# 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

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### 2 Answers

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):

`d=|52(0)+71(0)-3692|/sqrt(52^2+71^2)=3692/sqrt(7745)~~41.95`

**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.

AB=`sqrt(52^2+71^2)=88.0056`

Now, considering CD as the altitude, Area of the `Delta` =`1/2*88.0056*x`

So, `1/2*88.0056*x=1/2*52*71`

`rArr x=(52*71)/88.0056=41.95`

Therefore, the shortest distance from C to AB is **41.95 m.**