Question

if the length of $\overline{bf}$ is 32 inches, what is the length of $\overline{ch}$?

o a 67 in
o b 32 in
o c 16 in
o d 64 in

Explanation:

Step1: Recall the property of chords in a circle

In a circle, if a diameter is perpendicular to a chord, it bisects the chord. Also, chords that are equidistant from the center of a circle are congruent.

Step2: Analyze the given circle

Here, \(AD\) is a diameter. \(AD\perp BF\) at \(G\) and \(AD\perp EC\) at \(H\). Since \(OG = OH\) (given by the equal - segment marks on \(AD\)), chords \(BF\) and \(EC\) are equidistant from the center \(O\) of the circle.

Step3: Use the chord - equidistant property

By the property that chords equidistant from the center of a circle are congruent, \(BF=EC\). Given \(BF = 32\) inches. Since \(CH=\frac{1}{2}EC\) (because \(AD\perp EC\) and bisects \(EC\)) and \(BG=\frac{1}{2}BF\) (because \(AD\perp BF\) and bisects \(BF\)), and \(BF = EC\), we have \(CH=\frac{1}{2}EC=\frac{1}{2}BF\). But also, since \(BF\) and \(EC\) are congruent chords and the segments of congruent chords (bisected by the same - length perpendiculars from the center) are equal. So \(CH = BG\). And since \(BF = 32\) inches, and \(BG=\frac{BF}{2}\) (perpendicular from the center to a chord bisects the chord), but more directly, since \(BF\) and \(EC\) are congruent chords and the way they are bisected (by the same - structured perpendicular from the diameter), \(CH=\frac{EC}{2}\) and \(BF = EC\), so \(CH = 32\) inches.

Answer:

B. 32 in