triangle

Answer is B.

(1) is useless because we can derive the same from the diagram given.
But well, you can write down the equations and see that is not possible to extract "b" out of it.

So either answer is B or E.

(2) a = sqrt(84).

but bc = 5a (we can derive this from the diagram itself, no need to go to (1). Angle QPR = Angle EQR. Assume E is the point of intersection of PR and the line dropping from Q. As they are similar traingles we get 5/b = c/a)

The way to go about calculating b is simple. c^2 - b^2 = 25. -> (A)
And you know c in terms of "a" and "b" in which "a" is known. Hence you can reduce above equation just in terms of "b". Solve for b.

Calista.

stuart,
in questions like this should we assume that the point . is the center of the circle unless mentioned.

gmatguy16 wrote:stuart,
in questions like this should we assume that the point . is the center of the circle unless mentioned.

The . is the centre of the circle by convention, so yes, you can make that assumption (although on the actual GMAT they'll almost always explicitly state that as well).

Good question. Otherwise I don't think we will know the answer.