Final answer:
It is true that all chords of the outer circle that touch the inner circle are of equal length. This is because any such chord also acts as a tangent to the inner circle and all tangents from the same external point are equal. This can be demonstrated with simple geometric construction and application of the Pythagorean theorem.
Step-by-step explanation:
The statement that all chords of the outer circle that touch the inner circle are of equal length is true. Imagine any chord of the outer circle that touches the inner circle. This chord is also a tangent to the inner circle at the point of contact.
By the properties of tangents from the same external point, all tangents to a circle are equal in length. Therefore, if we draw two chords on the outer circle that touch the inner circle, the segments from the point of contact to the point where each chord intersects the outer circle will be equal in length.
Let's prove this with a simple construction: Let O be the center of both circles, r be the radius of the inner circle, and R be the radius of the outer circle. Let A and B be the points where one chord of the outer circle touches the inner circle and intersects the outer circle, respectively.
Similarly, let C and D be the points where another chord touches the inner circle and intersects the outer circle. Triangles OAB and OCD are right triangles with hypotenuse radii R and one leg of length r. By the Pythagorean theorem, the lengths AB and CD (which are the other legs of the triangles) are equal, thus proving that all such chords are of equal length.
If we consider the arc length, it is directly proportional to the radius of the circular path. However, since we are dealing with straight-line chords and not arcs in this context, this information is secondary to the proof.
Therefore answer is a. True.