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22 Oct 2019, 22:13
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Difficulty:
15%
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Question Stats:
85% (01:11) correct
15%
(01:19)
wrong
based on 53
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The figure above represents a triangular field. What is the minimum distance, in meters, that a person would have to walk to go from point A to a point on side BC?
A. 50
B. 60
C. 70
D. 80
E. 90
Attachment:
traingle.jpg [ 5.11 KiB | Viewed 2327 times ]
Archit3110
Major Poster
22 Oct 2019, 22:19
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Bunuel
drop a perpendicular D from A to side BC ; BC = 80 ; AC= 100 and AD =60
∆ 5:4:3 ;
IMO B ; 60
29 Oct 2019, 12:29
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The triangle has two sides with equal length. Their opposite angles must therefore also be of opposite length. We can conclude that this is an isosceles triangle.
The shortest way is a straight line from A to the line BC which is the height of the triangle. We call it D. D splits the triangle in two identical triangle, creating a 90° angle. This lets us apply the Pythagoras Theorem a^2 + b^2 = c^2.
In this case this would be a^2 + 80^2 = 100^2. As a result, solving for a, we get a = 60. The answer is B
The shortest way is a straight line from A to the line BC which is the height of the triangle. We call it D. D splits the triangle in two identical triangle, creating a 90° angle. This lets us apply the Pythagoras Theorem a^2 + b^2 = c^2.
In this case this would be a^2 + 80^2 = 100^2. As a result, solving for a, we get a = 60. The answer is B
