City B is 5 miles east of City A. City C is 10 miles south

Question

City B is 5 miles east of City A. City C is 10 miles south-east of city B. Which of the following is the closest to the distance from City A to City C?

A. 11
B. 12
C. 13
D. 14
E. 15

Bunuel · Accepted Answer

Refer to the diagram below: untitled.PNG From 45-45-90 right triangle BDC: BD=DC=5\sqrt{2} ; AC^2=AD^2+DC^2=(AB+BD)^2+DC^2=(5+5\sqrt{2})^2+(5\sqrt{2})^2\approx{(5+7)^2+7^2}=193 ; AC=\sqrt{193} \approx{14} . Answer: D.

beyondgmatscore · Answer

This is easy to do using coordinate geometry. Let B be at origin, then A would lie on X axis with coordinates (-5,0). Since C is south east of B, it will be along the line y=-x. and the coordinate of a point 10 units from origin on this line would be (5*2^0.5,-5*2^0.5) (calculated as follows - if the coordinates are k,-k then distance is given by 10 = (k^2+k^2)^0.5)

Now distance between A and C would be given by d = ((5*2^0.5)^2+(5*(2^0.5+1))^2))^0.5 which would be ~(195.7)^0.5 which is closest to 14 among the choices. so answer D

Now distance between A and C would be given by d = ((5*2^0.5)^2+(5*(2^0.5+1))^2))^0.5 which would be ~(195.7)^0.5 which is closest to 14 among the choices. so answer D

vikrantgulia · Answer

I am bit confuse with answer. Spoiler OA says answer is C and bunuel says answer is D.

Moreover, Bunuel how can you surely say that the BD=DC or BDC is an isoceles right angle triangle.

My conclusion to answer is C.

As C is in south-east direction with 10. SO BC=10 and AB=5

A. With Pythagoras theorem 10^2 +5^2 =sqrt(125) =11. So A is not an answer.
B. It is closer to 11 and more toward extreme
C. C is correct answer logically.
D. It is closer to 15 and more toward extreme
E. Sum of two side of triangle is always greater than 3rd side so 10+5=15 . So E is not an answer

Bunuel · Answer

Edited the OA, it's D, not C.

BDC is an isosceles triangle because south-east of city B means 45 degrees angle from AB.

aimtoteach · Answer

Hello  Bunuel ,

I tried it this way -

1) If we drop a perpendicular from B to AC (lets call the point X) we get 2 rt triangles - AXB and BXC
2) Triangle AXB has one side and 5 so we look for Pyth. Triplets and fill in 3 and 4 for AX and BX.
3)Triangle BXC - we have 2 sides now - so we again apply Pyth. theorem and find the side XC ( which is \sqrt{10^2-4^2} = ~9)

Side AC = 3+9 = 12

I know I am wrong but can you please point out where.

Thank you in advance.

roopika2990 · Answer

This is the same way I did it. Can you explain how the triangle made a 45 degree angle?

chetan2u · Answer

Hi,
where you are going wrong is by assuming the sides to be an integer..
hyp is 5, the other sides can be 1 and  \sqrt{99} 
and many more ways ...
Hope it helped

chetan2u · Answer

Hi  roopika2990 ,
the angle is 45 degree because it is given  south east  and south east will make an angle of 45 degree

Priyanka2018 · Answer

Bunuel  
triangle ABC is forming obtuse angle triangle so why cant we use the property of obtuse triangle
AB^2+ BC^2< AC^2

if we use this we get
125< AC^2
11.XX <AC
 AC has to be little greater than 11.xx
12 satisfy this condition
Can you please point out flaw in this approach.

Thanks in Advance

Bunuel · Answer

All answers greater than 12 also satisfy this, not only B.

rajatchopra1994 · Answer

My Explanation:

Since the position of city C is just given as southeast, its distance from A can vary from 15 (10 miles east of B - in a straight line) and (125)^1/2 south of city B (10 miles below B).

Since the 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.

This is the exam approach if you are short of time.

bumpbot · Answer

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City B is 5 miles east of City A. City C is 10 miles south

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City B is 5 miles east of City A. City C is 10 miles south

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