Shown in the diagram is a vacant lot where the corners are right angles. The lot is divided into two along EGAT-30 meters, EF-60 meters and ED- 40 meters. If the perimeter of AGEF and GBCDE are equal, how long is GB.

Since AF=30 and ED=40 we can say that BC=70 (since 30+40=70)

The perimeter of AGEF = AG + EG + EF + AF while
GBCDE = BG + EG + DE + CD + BC

P (AGEF) = P (GBCDE)
AG + EG + EF + AF = BG + EG + DE + CD + BC
We cancel the common EG
AG + EF + AF = BG + DE + CD + BC
We then substitute the values available to us
AG + 60 + 30 = BG + 40 + CD + 70
AG + 90 = BG + CD + 110
AG = BG + CD + 20

Let us then plot a point H, just extend line DE. We would then get that AG = AH + GH and CD = BG + GH

(AH + GH) = BG + (BG + GH) + 20
60 + GH = 2BG + GH + 20
We then cancel the common GH
60 = 2BG + 20
We subtract 20 from both sides
40 = 2BG
We divide both sides by 2
20 = BG

Final answer: BG is 20 meters long.

* please use the diagram attached as reference