Question

select the correct answer.
lyra and donna are testing the two - way radios they built for their high school science project. lyra goes to the top of a building that is 22 meters high with one of the radios and stands directly above the buildings entrance. donna walks out of the entrance, perpendicular to the building, with the other radio. if donna is more than 50 meters from the entrance, the connection between the radios breaks. what is the range of communication for the two radios to the nearest meter?
a. 50 meters
b. 55 meters
c. 62 meters
d. 72 meters

Explanation:

Step1: Identify the right triangle

The building height (22 m) and Donna's distance from entrance (50 m) are legs of a right triangle. The range is the hypotenuse.

Step2: Apply Pythagorean theorem

Let \( a = 22 \), \( b = 50 \), hypotenuse \( c=\sqrt{a^2 + b^2}=\sqrt{22^2 + 50^2}=\sqrt{484 + 2500}=\sqrt{2984}\approx54.63\approx55 \) meters. Wait, no—wait, the problem says "if Donna is more than 50 meters, connection breaks". So the maximum distance (range) is when Donna is at 50 meters? Wait, no, wait: Lyra is at height 22, Donna is moving perpendicular, so the distance between them is the hypotenuse of right triangle with legs 22 and \( x \) (Donna's distance from entrance). The connection breaks when \( x > 50 \), so the maximum \( x \) before breaking is 50, so we need to find the hypotenuse when \( x = 50 \). Wait, no—wait, the range is the distance between them when \( x = 50 \), because beyond that it breaks. So calculate \( c=\sqrt{22^2 + 50^2}=\sqrt{484 + 2500}=\sqrt{2984}\approx54.63\approx55 \)? Wait, but wait, maybe I mixed up. Wait, no: the building is 22 m high (vertical leg), Donna's distance from entrance is horizontal leg (50 m when connection is about to break). So the distance between them (range) is the hypotenuse. So \( c=\sqrt{22^2 + 50^2}=\sqrt{484 + 2500}=\sqrt{2984}\approx54.63\approx55 \). Wait, but the options: B is 55. Wait, but let me recalculate: 22 squared is 484, 50 squared is 2500, sum is 2984. Square root of 2984: 54.63, which is approximately 55. So the range is when Donna is at 50 meters, the distance between them is ~55 meters. So the range is 55 meters? Wait, but wait, maybe I made a mistake. Wait, no—wait, the problem says "the range of communication"—so when Donna is within 50 meters from entrance, the connection works. So the maximum distance (range) is the distance between them when Donna is at 50 meters from entrance. So using Pythagoras: \( c = \sqrt{22^2 + 50^2} = \sqrt{484 + 2500} = \sqrt{2984} \approx 54.63 \approx 55 \) meters. So the answer should be B? Wait, but wait, let me check again. Wait, 22 squared is 484, 50 squared is 2500, sum is 2984. Square root of 2984: let's calculate 54^2=2916, 55^2=3025. So 2984 is between 54^2 and 55^2. 2984 - 2916 = 68, 3025 - 2984 = 41. So closer to 55. So approximately 55 meters. So the range is 55 meters.

Answer:

B. 55 meters