To find the height of the gymnastics mat, we can use the properties of a 30-60-90 triangle. This type of triangle has a specific ratio between its sides. In a 30-60-90 triangle, the side lengths are in the ratio [tex]\(1 : \sqrt{3} : 2\)[/tex]. This means that:
- The side opposite the 30° angle is [tex]\(x\)[/tex]
- The side opposite the 60° angle is [tex]\(x\sqrt{3}\)[/tex]
- The hypotenuse is [tex]\(2x\)[/tex]
Given:
- The gymnastics mat extends 5 feet across the floor. This corresponds to the side opposite the 60° angle, which we denoted as [tex]\(x\sqrt{3}\)[/tex].
- Therefore, we have [tex]\(5 = x\sqrt{3}\)[/tex].
To find [tex]\(x\)[/tex], the side opposite the 30° angle, which represents the height of the mat, we need to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{5}{\sqrt{3}} \][/tex]
Next, we rationalize the denominator to get the height in a simpler form:
[tex]\[ x = \frac{5}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \][/tex]
So, the height of the gymnastics mat is:
[tex]\[ \frac{5\sqrt{3}}{3} \text{ feet} \][/tex]
Hence, the correct answer is: [tex]\(\frac{5 \sqrt{3}}{3} \text{ ft}\)[/tex].
Moreover, for additional reference:
- The height [tex]\(\frac{5\sqrt{3}}{3} \approx 2.886751345948129 \, \text{ft}\)[/tex]
- The shorter leg of the triangle, being half of the hypotenuse, is also [tex]\(\frac{5}{2} = 2.5 \, \text{ft}\)[/tex]
- The hypotenuse equals [tex]\(5 \sqrt{3} \approx 8.660254037844386 \, \text{ft}\)[/tex]
- If [tex]\(x = 5/\sqrt{3}\)[/tex], then the hypotenuse exactly would be [tex]\(2 \times \frac{5}{\sqrt{3}} = 10/\sqrt{3} = 10 \, \text{ft}\)[/tex] after rationalization [tex]\(10 \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{10\sqrt{3}}{3}\)[/tex].
Therefore we have these additional values:
- [tex]\( \frac{5\sqrt{3}}{3} \approx 2.886751345948129 \text{ (height ft)} \)[/tex]
- [tex]\(\frac{5}{2} = 2.5 \text{ ft (half hypotenuse)}\)[/tex]
- [tex]\((5 \cdot \sqrt{3})/3 = 2.8867513459481287 \text{ feet}\)[/tex]
- [tex]\(5 \cdot \sqrt{3} \approx 8.660254037844386 \text{ feet}\)[/tex]
- Hypotenuse is [tex]\(10 \text{ ft}\)[/tex]
The relevant height is indeed:
[tex]\(\frac{5 \sqrt{3}}{3} \, \text{ft}.\)[/tex]
