If you can see if you can on some of my

PQ // BC 

then

Angle P = Angle B, A is common angle the two triangles are similar

AP/AB  = AQ/AC

8/18 = 12/ AC

AC = 12 * 18/8 = 27 (your ans: D)

AQ + QC = AC

QC = AC - AQ = 27 - 12 = 15 (if it was required) 

tan60 = y/8

y = s tan60 = 8sqrt(3)  (B)

Next time, post each question separately. 

1. Use a straightedge to draw line a and label a point on the line as point P. 2. Construct a line perpendicular to line a through point P. 3. Label a point on this line as point R. 4. With the compass open to the desired side length of the square, place the compass point on point P and draw an arc on line a and an arc on PRâ†â†’ . 5. Label the points of intersection as points S and T. 6. Keeping the same compass width, place the compass on point T and draw an arc in the interior of â SPT to intersect the previously drawn arc. 7. Label the point of intersection as point Q. 8. Without changing the compass width, place the compass point on point S and draw an arc in the interior of â SPT . 9. Use the straightedge to draw QSÂŻÂŻÂŻÂŻÂŻ and QTÂŻÂŻÂŻÂŻÂŻ.