Question

Two vertical poles of length 10 and 12 ​feet, respectively, stand 17 feet apart. A cable reaches from the top of one pole to the some point on the ground between the poles and then to the top of the other pole. Express the amount of cable​ used, L, as a function of the distance of the cable from the 10​-foot ​pole, x.

Answer 1

Answer:

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433} ft$

Step-by-step explanation:

let length of calble = L

distance from 10-foot pole = x ft

Moving from the top of the 10-foot pole to point x, we have

From the diagram that

$x^{2} +10^{2}=L_{1} ^{2} \\x^{2} + 100 = L_{1} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}+100}=\sqrt{L_{1} ^{2}}\\ \sqrt{x^{2}+100}= L_{1}$

Moving from point x to the top of the 12-foot pole, we have

From the diagram that

$(17-x)^{2}+12^{2}=L_{2} ^{2} \\289-34x+x^{2} +144=L_{2} ^{2}\\x^{2} -34x+433=L_{2} ^{2}$

Take the square root of both sides

$\sqrt{x^{2}-34x+433}=\sqrt{L_{2} ^{2}} \\\sqrt{x^{2}-34x+433}=L_{2}$

Amount of cable used, L = L1 + L2

$L=\sqrt{x^{2}+100}+\sqrt{x^{2}-34x+433}$