In the above figure chord AB = Chord AC and chord BD intersect each other at point E. The prove that AB^2= AE × AC
Answers
Answer:
AB and CD are two chords of a circle which intersect each other at a right angle at point E if AE=6cm, EB=2cm and CE=3cm. What is the radius of the circle?
Swipe. Match. Chat. Download Tinder.
AB and CD are two chords of a circle which intersect each other at right angles at E. If AE = 6cm, EB = 2cm and CE = 3 cm, find the radius of the circle.
When two chords intersect, the product of the segments of either chord will be equal. Thus AE*EB = CE*ED, or 6*2 = 3*ED. So, ED = 12/3 - 4cm. The chord AB= 8cm and CD = 7 cm.
Draw a chord BC, and let the center be O. Angle BOC = 2 x angle BAC. But tan BAC = 3/6 or <BAC = tan (inverse) 0.5 = 26.56505118 so <BOC = 2x26.56505118 = 53.13010236. Let the radius be r. Apply the cosine rule: [r^2 + r^2 - CB^2]/2r.r = cos 53.13010236. [CB = {2^2+3^2}^0.5 = 13^0.5], or
[2r^2 - 13]/2r^2 = 0.6, or
[2r^2 - 13] = 1.2r^2, or
0.8r^2 = 13, or
r^2 = 13/0.8 = 16.25, or
r = 16.25^0.5 = 4.031128874 cm