triangles and more

1. First, we notice that the cosine of \(\angle ABC = 4/5\)(adjacent / hypotenuse).

With this, we can use the law of cosines to find the length of the median \(\overline {AM}\).

We get the equation:

\(\overline {AM}^2 = 4^2 + (2.5)^2 - 2*4*2.5* \cos(\angle ABC)\)

We combine like terms and get:

\(\overline {AM}^2 = 22.25 - 20* \cos(\angle ABC)\)

We realize we already have \(\cos( \angle ABC)\)= \(\frac45\)

We then substitute, and get:

\(\overline{AM}^2 = 22.25-20*\frac45 = 22.25 - 16 = 6.25\)

To get the length of the median \(\overline {AM}\), we just take the square root of this result, to get:

\(\overline{AM} = \sqrt{6.25} = 2.5\)

2. Please reupload your question, the side lengths of the rhombus appear to be missing from the problem.

3. Let's first start off with the internal angle measurement of a regular hexagon. Each angle is equal to:

\(180-\frac{360}6 = 120\) degrees. If we drop a perpendicular to the smaller hexagon from a point like A, then we get two 30-60-90 triangles(since the smaller hexagon has interior angle measurement of 120, and the angles must add up to 180, so 120 + 30 + 30 = 180).

Let's then define the length of a side of the larger hexagon as x, meaning that the midpoint segments have length \(\frac{x}2\). 

First, let's find the area of the larger hexagon. Because a regular hexagon is comprised of 6 equilateral triangles(they all have the same side length, with 60 degree interior angle measurements), the area is equal to the area of one of the equilateral triangles multiplied by 6. Given a side length x of an equilateral triangle, the area is equal to:

\((x^2*\sqrt{3})/4\), which can be derived from heron's formula or pythagoras' theorem( just name an arbitrary side of the equilateral triangle as x). With this in mind, the area of a regular hexagon with side length x would then be:

\((6*x^2*\sqrt{3})/4 = (3*x^2*\sqrt3)/2\)

Now that we have the area of our larger hexagon in terms of x, we need to find the area of the smaller hexagon in terms of x. We can accomplish this by dropping an altitude from point A to the smaller regular hexagon, forming a perpendicular bisector with that side.

We then have two 30 - 60 -90 right triangles, with the hypotenuse having length \(\frac{x}2\), and the 60 degree angle being at the angle bisector of \(\angle FAB\). The 60 degree leg(the side adjacent to the smaller hexagon) is then \(x/2 * \sqrt3/3\) (remember the ratios of a 30-60-90 triangle!).

We then have that the entire side of the smaller hexagon is \(2*x/2 * \sqrt3/3\)(because two 30-60-90 triangles make up the side of the smaller hexagon). The smaller side is then \(x* \sqrt3/3\). Using our area formula from earlier, we have that the area is equal to:

\(3*(x*\sqrt3/3)^2*\sqrt3/4 = x^2*\sqrt3/4\)

Our fraction is then equal to:

\((x^2*\sqrt3/4) / ((3*x^2*\sqrt3)/2) = (x^2*\sqrt3)/4 * 2/(3x^2*\sqrt3)\)

Cross multiplying and canceling out terms, we get:

\(2/12 = 1/6\) As our answer.

This method of finding the fraction is certainly more extreme in my opinion than other much faster solutions. One such way is just assuming the side length of the regular hexagon because the fraction will stay constant no matter the side length.

Thanks, jfan   !!!!

Here's another way to  find (3)

Call the side of the larger hexagon  = S

Using the Law of Cosines  we can find  the side of the smaller hexagon,s,  as

s^2  =  [(1/2)S]^2 + [(1/2)S]^2  - 2 [ (1/2)S (1/2)S ]  cos (120°)

s^2  =   (1/2)S^2  -(1/2) S^2 * ( -1/2)

s^2  = (3/4)S^2

s = (√3/2) S

Because  the  hexagons  are  similar, then  the  scale  factor  of the  smaller  hexagon to the larger hexagon =

    (√3/2) S

    _______ =     √3/2

         S

Then   the  area of  the  smaller  hexagon to the larger =  square  of the  scale factor  =  3/4