A footbridge is in the shape of an arc of a circle. The bridge is 7 ft tall and 29 ft wide. What is the radius of the circle that contains the bridge? Round your answer to the nearest tenth.

Question

contestada

Respuesta :

Newton9022 Newton9022 · Answer 1 · 2020-05-05T03:43:47+02:00

Answer:

18.5 ft (to the nearest tenth)

Step-by-step explanation:

The diameter of the circle passes through the center of the circle and divides the width of the circle into two equal parts.

We complete the circle and apply the Law of Intersecting Chords.

In the diagram, by this law:

AB X BC =DB X BE

7*x=14.5X14.5

x=14.5X14.5÷7

x=30.04

Therefore:

Length of the Diameter= 7+30.04 ft=37.04 ft

Radius =37.04/2= 18.5 ft

Answer Link

andromache andromache · Answer 2 · 2022-03-24T12:31:24+01:00

The radius of the circle should be 18.5 ft (to the nearest tenth).

Since there is the bridge is 7 ft tall and 29 ft wide.

Also,

here we apply the Law of Intersecting Chords.

So,

[tex]AB \times BC =DB \times BE\\\\7\times x=14.5\times 14.5\\\\x=14.5\times 14.5\div 7[/tex]

x=30.04

Now

Length of the Diameter= 7+30.04 ft=37.04 ft

So, Radius = 50% of 37.04 = 18.5 ft

Hence, The radius of the circle should be 18.5 ft (to the nearest tenth).

Learn more about radius here: https://brainly.com/question/21502832