A flagpole broke in a storm. It was originally 81 feet tall. 28 feet are still sticking straight out of the ground, where it sna
pped, but the remaining piece has hinged over and touches the ground some distance away. How far away is the end of the pole from the base of the pole along the ground?
1 answer:
Check out the attached image.
Point P = base of the pole
Point Q = point where the pole broke
Point R = point where the tip of the pole landed on the ground.
We're told that "28 feet are still sticking straight out of the ground", so PQ = 28 is the vertical leg of the right triangle. The horizontal leg is unknown, so we'll call it x, making PR = x. The hypotenuse is QR = 53 because 81-28 = 53. This is the left-over amount after taking off the 28 feet initially out of the ground from the 81 ft pole.
In short, we have these three values
a = x is one of the legs
b = 28 is the other leg
c = 53 is the hypotenuse
Use the Pythagorean Theorem to find x
a^2 + b^2 = c^2
x^2 + 28^2 = 53^2
x^2 + 784 = 2809
x^2 = 2809 - 784
x^2 = 2025
x = sqrt(2025)
x = 45
The end of the pole is 45 feet away from the base
Answer: 45 feet
Point P = base of the pole
Point Q = point where the pole broke
Point R = point where the tip of the pole landed on the ground.
We're told that "28 feet are still sticking straight out of the ground", so PQ = 28 is the vertical leg of the right triangle. The horizontal leg is unknown, so we'll call it x, making PR = x. The hypotenuse is QR = 53 because 81-28 = 53. This is the left-over amount after taking off the 28 feet initially out of the ground from the 81 ft pole.
In short, we have these three values
a = x is one of the legs
b = 28 is the other leg
c = 53 is the hypotenuse
Use the Pythagorean Theorem to find x
a^2 + b^2 = c^2
x^2 + 28^2 = 53^2
x^2 + 784 = 2809
x^2 = 2809 - 784
x^2 = 2025
x = sqrt(2025)
x = 45
The end of the pole is 45 feet away from the base
Answer: 45 feet
You might be interested in