Exercise 14 Page 772

Let's make a diagram to illustrate the given information.

Since the arc of the Sydney Harbour Bridge is part of a circle, we have that AB is a chord of this circle. Also, because CD is the perpendicular bisector of AB, we have that DC lies on a diameter of the circle. Let E be the other endpoint of this diameter.

Therefore, by the Segments of Chords Theorem the products of the lengths of the chord segments are equal. DC* CE = AC* CB Let's substitute the corresponding measures to find CE.

DC* CE = AC* CB

SubstituteValues

Substitute values

60* CE = 178* 178

▼

Solve for CE

Multiply

Multiply

60* CE = 31 684

DivEqn

.LHS /60.=.RHS /60.

CE = 31 684/60

UseCalc

Use a calculator

CE ≈ 528.1

Finally, we calculate the diameter of the circle by using the Segment Addition Postulate. d = DC + CE ⇒ d &= 60+528.1 ⇒ d &= 588.1 In consequence, the diameter of the circle containing the arc of the Sydney Harbour Bridge is approximately 558.1 meters.