In the diagram,ABC represents a horizontal triangular field.A path runs along the edge AB of the field.BC=52m,AC=71m, and angle ACB =90 Calculate the shortest distance from C to AB.Thanks
Without loss of generality, place C at the origin, A at (71,0) and B at (0,52).
The equation of the line through A and B is `y=-52/71 x+52` .
Rewrite in standard form as `52x+71y-3692=0` .
The distance from a point `(x_0,y_0)` to a line ax+by+c=0 is:
`d=|ax_0+by_0+c|/sqrt(a^2+b^2)` . Here a=52,b=71, and c=-3692 with the point at (0,0):
Thus the shortest distance to the line is approximately 42m.
In `Delta ABC` , `BC=52m,AC=71m` , and `angle ACB =90^o`
Considering AC as the altitude, Area of the `Delta` =`1/2*52*71` `m^2`
The shortest distance from C to AB is the altitude CD. Let it be x.
Now, considering CD as the altitude, Area of the `Delta` =`1/2*88.0056*x`
Therefore, the shortest distance from C to AB is 41.95 m.