A footbridge is in the shape of an arc of a circle. The bridge is 4.5 ft tall and 25 ft wide. What is the radius of the circle that contains the bridge? Round to the nearest tenth.

Assuming that the 36 feet wide line base is NOT a diameter, make a sketch

let the radius be x

then I see a right-angled triangle where

Step-by-step explanation:

(x-8)^2 + 18^2 = x^2

x^2 - 16x + 64 + 324 = x^2

-16x = -388

x = 24.25

draw a chord AB, your base , = 36

From the centre O draw a line OA , a radius

let the radius be x

so OA = x, the hypotenuse

from the centre draw a perpendicular to AB to hit AB at C

from C to the circle is 8 , so OC = x-8

OCA is a right - angled triangle.

jml angle lkj & angle mj = lm