A building is on fire and he fire department is called. They need to evauate people from the third floor and the second floor windows with the extension ladder attached to the back of the truck. The base of the ladder is 2m above the ground. The fully extended ladder is 25m in length.

q#1-If the fire truck parks so the base of the ladder is 7m from the building, will it reach the third-floor window which is 26m above ground?

q#2The second-floor window is 22m above the ground, by how much should the ladder be retracted so that it will just reach the second-floor window.

q#3If the ladder gets stuck in the fully extended position how much fartheer should the fire truck move away from the building to just reach the second floor window.

Guest Apr 27, 2020

#1**+1 **

Question #1: The base of the ladder is 7m from the building; the length of the ladder is 25m.

Pythagorean Theorem: 7^{2} + h^{2} = 25^{2} ---> 49 + h^{2} = 625 ---> h^{2} = 576 ---> h = 24.

Since the base of the ladder is 2 meters above the ground, it will reach 26 meters.

Question #2: The length of the ladder now is 'x', the hypotenuse. The base of the triangle is still 7m.

The ladder will have to reach a height of 20m (we need to subract the distance of the ladder from the ground.)

7^{2} + 20^{2} = x^{2} ---> 49 + 400 = x^{2} ---> 449 = x^{2} ---> x = 21.2 meters

---> retract the ladder 25m - 21.2m = 3.8m

Qustion #3: The length of the ladder is 25m; the height is 20m, find the base of the triangle.

b^{2} + 20^{2} = 25^{2} ---> b^{2} + 400 = 625 ---> b^{2} = 225 ---> b = 15 (or 8m farther than it was)

geno3141 Apr 28, 2020