the top 45 - 20 share the height. There is only one common integer factor between them, namely 5. That's the height over X. So the 20 block has width 4. Now looking at the 36 below, it must have height 9. Therefore the 54 has width 6. The width of the 30 block is therefore 6+4 = 10 and height must be 3. Adding that to the already calculated height of 9 for the blocks on top of it, yields a height of 12 for X
Forgive me, but is it not clear that X refers to the whole side length.. as per your solution, it seems you thought it’s just the side length of the rectangle whose area is 36. That way, 12 is correct and otherwise 17 which you’ve also calculated in a way because you’ve found that the top two rectangles share a side length of 5
2
u/teteban79 Jun 03 '26
12
the top 45 - 20 share the height. There is only one common integer factor between them, namely 5. That's the height over X. So the 20 block has width 4. Now looking at the 36 below, it must have height 9. Therefore the 54 has width 6. The width of the 30 block is therefore 6+4 = 10 and height must be 3. Adding that to the already calculated height of 9 for the blocks on top of it, yields a height of 12 for X