need to be convex so long as none of the lines going to the 'middle' doesn't  go outside  the polygon .

As it is a square and each area is a whole number  the 3 area 42+34+30 add up to 106 so the side length is either 11 (squared is 121), 12 (square is 144) or 13 (square is 169). subtract 106 from each of these gives 15, 38 or 63. Looking at the size of the shape compared to the other shapes it is bigger than 30 and smaller than 42 so the answer is 38. No algebra needed.

Cool solve, but in case anyone out there is unaware, if you need to quickly calculate this off the top of your head, the sum of opposing sides match. 42+30=34+38

Since the 4 regions have all an angle in common (next to the center) and since they all have 2 sides coming from the center of the sides of the global square, the sum of the diagonals are equal :
42 + 30 = x + 34
72 = x + 34
x = 72 - 34, so x = 38

From description, this is a perfect square.
If you take any random point inside the square and connect it to the middle of each side, you divide this square into four quadrilaterals.
Each pair of opposing quadrilaterals equals half the area of the square,
therefore
30+42=34+x > x=30+42-34 > x=38

One observation ... if you have a triangle that has a slanted dividing line to make two smaller triangles, but where the left side triangle is obtuse because the dividing line comes down to the left, then the height of the left triangle has to be taken from a line drawn from the intersection point at the bottom perpendicularly to the left side of the original triangle.  The apex of the original triangle continues to give the height of the smaller triangle on the right, but not of the left triangle, which is obtuse.  If you make the dividing line much more slanted, so that it is close to touching the left side of the original triangle, then the left area becomes tiny and the right area becomes very large.  They are not equal.

So many commenters are saying they are smarter because they intuited the answer.  It's actually smarter to show a proof.  Intuition doesn't show deeper understanding.

I liked this question and hated the AI PRESENTATION

Very clever to divide quads into triangles and use common altitudes to prove equal area triangles, setting up 4 equations and 4 unknowns. I'm sure someone discovered a proof or two off this. I'm double sure it showed up as a homework problem in a couple of my old HS geometry books as a student and later a teacher.

Opposite areas add to half the square’s area (to derive: divide each segment into 2 triangles using a line from the outside corner to the internal junction of the 4, give the junction co-ordinates, say, (a,l), use 1/2 base x height to get the area of each triangle, then add). This gives the missing area as (42 + 30) - 34 ie 38.

A little clarification would make this exercise more valid. Assuming the intersection of the quadrilaterals is exactly at each midpoint of each side of the square is also assuming the figure is drawn to scale. Neither of those stipulations were given as part of the exercise.

The diagonal totals are always equal.

So: 42 + 30 - 34 = 38

It took me a minute. I made an estimate: The unknown area of the figure must be greater than 30 and less than 42 (we compare it to the sizes of the other figures). So the area of the entire square is between 136 and 148. We take the square root and the side of the square is between 11.66 and 12.16. Therefore, our side must be 12. So the area of the entire square is 144. Subtracting the individual areas, we get that the figure has an area of 38.

I solved it in 5 seconds by just looking at the boundary lines.
If the 2 left pieces have the same vertical boundary the bottom piece is smaller by 42-34 = 8. Since the 2 right sides share that same vertical boundary the top piece is larger by 8 as well so 30+8 = 38

The differences between areas are a suggestion here.

a) 42 - 34 = 8 (top left - bottom left)
b) 34 - 30 = 4 (bottom left - bottom right)

So we can conclude that:

bottom right + results from point (a) => 30 + 8 = 38 OR
top left - result from (b) => 42 - 4 = 38

(+) or (-) sign is determined based on the location of the connection point between areas in a relation to the center of the whole square.

考え方..正方形の中の４つの４角形を４つが接している点とそれぞれの角とを結び..８つの
三角形に分けます。すると隣り合う三角形同士は底辺と高さが等しいので同面積に成ります。
拠って..対角の４角形同士を足すとこの正方形の１/２の面積に成ります。
式＝４２+３０＝３４+？　？＝７２－３４　？＝３８

2:39   did AI just say, "Noice" ?   Haha.