Area and fraction problems?

Question

1. b is not equal to a, then b over (a - b) - a over (b-a) =

Choices:
1. -1
2. 0
3. 1
4. b-a over a-b
5. a+b over a-b

2. 6 equilateral triangles with side 1 are joined to form a hexagon. A circle is circumscribed about the hexagon. What is the area of the shaded region in the figure below?

1. 3.14 - square root 3 over 2
2. 3.14 - 3(sq root 3 over 2)
3. 3.14(sq root 3) over 3
4. 2(3.14) - sq root 3 over 2
5. 2(3.14) - 3(sq root 3)

Thank you for any help. Please explain the reasoning/steps behind it if possible it's appreciated :].

Answer 1

1. b/(a-b) - a/(b-a) =
b/(a-b) + a/(a-b) =
(b + a) / (a - b)

That's the same as (a + b) / (a - b), your choice #5.

2. The equilateral triangles have height sqrt(3)/2 and base 1, so their area (1/2 * b * h) is sqrt(3)/4 each. Six of them turned into a hexagon have area 6*sqrt(3)/4 = 3*sqrt(3)/2.

The circle is circumscribed so it touches the pentagon at each vertex. The line from each vertex to the center of the pentagon and the center of the circle, is a side of one of the equilateral triangles, so it is length=1. This means the circle has radius 1 and therefore area (pi * r^2) = pi*1^2 = pi.

Assuming the shaded area is the portion of the circle NOT inside the hexagon, it's the area of the circle (pi) minus the area of the hexagon (3sqrt(3)/2).

pi - 3sqrt(3)/2 is your answer (1).