Asked by Isabel
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.
Answers
Answered by
Reiny
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
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
Answered by
Isabel
Thanks! But one question..um, where did you get the triangle?
Answered by
Reiny
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
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