Can anyone solve this?

If square C is 64 in^2, then its dimensions are 8 x 8.

If square D is 81 in^2, then its dimensions are 9 x 9.

Since D is bigger than C and I sits in that extra space, then the dimensions of I are 1 x 1.

Since H is smaller than C, specifically I smaller, than its dimensions are 7 x 7.

Then, H and C make up one side of B, so 7 + 8 = 15. The dimensions of B are 15 x 15.

Since E is bigger than D by I, then the dimensions of E are 10 x 10.

If we find the length of the right side (B + C + D), which is 15 + 8 + 9 = 32, then we know that F + E + D must also add up to 32. F + 10 + 9 = 32 || F = 13. The dimensions of F are 13 x 13.

Then, F is E + G, so 13 = 10 + G. G = 3. The dimensions of G are 3 x 3.

Lastly, since we know the side length of the whole square is 32, then A + F must equal 32. 32 = A + 13. A = 19. The dimensions of A are 19 x 19.

A = 19 x 19 = 361 in^2

B = 15 x 15 = 225 in^2

C = 8 x 8 = 64 in^2

D = 9 x 9 = 81 in^2

E = 10 x 10 = 100 in^2

F = 13 x 13 = 169 in^2

G = 3 x 3 = 9 in^2

H = 7 x 7 = 49 in^2

I = 1 x 1 = 1 in^2

Hope this helps!! :)

b. 153 students do not participate in sports and seventy-five 12th graders were surveyed

on monday rodney ate 1/4th of the sandwich = 3/12

on tuesday he ate 1/3rd of the sandwich = 4/12

therefore in total he has eaten = 3/12 + 4/12 = 7/12

thus we can say that he has eaten 7/12 of the sandwich on monday and tuesday alltogether.

happy to

pls mark as brainliest.