SOLUTION: two cars leave an intersection, one car travels north the other east. When the car traveling north had gone 12 miles the distance between the cars was 8 miles more than the distanc
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-> SOLUTION: two cars leave an intersection, one car travels north the other east. When the car traveling north had gone 12 miles the distance between the cars was 8 miles more than the distanc
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Question 283303: two cars leave an intersection, one car travels north the other east. When the car traveling north had gone 12 miles the distance between the cars was 8 miles more than the distance traveled by the car heading east. how far had the eastbound car traveled? Found 2 solutions by richwmiller, oberobic:Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! 12^2+b^2=(b+8)^2
144+b^2=b^2+16x+64
144=16x+64
80=16x
5=x
a classic 5,12,13 right triangle triplet
You can put this solution on YOUR website! This is an interesting application of the Pythagorean Theorem.
The car traveling north can be viewed as going up the y-axis 12 units.
The car going east can be be viewed as going to the right along the x-axis.
It has traveled 'd' miles.
The diagonal from that connects the two cars is the hypotenuse of a right triangle.
It's length is d+8, which is given.
.
12^2 + d^2 = (d+8)^2
144 + d^2 = d^2 +16d + 64
subtract d^2 from both sides
144 = 16d + 64
subtract 64 from both sides
80 = 16d
divide both sides by 16
5 = d
d = 5
.
Check to ensure diagonal 'c' is 5+8 =13.
c^2 = 12^2 + 5^2
c^2 = 144 + 25
c^2 = 169
So, c = 13
Correct.
.
Answer:
The eastbound car has travelled 5 miles.
.
Done.
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