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 17 feet less than the height of the pole. If the wire is to be 8 feet longer than the height of the pole, what is the height of the pole?
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 17 feet less than the height of the pole. If the wire is to be 8 feet longer than the height of the pole, what is the height of the pole?
You can get everything in terms of a single unknown and use Pythagoras' Theorem.
The vertical leg of the right triangle ............... H the height of the pole
The horizontal leg of the right triangle ........... H – 17 the distance to the anchor point
The hypotenuse of the right triangle .............. H + 8 the length of the wire
P's Theorem c2 = a2 + b2 (H + 8)2 = H2 + (H – 17)2
H2 + 16H + 64 = H2 + (H2 – 34H + 289)
H2 + 16H + 64 = H2 + H2 – 34H + 289 = 2H2 – 34H + 289
combine like terms H2 – 2H2 + 16H + 34H + 64 – 289 = 0
–H2 + 50H – 225 = 0
multiply both sides by –1 H2 – 50H + 225 = 0
factor (H – 5) (H – 45) = 0
the solutions are H = 5 and H = 45
H cannot be 5 in real life because H – 17 (the distance to the anchor point) would be a negative number.
Therefore H = 45 the height of the pole
.
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 17 feet less than the height of the pole. If the wire is to be 8 feet longer than the height of the pole, what is the height of the pole?
You can get everything in terms of a single unknown and use Pythagoras' Theorem.
The vertical leg of the right triangle ............... H the height of the pole
The horizontal leg of the right triangle ........... H – 17 the distance to the anchor point
The hypotenuse of the right triangle .............. H + 8 the length of the wire
P's Theorem c2 = a2 + b2 (H + 8)2 = H2 + (H – 17)2
H2 + 16H + 64 = H2 + (H2 – 34H + 289)
H2 + 16H + 64 = H2 + H2 – 34H + 289 = 2H2 – 34H + 289
combine like terms H2 – 2H2 + 16H + 34H + 64 – 289 = 0
–H2 + 50H – 225 = 0
multiply both sides by –1 H2 – 50H + 225 = 0
factor (H – 5) (H – 45) = 0
the solutions are H = 5 and H = 45
H cannot be 5 in real life because H – 17 (the distance to the anchor point) would be a negative number.
Therefore H = 45 the height of the pole
.
