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

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.

Step-by-step 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.    

step-by-step :

answer: 94

step-by-step explanation: 79+91+92+94+94 = 450 divided by 5 (since there are 5 number that are being added) = 90