Question

A footbridge is in the shape of an arc of a circle. The bridge is 3.3 ft tall and 23 ft wide. What is the radius of the circle that contains the bridge?

Answer 1

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:

`r=(h)/(2)+(w^2)/(8h)`

where h is the height of the arc and w is its width.

This formula is derived from the intersecting chords theorem:

`a\cdot a=b\cdot c`

Since in our case a is half of the width of the arc and b its height:

`(w)/(2)\cdot(w)/(2)=h\cdot c`
`(w^2)/(4)=h\mathrm{}c`

dividing both sides by h:

`(w^2)/(4h)=c`

since the diameter of the circle is b+c, or in this case h+c:

`d=h+(w^2)/(4h)`

since the radius is half the diameter:

`r=(h)/(2)+(w^2)/(8h)`

Now, let's plug the data we were given into this formula:

`r=(3.3)/(2)+(23^2)/(8\cdot3.3)`
`r=1.65+(529)/(26.4)=1.65+0.0378=21.6878`

So the radius of the circle will be 21.6878 ft.