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A circle is shown. 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, and the length of A R is x + 2.

What is the length of the shorter of the two chords shown?

13 units
16 units
18 units
19 units

User AshleyF
by
8.6k points

2 Answers

1 vote

Answer:

A

Explanation:

just took the test on edg

User Rashawn
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8.1k points
3 votes

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.

A circle is shown. Chords Q S and R T intersect at point A. The length of Q A is 9, the-example-1
User John Zhang
by
7.7k points

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