You are not logged in.
Pages: 1
This is a problem that I've had a bit of a headache over, find the length FG.
If possible, please only use similarity, congruency, pythagoras, and trigonometry, to do this, not coordinate geometry or anything too advanced. Thanks!
Offline
I think this way works.
We can tell length BE by Pythagoras, because we know AB and AE. So BE = √(6²+4.5²) = 7.5.
We know EF (by Pythagoras and also because it's given), so taking that away gives that BF = 5.
Triangles ABG and CFG are similar, because they can be shown to have two matching angles, and we also know that each side of ABG is 1.5 times as big as those of CFG.
BF = BG+FG, and because those two triangles are similar that means that BF = 2.5FG.
Therefore, 2.5FG = 5 --> FG = 2cm.
Why did the vector cross the road?
It wanted to be normal.
Offline
How are FGC and BGC similar? Could you please provide a proof?
Offline
I never said they were, I said that ABG and CFG were similar. You get ABG by enlarging CFG by a scale factor of -1.5, with the point of enlargement at G.
You can show this by the X and Z rules of parallel lines.
Why did the vector cross the road?
It wanted to be normal.
Offline
Thanks mathsy!
Offline
Pages: 1