Two vertical poles of lengths 6 feet and 8 feet stand 10 feet apart.

Question

A cable reaches from the top of one pole to some point of the ground between the poles and then to the top of the other pole. Where should this point be located to use 18 feet of cable?

Answer 1

Let x equal the distance from the point that the cable reaches the ground to the base of the 6-foot pole.

 

Then the distance from the 8-ft pole is 10 - x.

 

The two triangles will share a point where the cable reaches the ground.

 

The hypotenuses will sum up to 18 ft.

 

√[6²+x²] + √[8²+(10-x)²] = 18

 

√[36+x²] + √[64+100-20x+x²] = 18

 

√[36+x²] + √[164-20x+x²] = 18

 

√[36+x²] - 18 = √[164-20x+x²]

 

(36 + x²) -36√[36 +x²] + 324 = 164 - 20x +x²

 

360 - 36√[36 +x²] = 164 - 20x

 

196 +20x = 36√[36 +x²]

 

196² + 7840x + 400x² = 1296(36 + x²)

 

896x² - 7840x + 8240 = 0

 

56x² - 490x + 515 = 0

 

Using the quadratic formula, x = 1.22 ft or 7.53 ft.