A telephone pole has a wire attached to its top that is anchored to the ground. The distance from the bottom of the pole to the anchor point is 63 feet less than the height of the pole. If the wire is to be 9 feet longer than the height of the pole, what is the height of the pole?

Answer:Height of pole =108 feetStep-by-step explanation:Let height of the pole be= AB= [tex]x[/tex] feetDistance from bottom of the pole to anchor point =BC= [tex]x-63[/tex] feetLength of wire =AC= [tex]x+9[/tex] feetApplying Pythagorean theorem.In the given [tex]\triangle ABC[/tex][tex]AC^2=AB^+BC^2[/tex]Plugging in values of each.[tex](x+9)^2=x^2+(x-63)^2[/tex]Expanding the binomials using identities.[tex]x^2+18x+81=x^2+x^2-126x-3969\\[/tex]Combining like terms.[tex]x^2+18x+81=2x^2-126x+3969[/tex]Subtracting [tex]x^2[/tex] from both sides.[tex]x^2-x^2+18x+81=2x^2-x^2-126x+3969[/tex][tex]18x+81=x^2-126x+3969[/tex]Subtracting [tex]18x+81[/tex] from both sides.[tex](18x+81)-18x-81=x^2-126x+3969-18x-81[/tex][tex]0=x^2-144x+3888[/tex]We get the quadratic equation to solve.[tex]x^2-144x+3888=0[/tex]Solving quadratic using formula:[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]Plugging in values.[tex]x=\frac{-(-144)\pm\sqrt{(-144)^2-4(1)(3888)}}{2(1)}[/tex][tex]x=\frac{144\pm\sqrt{20736-15552}}{2}[/tex][tex]x=\frac{144\pm\sqrt{5184}}{2}[/tex][tex]x=\frac{144\pm72}{2}[/tex][tex]x=\frac{144\pm 72}{2}[/tex][tex]x=\frac{144+72}{2}[/tex] and [tex]x=\frac{144-72}{2}[/tex][tex]x=\frac{216}{2}[/tex] and [tex]x=\frac{72}{2}[/tex][tex]x=108[/tex] and [tex]x=36[/tex] Since distance from pole to anchor point is 63 feet less than the height of pole, thus the height of the pole has to be =108 feet