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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?

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.

View image DespinaQ770431
View image DespinaQ770431