A bridge is the shape of an arc of a circle. The bridge is 8 feet tall and 36 feet wide. What is the radius of the circle that contains the bridge?
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Hint:
If $c$ is the chord of the arc and $h$ its height (sagitta) than half of $c$, and $r-h$ are the sides of a rectangular triangle with hypotenuse the radius $r$ of the circle:
$$ (r-h)^2+\left(\frac{c}{2}\right)^2=r^2 $$ In the figure: $$ CB=h \qquad CD=c/2\qquad AD=r $$
Emilio Novati
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Another hint...if the radius is $r$ then by the intersecting chords theorem $$8(2r-8)=18^2$$
David Quinn
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