Math Puzzle: Go to great lengths

If the rectangle with side length 8 has height a, its area is A = 8a. Now the dimensions of the other rectangles can be determined successively. We have B = C = a/2 × 16, D = a/3 × 24, E = 4a/3 × 6, F = 4a/15 × 30, G = 8a/5 × 5, and H = 8a/35 × 35. The square therefore has a side length of 35.

Here are some additional calculations for ways to determine the above. For these calculations, all horizontal lengths are called widths and all vertical lengths are called heights.

Formula for B and C
8a = a/2 × x (where x is the height)
8a × 2/a = x
8 × 2 = 16 = x, the height of B and C

So the area of B and C is equal to a/2 × 16, where the width = a/2 and height = 16.

Formula for D
D’s height is the height of A (8) plus the height of B (16), so:
8a = (16 + 8) × x (where x is the width)
8a = 24 × x
(8a)/24 = x
a/3 = x, the width of D

So the area of D is 24 × a/3, where width = a/3 and height = 24.

Formula for E
The width of E is the width of D (a/3) and A (a), so width = a + a/3 = (3a)/3 + a/3 = (4a)/3
8a = (4a)/3 * x (where x is the height)
x = 8a/(4a/3) = 8a × 3/4a
x = 24a/4a = 6

So the area of E is (4a)/3 × 6, where width = (4a)/3 and height = 6.

Formula for F
The height of F is the height of A (8) plus the height of C (16) plus the height of E (6) = 30
8a = 30x
x = 8a/30 = 4a/15

So the area of F is (4a)/15 × 30, where width = 4a/15 and height = 30.

Formula for G
The width of G is width of D (a/3) plus the width of A (a) and the width of F ((4a)/15), which equals (5a)/15 + (4a)/15 + 15a/15 = (24a)/15
8a = (24a)/15 × x, where x is the height of G
x = 8a/((24a)/15) = 8a × 15/(24a) = 120a/24a = 5

So the area of G is 5 × (24a)/15, where the width = (24a)/15 and the height = 5.

Then, finally, H, where we really only care about the height, and the height can be found by summing the heights of G (5), D (24) and E (6)—which equals 35.