0

# Find the length of each.

two straight wires are attached at a point 24 feet above the base of a vertical pole standing on level ground. One wire is 5 feet longer than the other reaches a point on the ground 11 feet farther from the base. Find the length of each.

### 2 Answers by Expert Tutors

Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
0
L^2 = x^2 + 24^2

(L+5)^2 = (x+11)^2 + 24^2

L^2 + 10L + 25 = x^2 + 22x + 11^2 + 24^2

From this subtract 1st equation

10L + 25 = 22x + 121

22x = 10L - 96

11x = 5L - 48

x = (5L - 48)/11

L^2 = ((5L - 48)/11)^2 + 24^2

121 L^2 = (5L - 48)^2 + 121*24^2

121 L^2 = 25L^2 - 480L + 4*24^2 + 121*24^2

96 L^2 + 480L - 125*24^2 = 0

L^2 + 5 L - 750 = 0

25 + 3000 = 3025 = 55^2

L = (-5 ± 55)/2 = -30, 25

L = 25

L + 5 = 30
Arthur D. | Effective Mathematics TutorEffective Mathematics Tutor
5.0 5.0 (7 lesson ratings) (7)
0
two right triangles are formed
the smaller triangle has the following measurements:         ________
let x=hypotenuse, one leg is 24 feet, and the other leg is √x^2-24^2
the larger triangle has the following measurements:               ________
let x+5=hypotenuse, one leg is 24 feet, and the other leg is √x^2-24^2 + 11
use the Pythagorean Theorem on the larger triangle:
_______
(x+5)^2=24^2+(√x^2-576 + 11)^2
_______
x^2+10x+25=576+x^2-576+22(√x^2-576 + 121
_______
10x+25=22√x^2-576  + 121
_______
10x-96=22√x^2-576    note: you have x^2 on both sides and 576-576=0
now square both sides
100x^2-1920x+9216=484(x^2-576)
100x^2-1920x+9216=484x^2-278,784
384x^2+1920x-288,000=0
divide all terms by 384(if you did not know this, divide by 2 repeatedly and then divide by 3)
x^2+5x-750=0
(x+30)(x-25)=0
x-25=0
x=25 feet and x+5=30 feet are the lengths of the wires
the triangles have measurements of 25 ft, 24 ft , and 7 ft and 30 ft, 24 ft and 18 ft