Answer:
Problem
The following figure is made up of equilaterals triangles and squares of side 24. What's the area of the square that forms connecting the centers of the four squares?The image isn't accurate at all
What i've done so far
I've tried many things and i just got this: -Two of the sides of the red square are paralell to the shared side of the two equilateral triangles, that makes the two small-triangles (up and down) equilaterals and similars to the 2 original ones.
-The other 2 small-triangles (left and right) are congruent to the others (they're equilaterals, and we can prove it playing with the angles).
-We can replace the small triangles (left and right) with the other two (up and down), so the area will be the area of the 2 original triangles and the other four regions, that equals 1 small square. By pithagoras, the height of the triangle of side 24 is
242−122−−−−−−−−√ = 576−144−−−−−−−−√ = 432−−−√
Step-by-step explanation:
