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? Round your answer 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
(x-8)^2 + 18^2 = x^2
x^2 - 16x + 64 + 324 = x^2
-16x = -388
x = 24.25
Thanks! But one question..um, where did you get the triangle?
draw a whole circle
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.
so ...
see above
To find the radius of the circle that contains the bridge, we can use the information given: the bridge is 8 feet tall and 36 feet wide.
Since the bridge has a shape that is an arc of a circle, we know that the highest point of the arc, which is the center of the circle, is located halfway between the top of the bridge. In this case, the highest point of the arc is 8/2 = 4 feet above the base of the bridge.
Now, we have a right triangle with one side measuring half the width of the bridge (18 feet), the other side measuring the height of the arc (4 feet), and the hypotenuse representing the radius of the circle.
Using the Pythagorean theorem, we can solve for the radius (r):
r^2 = (18)^2 + (4)^2
r^2 = 324 + 16
r^2 = 340
Taking the square root of both sides:
r ≈ √340
r ≈ 18.4
Therefore, the radius of the circle that contains the bridge is approximately 18.4 feet.