IDNLearn.com is designed to help you find the answers you need quickly and easily. Get accurate and detailed answers to your questions from our dedicated community members who are always ready to help.
Sagot :
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]
- 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]
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Accurate answers are just a click away at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.