Question

two cars leave the same parking lot, with one heading north and the other heading east. after several minutes, the eastbound car has traveled 6 kilometers. if the two cars are now a straight-line distance of 7 kilometers apart, how far has the northbound car traveled? if necessary, round to the nearest tenth. kilometers

Explanation:

Step1: Identify the problem type

This is a right - triangle problem where the east - bound and north - bound distances are the legs of a right triangle, and the straight - line distance between the cars is the hypotenuse. We can use the Pythagorean theorem, which states that for a right triangle with legs of length \(a\) and \(b\) and hypotenuse of length \(c\), \(a^{2}+b^{2}=c^{2}\). Let the distance traveled by the north - bound car be \(x\) (this is one leg, \(a = x\)), the distance traveled by the east - bound car is \(b = 6\) kilometers, and the hypotenuse \(c = 7\) kilometers.

Step2: Apply the Pythagorean theorem

Substitute the known values into the Pythagorean theorem: \(x^{2}+6^{2}=7^{2}\).
First, calculate \(6^{2}=36\) and \(7^{2}=49\). So the equation becomes \(x^{2}+36 = 49\).
Then, subtract 36 from both sides of the equation: \(x^{2}=49 - 36\).
\(49-36 = 13\), so \(x^{2}=13\).

Step3: Solve for x

Take the square root of both sides: \(x=\sqrt{13}\).
\(\sqrt{13}\approx3.6\) (rounded to the nearest tenth).

Answer:

\(3.6\)