A footbridge is in the shape of an arc of a circle. The bridge is 6ft tall and 28 ft wide. What is the radius of the circle that contains the bridge?

Question

The footbridge is only on the top of the circle..

Answer 1

Draw a circle on the x-y plane. Label the points where y axis intersects your circle, top point "A", and bottom point "B". At some point above the x axis, draw a horizonal chord and label the left intersection point "C" and the right point "D". This represents the length of the footbridge. Where the y axis and chord CD intersect, label point "E". The measure of segment AE is 6, the height of the footbridge. Since chord AB bisects chord CD, the measures of CE is 14, and DE is 14.

Now you have two chords that intersect inside the circle, so we use AE * BE = CE * DE. Let BE = x

6 * x = 14 * 14

6x = 196

x = 32 2/3

AB is the diameter of the circle, so AE + BE = 6 + 32 2/3 = 38 2/3

The radius is half the diameter, so radius = 19 1/3