Question

in circle o above, ae ⊥ bo, ae is $\frac{3}{4}$bd, and the radius is 16r. what is the approximate length of co in terms of r? (a) 10.6r (b) 15.9r (c) 16.1r (d) 17.2r (e) 17.9r

Explanation:

Step1: Recall circle - related properties

Let the radius of the circle \(O\) be \(R = 16r\). Then \(BD=2R = 32r\), and since \(AE=\frac{3}{4}BD\), we have \(AE=\frac{3}{4}\times32r = 24r\). Because \(AE\perp BO\), \(AC = CE=\frac{AE}{2}=12r\) (a perpendicular from the center of a circle to a chord bisects the chord).

Step2: Apply the Pythagorean theorem in right - triangle \(ACO\)

In right - triangle \(ACO\), by the Pythagorean theorem \(AC^{2}+CO^{2}=AO^{2}\). We know that \(AO = R=16r\) and \(AC = 12r\). Substitute these values into the Pythagorean formula: \((12r)^{2}+CO^{2}=(16r)^{2}\).

Step3: Solve for \(CO\)

Expand the equation: \(144r^{2}+CO^{2}=256r^{2}\). Then \(CO^{2}=256r^{2}-144r^{2}=112r^{2}\). So \(CO=\sqrt{112r^{2}}=\sqrt{16\times7}r = 4\sqrt{7}r\approx4\times2.65r=10.6r\).

Answer:

A. \(10.6r\)