# 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.

Update:

FIND PART B) REMEMBER TO GIVE RATIOS except 1:1 or 25:9 and 9:25

Relevance

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

Likewise:

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...

• 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