Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Let's draw the situation presented in the problem:
In order to determine the radius of the circle that contains the bridge, we'll use the following formula:
[tex]r=\frac{h}{2}+\frac{w^2}{8h}[/tex]where h is the height of the arc and w is its width.
This formula is derived from the intersecting chords theorem:
[tex]a\cdot a=b\cdot c[/tex]Since in our case a is half of the width of the arc and b its height:
[tex]\frac{w}{2}\cdot\frac{w}{2}=h\cdot c[/tex][tex]\frac{w^2}{4}=h\mathrm{}c[/tex]dividing both sides by h:
[tex]\frac{w^2}{4h}=c[/tex]since the diameter of the circle is b+c, or in this case h+c:
[tex]d=h+\frac{w^2}{4h}[/tex]since the radius is half the diameter:
[tex]r=\frac{h}{2}+\frac{w^2}{8h}[/tex]Now, let's plug the data we were given into this formula:
[tex]r=\frac{3.3}{2}+\frac{23^2}{8\cdot3.3}[/tex][tex]r=1.65+\frac{529}{26.4}=1.65+0.0378=21.6878[/tex]So the radius of the circle will be 21.6878 ft.


We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.