Question

during the summer heat, a 6-mi bridge expands 6 ft in length. assuming the bulge occurs straight up the middle, how high is the bulge?
select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
a. using radicals, b is exactly . the b, up to three decimal places, is approximately .
b. the b, is exactly . no approximation is necessary.

Explanation:

Step1: Convert units

First, convert the length of the bridge from miles to feet. Since \(1\) mile \( = 5280\) feet, a \(6 -\)mile bridge has a length \(L=6\times5280 = 31680\) feet. After expansion, the new length is \(L'=31680 + 6=31686\) feet.
The half - length of the original bridge \(a=\frac{31680}{2}=15840\) feet, and the half - length of the expanded bridge \(c=\frac{31686}{2}=15843\) feet.

Step2: Apply the Pythagorean theorem

We can consider a right - triangle, where the hypotenuse is the half - length of the expanded bridge \(c\), one leg is the half - length of the original bridge \(a\), and the other leg is the height of the bulge \(b\). According to the Pythagorean theorem \(c^{2}=a^{2}+b^{2}\), so \(b=\sqrt{c^{2}-a^{2}}\).
Substitute \(a = 15840\) and \(c = 15843\) into the formula:

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$$b=\sqrt{15843^{2}-15840^{2}}=\sqrt{(15843 + 15840)(15843 - 15840)}=\sqrt{31683\times3}=\sqrt{95049}= 308.3$$

(approximate value, calculated as \(b=\sqrt{15843^{2}-15840^{2}}=\sqrt{15843\times15843-15840\times15840}=\sqrt{(15840 + 3)^{2}-15840^{2}}=\sqrt{15840^{2}+6\times15840 + 9-15840^{2}}=\sqrt{6\times15840+9}=\sqrt{95040 + 9}=\sqrt{95049}\approx308.3\))

Answer:

A. Using radicals, \(b\) is exactly \(3\sqrt{10561}\). The \(b\), up to three decimal places, is approximately \(308.3\)