A footbridge is in the shape of an arc of a circle. the bridge is 10 ft tall and 27 ft long, horizontally. what is the radius of the circle that contains the bridge? round your answer to the nearest tenth.
see the picture attached to better understand the problem
we know that the bridge is 10 ft tall and 27 ft long, horizontally so BC=10 ft DE=27 ft---------> DC=27/2----> 13.5 ft DA=r (radius of the circle) AC=r-10
Applying the Pythagoras Theorem DA²=DC²+AC² ------> r²=13.5²+(r-10)²----> r²=182.25+r²-20r+100 20r=282.25--------> r=14.11 ft
