The shorter length of the chord(QS) is 13 units, if the chords Q S and R T intersect at point A and the length of Q A is 9, the length of A S is 4 x, the length of T A is 12, and the length of A R is x + 2.
Explanation:
The given is,
Chords Q S and R T intersect at point A
The length of Q A is 9
The length of A S is 4 x
The length of T A is 12
The length of A R is x + 2
For the above question diagram is missing, so i attach the diagram.
Step:1
From the Chord theorem,
It states that the products of the lengths of the line segments on each chord are equal.
Chords theorem for the given diagram (Ref attachment),
(QA × AS) = (TA × AR)........................(1)
From the given values equation (1) becomes,
(9 × 4x) = (12 × (x+2))
36x = 12x + 24
36x - 12x = 24
24x = 24
x = 1
From the x values, chord lenths are
For the QS,
QS = QA + QS
= 9 + 4x
= 9 + 4(1)
QS = 13 units
For the RT,
RT = RA + TA
= (x+2) + 12
= (1+2) + 12
= 3 +12
RT = 15 units
Result:
The shorter length of the chord(QS) is 13 units.