Question

The bottoms of two vertical poles are 12 feet apart and are on a region of flat ground. One pole is 6 feet tall and the other is 15 feet tall. How long, in feet, is a wire stretched from the top of one pole to the top of the other pole?

Answer 1

Answer:

The length of the wire is 15 ft.

Step-by-step explanation:

The solution of this exercise comes from an application of the Pythagorean theorem. The explanation here is complemented with the figure attached.

In the figure the segments AB and CD represents the vertical poles, where the length of AB is 6 ft and the length of CD is 15 ft. We want to find the length of the segment BD, that represents the stretched wire. The length of the segment AC is 12 ft, which is the distance between the poles.

If we draw an imaginary line from A perpendicular to DC, we obtain a rectangle ABEC, and a right triangle BED. Then, the length of BE is 12 ft. Moreover, the length of CE is 6 ft, because is equal to the length of AB. Hence, the length of DE is 9 ft, because DE = DC-EC.

As we want to find the length of the hypotenuse BD of the right triangle BED, and we already have the lengths of the other two sides, we only need to apply the Pythagorean theorem. This is

$BD^2 = BE^2 + ED^2 = 12^2+9^2= 144+81=225.$

Then, taking square roots in both sides: BD=15 ft.