Two vertical poles of lengths 11 ft and 14 ft stand 16 ft apart. A cable reaches from the top of one pole to some point on the ground between the poles and then to the top of the other pole. Where should this point be located to use 30 ft of cable?

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rejkjavik rejkjavik · Answer 1 · 2018-08-22T20:38:04+02:00

We are given

total length of cable is 30 feet

and this cable is on hypotenuse sides of the triangles

so, we can draw triangle as

Let's assume those hypotenuse as 'a' and 'b'

so, we will have

[tex] a+b=30 [/tex]

now, we can find 'a' and 'b' from triangle

Smaller triangle:

[tex] a=\sqrt{x^2+11^2} [/tex]

Larger triangle:

[tex] b=\sqrt{(16-x)^2+14^2} [/tex]

we know that

[tex] a+b=30 [/tex]

so, we can plug this value

and we get

[tex] \sqrt{x^2+11^2}+\sqrt{(16-x)^2+14^2} =30 [/tex]

now, we can solve for x

[tex] \left(\sqrt{x^2+11^2}\right)^2=\left(30-\sqrt{\left(16-x\right)^2+14^2}\right)^2 [/tex]

[tex] 32x-1231=-60\sqrt{x^2-32x+452} [/tex]

[tex] 1024x^2-78784x+1515361=3600x^2-115200x+1627200 [/tex]

now, we can use quadratic formula

we get

[tex] x=\frac{-\sqrt{36416^2-1152389056}+36416}{5152},\:x=\frac{\sqrt{36416^2-1152389056}+36416}{5152} [/tex]

[tex] x=4.510,x=9.627 [/tex]

so,

The point should be located 4.510 feet , 9.627 feet from the smaller pole to use 30 feet cable............Answer