Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Saturday, February 14, 2015
Geometry Problem 1083: Semicircle, Diameter, Perpendicular, 90 Degrees, Tangent Line, Metric Relations
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through D draw line perpendicular to AB, end on F;
ReplyDeleteConnect GC,
triangle GDF similar to triangle DGE, ( angle AGD = angle GCD)
a is radius of circle
so X^2 = 2a(a+b)
http://s29.postimg.org/m7r8fv3hz/pro_1083.png
ReplyDeleteDraw lines and points per sketch
Lot O is the center of the circle, we have FO⊥AE and GO⊥AB => AFOG is a square
So CD= 2.AF= 2a
We also have ∠ (HDG) =∠ (BGC) and ∠ (CDG)= ∠ (BGC)
so GD is an angle bisector of angle CDH
DH=GK= a + b
Relation in the right triangle DGC give DG^2= DK.DC
or x^2=2a.(a+b)
May I point out a typo ?
ReplyDeleteInstead of DH=GK= a + b, should it not be DH=DK= a + b ?
Thanks for the correction.
ReplyDeletethe corrected statement is " DH=DK=a+b"
Peter Tran
https://www.dropbox.com/s/uggeh3qzew0fdld/1083.fig?dl=0
ReplyDeleteComplete the rectangle BAEJ and semicircle to full circle.
Let EJ and diameter GOK intersect at H.
Then x^2 = GD^2 = GK.GH = 2a(a + b).
Let H be on AG such that HD//BC//GO where O is the centre of the semicircle CGD
ReplyDeleteAGOF is a square of side a and the radius of the semicircle is a
Tr. a CDG and GDH are similar hence
2a/x = x/(a+b) from which the result follows
Sumith Peiris
Moratuwa
Sri Lanka
Further if DE = c and DF = y then it can be shown that
ReplyDeletey^2 = 2ac
Moreover if J is the food of the perpendicular from G to CD,
ReplyDeleteCJ = a-b and GJ^2 = a^2 - b^2