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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
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