The answer is also not 1200

I like this problem, because at first it appears you don't have enough information, however...

If you let $L$ be the length of one side of the square and $x$ be the height of the rectangle on the top, then...

• $2L + 2x = 400$
• $L + 2(L-x) = 400$ (by symmetry, since this must hold for both of the bottom rectangles)

Adding these two:

$5L = 800$

$L = 160$

So, Perimeter $= 4*L = \boxed{640}$

5 lengths of the side will cover the perimeter of green and red (or green and blue) 5 lengths = 800. 4 lengths = 640

"Square's PeriMeter of Three Rectangles"-Soln

Chito, as translated digitally by Tib, offers Two Solutions

The First is Geometrical as such

Let the Large/ OverAll Square . . | |

Be Divided Equally into a 'Grid' of Smaller Squares Making Four Columns and Four Rows( of Squares)

. . . . . | | | | | | | | | | | | | | | | | | | |

Then the Top/ Horizontal/ Green Rectangle will be Composed of . . . . . | | | | |

The FiRst Row and aLL Columns of Squares and will have a Perimeter of 400

Where Pg : PeriMeter of the Green Rectangle lg : Length of the Green Rectangle and wg : Width of the Green Rectangle

Pg = 2lg + 2wg 400 = 2(160) + 2(40) SiNce lg = 4(wg)

Psquare = 4 lsquare where lsquare = lg = 4 160 = 640

And the Dimensions of the Remaining Vertical Rectangles, Assuming they are Identical, are Pblue = 2 lblue+ 2 wblue 400 = 2(120) + 2(80) SiNce lb = lg - wg lb = 160 - 40 = 120 . . . | | | | | | | | | = Left/ Vertical #1/ Blue Rectangle

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The Second Solution is moreso Algebraic in that

Pgreen = 2 lgreen + 2 wgreen 400 = 2lg + 2wg ---> wg = 200 - lg

Pblue = Pred = 2 lblue + 2 wblue 400 = 2( lg - wg ) + 2( lg/2 ) 400 = 2 lg - 2 wg + lg 400 = 3 lg - 2 wg 400 = 3 lg - 2( 200 - lg ) 800 = 5 lg ---> lg = 160 wg = 200 - 160 = 40 and wblue = lg/2 = 160/2 = 80 and lblue = lgreen - wgreen = 160 - 40 = 120 And so Psquare = lgreen + 2 lblue + 2 wgreen + 2*wblue = 160 + 2(120) + 2(40) + 2(80) = 640

*OG Formatting Shown on "FaceBook"-Comment of the "Brilliant.org"-FBpage's Post of Said Problem