Thursday, June 23, 2011

Ant in a field

An ant, located in a square field, is 13 meters from one of the corner posts of the field, 17 meters from the corner post diagonally opposite that one, and 20 meters from a third corner post.  Find the area of the field.  Assume the land is flat.
Square field, containing an ant, as described above.

1 comment:

  1. Label the vertices of the square A, B, C, D. The ant is at point P, with PB = 13, PC = 20, PD = 17.
    Rotate triangleCDP 90° counterclockwise about C, so that D goes to B, and P goes to Q.
    By definition, angleQCP = 90°. Hence, by Pythagoras' Theorem, PQ = 20root 2.
    Also, since trianglePQC is isosceles, angleCPQ = 45°.
    Square field, containing an ant, as described above.

    Applying the law of cosines (also known as the cosine rule) to triangleBQP
    172 = 132 + (20root 2)2 − 2 · 13 · 20root 2 · cos QPB.
    Simplifying, we find cos QPB = 17root 2 / 26.
    Then sin2 QPB = 1 − cos2 QPB = 49/338.
    Hence sin QPB = 7root 2 / 26. (We will need this result below.)

    We have angleCPB = angleQPB + angleCPQ = angleQPB + 45°.
    Hence cos CPB = cos (QPB + 45°)
    = cos QPB · cos 45° − sin QPB · sin 45°, by trigonometric identity cos(a + b) = cos a · cos b − sin a · sin b
    = (cos QPB − sin QPB) / root 2
    = (10root 2 / 26) / root 2
    = 5/13

    Applying the law of cosines to triangleCPB
    BC2 = 202 + 132 − 2 · 20 · 13 · (5/13) = 369.

    Therefore the area of the field is 369 m2.

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