Question

a telephone pole is 30 feet tall. a cable attached to the top of the pole is anchored to the ground 15 feet away from the base of the pole.

1. what is the length of the cable to the nearest tenth of a foot?

a. 26
b. 30.7
c. 32.2
d. 33.5

1. if a 35 - foot cable were run from the top of the pole and anchored to the ground at a distance from the pole, about how far away from the pole would it be anchored?

a. 16 ft
b. 18 ft
c. 30 ft
d. 38 ft

1. if the telephone pole were 2 feet taller, and the cable was still anchored to the ground 15 feet away from the pole, how would the change in pole height affect the length of the cable?

a. the cable would be exactly 2 feet longer.
b. the length of the cable would not change.
c. the cable would be about 1.8 feet longer.
d. the cable would be about 2.2 feet longer.

Explanation:

2.

Step1: Apply Pythagorean theorem

The Pythagorean theorem is \(a^{2}+b^{2}=c^{2}\), where \(a = 15\), \(b=30\), and \(c\) is the length of the cable.
$$15^{2}+30^{2}=c^{2}$$

Step2: Calculate \(15^{2}+30^{2}\)

\(15^{2}=225\), \(30^{2} = 900\). Then \(15^{2}+30^{2}=225 + 900=1125\). So \(c^{2}=1125\).

Step3: Find \(c\)

\(c=\sqrt{1125}\approx33.5\)

Step1: Apply Pythagorean theorem

Let the distance from the pole be \(x\). The Pythagorean theorem gives \(x^{2}+30^{2}=35^{2}\) (since the height of the pole \(a = 30\), cable length \(c = 35\)).
$$x^{2}+900 = 1225$$

Step2: Solve for \(x^{2}\)

\(x^{2}=1225 - 900=325\)

Step3: Find \(x\)

\(x=\sqrt{325}\approx18\)

Step1: Original cable length

Original \(c_{1}=\sqrt{15^{2}+30^{2}}=\sqrt{1125}\approx33.5\)

Step2: New height and new cable length

New height \(b = 32\). New \(c_{2}=\sqrt{15^{2}+32^{2}}=\sqrt{225 + 1024}=\sqrt{1249}\approx35.3\)

Step3: Calculate the difference

\(c_{2}-c_{1}\approx35.3-33.5 = 1.8\)

Answer:

D. 33.5

3.