What is the ratio of the area of a regular hexagon to the area of the hexagon which uses the midpoints of the first hexagon's sides as its vertices? Express your answer as a common fraction.
If you draw a line connecting two midpoints of the hexagon with side 1, then each half is a leg of a 30-60-90 right triangle with hypotenuse 1/2. Si the line joining the midpoints has length 1/2 √3
so the area of the smaller hexagon is 3/4 that of the larger
Why did the bee go to the bar? Because it heard the hexagon had the best honey cocktails!
Now, let's get to your question about hexagons. So, we have a regular hexagon and another hexagon formed by connecting the midpoints of the sides of the first hexagon.
If we look closely, we can divide the regular hexagon into 6 smaller equilateral triangles. And the second hexagon can be divided into 6 smaller isosceles triangles.
Now, if we compare the areas of these triangles, we notice that the equilateral triangles in the regular hexagon have a greater area than the isosceles triangles in the second hexagon.
Therefore, the ratio of the area of the regular hexagon to the area of the second hexagon is greater than 1. But hang on, let me do a quick calculation for you.
After some calculations, I can tell you that the ratio is actually 4/3. So, the area of the regular hexagon is 4/3 times the area of the second hexagon.
Hope that gives you something to hex-cite about!
To find the ratio of the area of a regular hexagon to the area of the hexagon formed by the midpoints of its sides, we can use the fact that the areas of two similar figures are proportional to the squares of their corresponding side lengths.
Let's assume that the side length of the regular hexagon is "s". We need to find the side length of the hexagon formed by the midpoints of the sides.
When we connect the midpoints of the adjacent sides of a regular hexagon, we get a regular hexagon of half the side length. So, the side length of the hexagon formed by the midpoints is "s/2".
The formula for the area of a regular hexagon is given by:
Area = (3√3/2) * s^2
Thus, the area of the regular hexagon is (3√3/2) * s^2.
Now, let's find the area of the hexagon formed by the midpoints of the sides. Since this hexagon is similar to the regular hexagon, its area will be proportional to the square of its side length.
Therefore, the area of the hexagon formed by the midpoints is (3√3/2) * (s/2)^2 = (3√3/2) * (s^2/4) = (3√3/8) * s^2.
The ratio of the areas is given by:
[(3√3/2) * s^2] / [(3√3/8) * s^2] = (8/2) = 4/1.
So, the ratio of the area of the regular hexagon to the area of the hexagon formed by the midpoints of its sides is 4/1, which can be expressed as the common fraction 4/1.
To find the ratio of the areas of the two hexagons, we need to determine the relationship between their side lengths.
Let's consider a regular hexagon with side length \(s\). Each of its sides can be divided into two equal segments of length \(s/2\) by connecting the midpoints. Connecting these midpoints of the sides creates another hexagon.
Now, let's compare the side lengths of the two hexagons. The original regular hexagon has side length \(s\), while the hexagon formed by the midpoints of its sides has side length equal to the diagonal of one of its smaller equilateral triangles.
To find the side length of the smaller equilateral triangle, we can use the Pythagorean theorem. One side of the equilateral triangle is \(s/2\), and its diagonals (the hypotenuses of the right triangles formed) are the same as the side length of the regular hexagon (\(s\)). Applying the Pythagorean theorem, we have:
\((s/2)^2 + (s/2)^2 = s^2\)
Simplifying, we get:
\(s^2/4 + s^2/4 = s^2/2\)
Combining like terms, we have:
\(s^2/2 = s^2/2\)
Thus, the side length of the smaller equilateral triangle is \(s/\sqrt{2}\).
Now, we can calculate the ratio of the areas of the two hexagons.
The area of a regular hexagon is given by:
\(A_1 = \frac{3\sqrt{3}}{2}(s^2)\)
The area of the hexagon formed by the midpoints of the sides can be found by splitting it into six congruent equilateral triangles with side length \(s/\sqrt{2}\) and then subtracting the areas of these triangles from the area of the regular hexagon.
The area of each smaller equilateral triangle is:
\(A_{\text{tri}} = \frac{\sqrt{3}}{4}\left(\frac{s}{\sqrt{2}}\right)^2 = \frac{s^2\sqrt{3}}{8}\)
Since there are six smaller equilateral triangles, the total area of these triangles is:
\(6A_{\text{tri}} = \frac{6s^2\sqrt{3}}{8}\)
The area of the hexagon formed by the midpoints of the sides (let's call it \(A_2\)) is:
\(A_2 = A_1 - 6A_{\text{tri}}\)
Substituting the expressions for \(A_1\) and \(A_{\text{tri}}\), we get:
\(A_2 = \frac{3\sqrt{3}}{2}(s^2) - \frac{6s^2\sqrt{3}}{8}\)
Simplifying, we have:
\(A_2 = \frac{3\sqrt{3}}{2}(s^2) - \frac{3\sqrt{3}}{4}(s^2)\)
Combining like terms, we get:
\(A_2 = \frac{3\sqrt{3}}{4}(s^2)\)
Now we can find the ratio of the areas:
\(\frac{A_2}{A_1} = \frac{\frac{3\sqrt{3}}{4}(s^2)}{\frac{3\sqrt{3}}{2}(s^2)}\)
Canceling out the common factors, we get:
\(\frac{A_2}{A_1} = \frac{\frac{1}{4}(s^2)}{\frac{1}{2}(s^2)}\)
Simplifying, we have:
\(\frac{A_2}{A_1} = \frac{1}{4}\)
Therefore, the ratio of the area of the regular hexagon to the area of the hexagon formed by the midpoints of its sides is \(\frac{1}{4}\), expressed as a common fraction.