Question

in circle o above, ae ⊥ bo, ae is $\frac{5}{6}$bd, and the radius is 12r. what is the approximate length of co in terms of r? (a) 3.6r (b) 4.9r (c) 5.7r (d) 6.1r (e) 6.6r

Explanation:

Step1: Find lengths of AE and BD

Since the radius of the circle \(O\) is \(12r\), the diameter \(BD = 2\times12r=24r\). Given that \(AE=\frac{5}{6}BD\), then \(AE=\frac{5}{6}\times24r = 20r\).

Step2: Use the property of perpendicular - chord

Since \(AE\perp BO\), \(AC = CE=\frac{AE}{2}=10r\) (a perpendicular from the center of a circle to a chord bisects the chord). Let \(OC = x\) and the radius \(OA = 12r\).

Step3: Apply the Pythagorean theorem

In right - triangle \(OCA\), by the Pythagorean theorem \(OA^{2}=AC^{2}+OC^{2}\). Substitute \(OA = 12r\) and \(AC = 10r\) into the formula: \((12r)^{2}=(10r)^{2}+x^{2}\).

Step4: Solve for \(x\)

Expand the equation: \(144r^{2}=100r^{2}+x^{2}\). Then \(x^{2}=144r^{2}-100r^{2}=44r^{2}\). So \(x=\sqrt{44}r\approx6.6r\).

Answer:

(E) \(6.6r\)