City B is 5 miles east of city A. City C is 10 miles southeast of city B.
Which of the following is the closest to the distance from city A to city C?
a) 11 miles
b) 12 miles
c) 13 miles
d) 14 miles
e) 15 miles
OA not available
Distance from City A to City C
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Hi, Here is my approach..
Since the position of city C is just given as south east, its distance from A can vary from 15( 10 miles east of B - in straight line) and (125)^1/2south of city B (10 miles below B).
Since shortest distance is asked, eliminate 15 (because this is the longest), and 11(because shortest can only be (125)^1/2.
With no additional information provided, I would go for the next shortest distance in the list - 12 miles...
IMO B
Since the position of city C is just given as south east, its distance from A can vary from 15( 10 miles east of B - in straight line) and (125)^1/2south of city B (10 miles below B).
Since shortest distance is asked, eliminate 15 (because this is the longest), and 11(because shortest can only be (125)^1/2.
With no additional information provided, I would go for the next shortest distance in the list - 12 miles...
IMO B
Maybe I am slow to pick this up...could anyone please elaborate why the minimum distance is 125^1/2.
Am I assuming correct that angle ABC is 135 degrees and the minimum should be hyp AC formed by the extended
right angled triangle. Kindly correct my thinking.
Am I assuming correct that angle ABC is 135 degrees and the minimum should be hyp AC formed by the extended
right angled triangle. Kindly correct my thinking.
if C is directly below B (i.e south of B), then ABC will be a right angled triangle and AC will be the hypotenuse.
Hence the min. distance will be sqrt(125)
However, since C is south east of B ... ABC is not rt angled triangle... and AC shld be > sqrt(125) which is 12
Hence the min. distance will be sqrt(125)
However, since C is south east of B ... ABC is not rt angled triangle... and AC shld be > sqrt(125) which is 12
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Wow, this one is hard. I've noticed with any direction questions (i.e. city is north, south, etc) it is best to draw a coordinate system.
Also, with the isosceles right triangle, it would be easier to just use the rules of 45-45-90 triangles and know that 10 = x sq rt 2, solve for x to get 5 sq root 2.
I found this:
https://books.google.com/books?id=JpeDFS ... 3F&f=false
Also, with the isosceles right triangle, it would be easier to just use the rules of 45-45-90 triangles and know that 10 = x sq rt 2, solve for x to get 5 sq root 2.
I found this:
https://books.google.com/books?id=JpeDFS ... 3F&f=false
Monika
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Haha, sorry, I've never seen that question so I didn't have the book solution.
I don't know any other way to solve it...but when I originally looked at it, I definitely knocked out E (15 miles) because for the rules of a triangle, we know that (10-5) < x < (10 +5) so 5 < x < 15 (x being the length of the other side of the triangle).
I also knocked out A with the same reasoning as others used in this post in that if it was a right triangle, the third side would be the sq root of 125 (which is a little bigger than 11 since sq root of 121 = 11). However, the side of the triangle is longer than it would be for a right triangle so it should be greater than 11.
That still left me with B, C, or D.
Hopefully someone else has other ideas on other ways to solve besides the book solution!
I don't know any other way to solve it...but when I originally looked at it, I definitely knocked out E (15 miles) because for the rules of a triangle, we know that (10-5) < x < (10 +5) so 5 < x < 15 (x being the length of the other side of the triangle).
I also knocked out A with the same reasoning as others used in this post in that if it was a right triangle, the third side would be the sq root of 125 (which is a little bigger than 11 since sq root of 121 = 11). However, the side of the triangle is longer than it would be for a right triangle so it should be greater than 11.
That still left me with B, C, or D.
Hopefully someone else has other ideas on other ways to solve besides the book solution!
Monika