Final answer:
The length of EF is twice the length of BP, which is 6 cm. Since BP is half the radius of the larger circle and the triangles formed are similar, EF which is twice as long as BP must be 12 cm.
The correct answer is a.
Step-by-step explanation:
The student's question involves finding the length of line segment EF in a circle. Given that OP is 6 cm, which is half of the radius, we can infer that OP is the radius of the smaller, inscribed circle within the larger circle with radius 12 cm. This forms two identical right-angled triangles within the larger circle. The line AB is the diameter of the larger circle which is twice the radius, therefore AB = 24 cm.
As OP is the radius of the smaller circle and also a segment from the center to the perimeter, that means that EF is a line tangent to the smaller circle, perpendicular to OP, creating a right-angled triangle OPE. By the Pythagorean theorem, we can calculate the length of EF. Since OB (the radius of the large circle) is 12 cm, and OP is 6 cm, the length of BP, which is the other leg of the right-angled triangle OBP, is also 6 cm (half of OB).
Because EF passes through P and is tangent to the smaller circle, it forms right angles at P. Hence, the triangles OPE and OPB are similar, and EF is twice as long as BP because both are the hypotenuse of similar right triangles with halved dimensions. Therefore, EF is twice 6 cm, which is 12 cm.