Question

sum 8 geometry
special segments
what is the length of the shorter of the two chords shown ?
19 units
13 units
18 units
16 units

Explanation:

Step1: Apply Chord Segment Theorem

When two chords intersect in a circle, the products of the lengths of their segments are equal. So, \( 9\times4x = 12\times(x + 2) \).

Step2: Solve for \( x \)

Expand the equation: \( 36x = 12x + 24 \). Subtract \( 12x \) from both sides: \( 24x = 24 \). Divide by 24: \( x = 1 \).

Step3: Find Chord Lengths

For chord \( QS \): \( 9 + 4x = 9 + 4(1) = 13 \)? Wait, no, wait. Wait, chord \( QT \) (wait, no, the chords are \( QR \) and \( TS \)? Wait, no, the two chords are \( QT \) (wait, no, the chords are \( QS \) and \( TR \)? Wait, no, the chords are \( QS \) (with segments 9 and \( 4x \)) and \( TR \) (with segments 12 and \( x + 2 \)). Wait, no, when two chords intersect at \( A \), so chord 1: segments 9 and \( 4x \), chord 2: segments 12 and \( x + 2 \). So lengths: chord 1 length is \( 9 + 4x \), chord 2 length is \( 12 + (x + 2) = x + 14 \). Wait, no, wait, the chord with segment 9 and \( 4x \): length is \( 9 + 4x \). The other chord: segment 12 and \( x + 2 \), length is \( 12 + x + 2 = x + 14 \). Now, we found \( x = 1 \). So chord 1 length: \( 9 + 4(1) = 13 \)? Wait, no, wait, 9 + 4x: 9 + 4*1 = 13? Wait, no, 9 + 4x is 13? Wait, but the other chord: 12 + (x + 2) = 12 + 3 = 15? Wait, no, I must have messed up the segments. Wait, no, the two chords are \( QS \) (with parts 9 and \( 4x \)) and \( TR \) (with parts 12 and \( x + 2 \)). Wait, no, when chords intersect, the formula is \( a\times b = c\times d \), where \( a \) and \( b \) are segments of one chord, \( c \) and \( d \) segments of the other. So \( 9\times4x = 12\times(x + 2) \). So 36x = 12x + 24 → 24x = 24 → x = 1. Then, the length of the chord with segments 9 and \( 4x \) is \( 9 + 4x = 9 + 4 = 13 \)? Wait, no, 9 + 4x is 13? Wait, but the other chord: 12 + (x + 2) = 12 + 3 = 15? Wait, that can't be. Wait, no, maybe I mixed up the segments. Wait, the chord with segment 9 and \( 4x \): 9 and \( 4x \) are parts of chord \( QS \), so \( QA = 9 \), \( AS = 4x \), so \( QS = QA + AS = 9 + 4x \). The other chord: \( TA = 12 \), \( AR = x + 2 \), so \( TR = TA + AR = 12 + x + 2 = x + 14 \). Now, with \( x = 1 \), \( QS = 13 \), \( TR = 15 \). Wait, but the options include 13, 18, 19, 16. Wait, maybe I made a mistake. Wait, wait, maybe the chords are \( QS \) and \( TR \), but maybe the segments are \( 9 \) and \( 4x \) for one chord, and \( 12 \) and \( x + 2 \) for the other. Wait, no, the theorem is \( QA \times AS = TA \times AR \). So \( 9 \times 4x = 12 \times (x + 2) \). So 36x = 12x + 24 → 24x = 24 → x = 1. Then, \( AS = 4x = 4 \), so \( QS = 9 + 4 = 13 \). \( AR = x + 2 = 3 \), so \( TR = 12 + 3 = 15 \). Wait, but 13 is an option. Wait, but maybe the chords are \( QS \) (length 13) and \( TR \) (length 15). So the shorter chord is 13. Wait, but let's check again. Wait, maybe I messed up the chord labels. Wait, the chord with segment 9 and \( 4x \): 9 + 4x = 13, the other chord: 12 + x + 2 = 15. So 13 is shorter. So the shorter chord length is 13.

Answer:

13 units