Hard problem on geometry Help!?
Question and everything you need at this link to solve it.
Please explain in very detail and how you proved everything.
FIND PART B) REMEMBER TO GIVE RATIOS except 1:1 or 25:9 and 9:25
- Andy JLv 71 decade agoFavorite Answer
And because the triangles are similar and have ratio of areas of 9:25, their corresponding sides have ratio √9 : √25 or 3:5. Thus:
AG / CG = 3/5
AG / 10 = 3/5
AG = 6
GB / GE = 3/5
GB / 15 = 3/5
GB = 9
Again, because of similarity between triangles AGF and CGB:
GB / GF = CG / AG
I think you can figure it from here...
- 1 decade ago
The actual unreduced area ratio for AGB : GCE is 27/75
BG/EG = GA/GC
x/15 = y/10
y = 10x/15
The area of triangle of AGB is 27, then (1/2)*10x^2/15 = 27
x = GB = 9 and y = AG = 6
For triangle AGF you do the same, the sides AG and EG are congruent, so are GF and GC
AG/EG = GF/GC
GF = 10*AG/15 = 4
FE = 15 - 4 = 11
This isn't really relevant but if you look at the drawing there's one angle indicated that is perpendicular this means that all the adjacent angles next to it are 90.
I will pick the triangles AGF and ACB giving 12/27 = 4/9