Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 823.
Sunday, November 18, 2012
Problem 823: Tangent Circles, Diameter, Perpendicular, Chord, Secant, Triangle, Area, Tangency Point
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By the power of circle,
ReplyDeleteQA*QB = QE*QF = 3*(3+5) = 24
Yet QA*QB = QC*QB = (CD*QB)/2 = area of triangle BCD,
so the answer is 24.
Let OA = R, QA = r
ReplyDeleteArea of ΔBCD = r.(r + 2R)= QA.QB = QE.QF = (3)(8)= 24
Area of triangle BCD is (2R+r)×r
ReplyDeleteSo if WE try to find this value of Area WE need a relationship between R and r
start with triangle QFO
WE have R^2=64+(R+r)^2-16(R+r)cosQ
the same Way with traingle OEQ
R^2=9+(R+r)^2-6(R+r)cosQ
WE simplify cosQ by divieded two equation we get:
8/3=64+2Rr+r^2/9+2Rr+r^2
Finaly get r^2+2Rr=24
That means r(r+R)=24
That is Area of traingle BCD