Maths Fun(?)

-1. Draw a regular hexagon then draw the six "alternate" vertices. (Skip one vertex - 1 to 3, 2 to 4 etc). If the area of the original hexagon is 6, what is the area of the internal hexagon that is formed?

-2. Inscribe a square in a circle of radius R. Draw an arc from one corner of the square with radius equal to the side of the square. The area formed between this arc and the circle is a crescent. Find the area of this crescent in terms of R.

-3. In the equations below, replace each letter with a number from 1 to 9, using each number only once:

a/b + c/12 = 1 1/12
d/e + 5/f = 7/8
g/h + 3/i = 1 1/2

2. 1\2*R^2

3.

2/3 + 5/12 = 1 1/12
1/4 + 5/8 = 7/8
7/6 + 3/9 = 1 1/2

1. Draw a regular hexagon then draw the six "alternate" vertices.
(Skip one vertex - 1 to 3, 2 to 4 etc). If the area of the original hexagon is 6,
what is the area of the internal hexagon that is formed?

So I don't have to type "sqrt(3)"!):
u = sqrt(3)
v = 1/u (ratio radii outer:inner)

R = outer radius
Pentagon area: 3u(R^2) / 2

Given: area = 6 ; soooooo:
3u(R^2) / 2 = 6
R = sqrt(4 / u) : ~1.5196....

r = inner radius
r = R * v : ~.8773....
area = 3u(r^2)/2
area = (drum rolls!) 2 exactement.

CCb - very nice!

Denis - good solve. It can be done by similar triangles. The inner hexagon contains 6 identically-sized equilateral triangles and the outer hexagon contains these 6 plus 6 more plus 12 halves for a total of 18 triangles. Thus, the area of the outer hexagon is 3 times the area of the inner one.

manxman - if I read your answer correcly (one over 2 R-squared) then we beg to differ. What's your working?

Sorry, checked my work over and got R^2

Great solve!