Question
The secant CD intersects the circle at E and CF is
tangent to the circle at F. If the length of the secant CD is equal to 20 cm and the length of tangent CF is 15 cm, find the length of the chord DE.Solution
CF is the tangent segment, which is 15 cm. CD is the secant segment, which is 20 cm. DE is the chord we want to find. According to the Power of a Point theorem: CF2 = CD Ă— CE Since CD = CE + ED and CE =CD-DE, we can rewrite the equation as 152 = 20 Ă— (20- DE) Now, 225 /20=20-DE 11.25 =20-DE DE =20-11.25 =8.75 The length of the chord DE is 8.75cm,
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