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