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16 Oct 2017, 10:12
Kudos
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D
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Difficulty:
35%
(medium)
Question Stats:
74% (02:01) correct
26%
(02:01)
wrong
based on 239
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City B is 8 miles east of City A. City C is 6 miles north of city B. City D is 16 miles east of city C, and city E is 12 miles north of city D. What is the distance from city A to city E?
(A) 10 miles
(B) 20 miles
(C) 24 miles
(D) 30 miles
(E) 42 miles
(A) 10 miles
(B) 20 miles
(C) 24 miles
(D) 30 miles
(E) 42 miles
16 Oct 2017, 12:02
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Bunuel
there are ONLY two sides movement and we have to find the distance between FIRST and LAST point..
FINALLY we have a RIGHT angled triangle with one sides corresponding to EAST movement and second side corresponding to NORTH movement. The HYPOTENUSE is the distance we are looking for.
so side X = 8(east)+16(east)=24
side Y = 6(N)+12(N)=18
if you look at the sides they are 18:24 0r 6*3:6*4 that is in ratio 3:4 so hypotenuse will be in multiple of 5 as 3:4:5 right angle triangle
so distance = 6*5=30
If you do not this you can find hyp by \(\sqrt{18^2+24^2}=\sqrt{900}=30\)
16 Oct 2017, 16:46
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Bunuel
Am I missing something here?
I just figured out the lengths of the hypotenuse of each right triangle and added them.
Both are right triangles whose legs head first east (to the right) and then north (up).
Their hypotenuses have identical slopes of 3/4.
They are both 3x-4x-5x right triangles.
Right ∆ ABC has legs, in miles, of distance 6 and 8. It follows the 3x: 4x: 5x ratio. Its hypotenuse, to keep with the ratio, is 6: 8: 10 miles*
Right ∆ CDE has legs, in miles, of 12 and 16. That's 3(4) and 4(4). So hypotenuse is 5(4) = 20 miles*
Total distance from A to E: 10 + 20 = 30 miles
Answer D
Is this method sound? I can't see why not...
*Or use Pythagorean theorem for each
∆ ABC:
\(6^2 + 8^2 = h^2\)
\(36 + 64 = h^2\)
\(h^2 = 100\)
\(h = 10\)
∆ CDE:
\(12^2 + 16^2 = h^2\)
\(144 + 256 = h^2\)
\(h^2 = 400\)
\(h = 20\)
