The path described by the limit of the radar is the circle 50 km circle, and
the path of the ship is the 60 km dashed straight line.
The shortest distance between the ship and Point O, [tex]\overline {OP}[/tex] is (C) 40 km
Reasons:
The tracking distance of the radar, r = 50 km
Distance the ship moves while within the reach of the radar = 60 km.
Required:
The shortest distance between the ship and the Point O
Solution:
The closest distance between the path of the ship and the Point O is given
by the perpendicular distance from the Point O to the path of the ship
which is given by Pythagoras's theorem and that the perpendicular line
from O to the path of the ship bisects the length of the ship's path.
Let A and B represent the points where the path of the ship intersects with
the circumference of the radar, and let P represent the point the
perpendicular from O intersects AB, we have;
[tex]\overline {AO}[/tex] = The radius of the radar = 50 km
[tex]\overline {AB}[/tex] = [tex]\overline {AP}[/tex] + [tex]\overline {PB}[/tex] (segment addition postulate)
[tex]\overline {AB}[/tex] = 60 (given)
[tex]\overline {AP}[/tex] + [tex]\overline {PB}[/tex] = 60 (transitive property)
[tex]\overline {AP}[/tex] = [tex]\overline {PB}[/tex] (definition of [tex]\overline {AB}[/tex] bisected by [tex]\overline {OP}[/tex])
∴ [tex]\overline {AP}[/tex] + [tex]\overline {PB}[/tex] = [tex]\overline {AP}[/tex] + [tex]\overline {AP}[/tex] = 2× [tex]\overline {AP}[/tex] = 60
[tex]\overline {AP}[/tex] = 30
[tex]\overline {AO}[/tex]² = [tex]\overline {OP}[/tex]² + [tex]\overline {AP}[/tex]² (Pythagoras's theorem)
[tex]\overline {OP}[/tex]² = [tex]\overline {AO}[/tex]² - [tex]\overline {AP}[/tex]²
∴ [tex]\overline {OP}[/tex]² = 50² - 30² = 1600
[tex]\overline {OP}[/tex] = √(1600) = 40
[tex]\overline {OP}[/tex] = 40
The shortest distance between the ship and Point O, [tex]\overline {OP}[/tex] is (C) 40 km
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