Two streets intersect at a 30-degree angle. at the intersection, there are four crosswalks formed that are the same length. what type of quadrilateral is formed by the cross walks

I am attaching a sketch for reference, the quadrilateral that is formed is the red area ABDC. First of all, we have that the quadrilateral has all of its sides equal. We also know that the crosswalks are parallel in pairs, because they are both normal to the same two parallel lines (that define a road in our sketch). AB is parallel to CD this way and AC is parallel to BD (in Euclidean Geometry, two lines normal to the same line are parallel). Thus, ABDC is a parallelogram with equal sides. It is either a square or a rhombus. It cannot be a square since one of its angles is 30 degrees, thus it is a rhombus (also called an equilateral quadrilateral).

It is 198,87032,750 + 375 = 33145 so, we take the 6% and we make it a decimal (0.6) and multiply that, then that is your answer