Question

question 4 of 10
fill in the blank. given ( odot r ) below, you can conclude that ( overline{ra} ) is congruent to

a. ( overline{mu} )
b. ( overline{af} )
c. ( overline{re} )
d. ( odot r )

Explanation:

Step1: Recall the property of chords in a circle

In a circle, if two chords are equidistant from the center of the circle, then the chords are congruent. Here, \(WA = 11.2\) and \(EU=11.2\). The segments \(RA\) and \(RE\) are the distances from the center \(R\) to the chords \(WF\) and \(MU\) respectively. But we are asked about the congruence of \(\overline{RA}\) and another segment. Wait, no, looking at the problem again - actually, in a circle, if a line from the center is perpendicular to a chord, it bisects the chord. But here, we note that the lengths of the parts related to the chords: Since \(WA = EU = 11.2\) (the non - center part of the chord - related segments). But actually, the key is that in a circle, if two chords are congruent (here \(WF\) and \(MU\) are congruent as they are cut into segments of equal length by the perpendiculars from the center), and \(RA\perp WF\), \(RE\perp MU\). But wait, no, the problem is about the congruence of \(\overline{RA}\) and \(\overline{RE}\). Because the chords \(WF\) and \(MU\) are congruent (since \(WA=\frac{WF}{2}\), \(EU = \frac{MU}{2}\) and \(WA = EU\)), and the perpendicular from the center to a chord: If two chords in a circle are congruent, then the distances from the center of the circle to the chords are congruent.

Answer:

C. \(\overline{RE}\)