Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below for more details.
http://s21.postimg.org/5i7u2r6nr/pro_1158.pngLet x= O4MSince O2O3E is a right triangle => O3E^2= 9^2-1^2= 80Since O4EO3 is a right triangle => O3O4=Sqrt((x+4)^2+80)We have O4C=O4D => x+10= sqrt((x+4)^2+80) +4Solve this equation we get x= 15In right triangle CBF we have MB^2= MC. MF= 10 .x 40= 400So MB=20 and AB=40
Let O2O4 meet AB at C the point of tangency and mid point of AB. Let O1D be an altitude of Tr. O1O2O4 and let O1D = hWe need to use Pythagoras thrice h^2 = 9^2 -1 = 80 .... (1) from Tr. O1O2DIf the radius of O4 = r then(r-4)^2 = h^2 + (r-6)^2 from which upon simplifying and using (1) r = 25 (Tr. O1O4D)Now use Pythagoras on Tr. ACO4(AB/2)^2 = r^2 -(r-10)^2 = 20r-100 = 400So AB = 40Sumith PeirisMoratuwaSri Lanka