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City B is 5 miles east of the city A, city C is 10 miles

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City B is 5 miles east of the city A, city C is 10 miles [#permalink]

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New post 02 Feb 2008, 11:22
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1. City B is 5 miles east of the 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
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New post 02 Feb 2008, 11:31
E 15 miles....



AC >= 5 + 10

What is the OA ?
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New post 02 Feb 2008, 11:32
rishi2377 wrote:
1. City B is 5 miles east of the 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


lets drop a line from c to the base and called it k.
AC^2 = BK^2 + CK^2
10^2 = 2 BK^2 (since BK=CK)
BK = 10/sqrt = 5 sqrt2
CK = 5 sqrt2
AK = AB+BK = 5 + 5 sqrt2

AC = sqrt [CK^2 + AK^2)
AC = sqrt [(5+5 sqrt2)^2 + (5 sqrt2)^2]
AC = sqrt [25+ 2x5x5 sqrt2 + 50]
AC = sqrt [75 + 50 sqrt2]
AC = sqrt [75 + 50 x 1.41]
AC = sqrt [75 + 71]
AC = sqrt [146]
AC = approx 12
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Last edited by GMAT TIGER on 02 Feb 2008, 11:36, edited 1 time in total.
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Re: PS: [#permalink]

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New post 02 Feb 2008, 11:35
GMAT TIGER, why is triangle ABC necessarily a right triangle?
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New post 02 Feb 2008, 11:38
solaris1 wrote:
GMAT TIGER, why is triangle ABC necessarily a right triangle?


not triangle ABC but triangle ACO, when we drop a line from C to the base, which is lets say o.

remember : southeast, which makes the two angles 45 degree each and one 90 degree.
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Re: PS: [#permalink]

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New post 02 Feb 2008, 11:41
GMAT TIGER wrote:
solaris1 wrote:
GMAT TIGER, why is triangle ABC necessarily a right triangle?


remember : southeast, which, when we drop a line from c to ther base (o), makes the two angles 45 degree each and one 90 degree.



But B is east of city A and since city C is south east of city B will make the angle ABC an obtuse angle....No ?
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Re: PS: [#permalink]

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New post 02 Feb 2008, 12:07
solaris1 wrote:
GMAT TIGER, why is triangle ABC necessarily a right triangle?


First, if you go to southeast fr5om city B resulting triangle will not be right triangle.
and second, answer is neighter 15 miles nor 12.
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New post 02 Feb 2008, 12:15
GMAT TIGER wrote:
rishi2377 wrote:
1. City B is 5 miles east of the 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


lets drop a line from c to the base and called it k.
AC^2 = BK^2 + CK^2
10^2 = 2 BK^2 (since BK=CK)
BK = 10/sqrt = 5 sqrt2
CK = 5 sqrt2
AK = AB+BK = 5 + 5 sqrt2

AC = sqrt [CK^2 + AK^2)
AC = sqrt [(5+5 sqrt2)^2 + (5 sqrt2)^2]
AC = sqrt [25+ 2x5x5 sqrt2 + 50]
AC = sqrt [75 + 50 sqrt2]
AC = sqrt [75 + 50 x 1.41]
AC = sqrt [75 + 71]
AC = sqrt [146]
AC = approx 12


i missed to include the red part above.

AC = sqrt [(5+5 sqrt2)^2 + (5 sqrt2)^2]
AC = sqrt [(25+ 2x5x5 sqrt2 + 50) + 50]
AC = sqrt [(75 + 50 sqrt2) + 50]
AC = sqrt [125 + 50 x 1.41]
AC = sqrt [125 + 71.5]
AC = sqrt [196.5]
AC = approx 14

so D.
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New post 02 Feb 2008, 12:51
the answer should be 14

I have attached a figure.

AB=5
BC=10
AC=??
L CBD = 45 deg (south-east) and triangle CBD is 45-45-90

so, BD = CD = \(sqrt(100/2) = sqrt50\)

now consider triangle ACD, we have \((AB+BD)^2 + CD^2 = AC^2\)

\((5+sqrt50)^2 + (sqrt50)^2 = AC^2\)
\(125 + 10sqrt50 = AC^2\) ; sqrt50 ~ 7
125+70=AC^2; AC = sqrt(195) ~ 14

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Re: PS: [#permalink]

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New post 02 Feb 2008, 13:35
Thanx for explanation, OA is 14...This is a good question.
Re: PS:   [#permalink] 02 Feb 2008, 13:35
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