GEOMETRY - Areas of Triangles

Question

A triangular lot ABC has side AB = 48m and side AC = 100m. The height of triangle ABC = 96m represented by CD, with D being a point on line AB. If a fence is built perpendicular to side AB so that the area of the lot is equally divided, how far from point A should the perpendicular be drawn?

Answer 1

We have two similar triangles formed: one with base 28 and height 96 and the other with base x from A and height h. to know the sides of the new triangle A, D',C' we need to know the distance from A of the actual triangle, using Pythagoras, we have that AD=28 m. now we make the proportion: (28/96)=(AD'/C'D'), and the other equation which is (AD'*C'D')/2= 1152 m2. =>28*C'D'=96*AD' ; plugging C'D' = 96*AD'/28 into the second equation we obtain AD'^2*96/(28) = 2304 => AD'^2=2304*28/96 = 672

Then AD' = 25.92 m ≈ 26 m      h= 88.86 Area = 1151.62 which is original area times 0.5.

Answer 2

Christy, a segment that goes from a vertex of a triangle and is perpendicular to its opposite side is called...? 

Yes!  The height.

So, if segment CD is the height of ΔABC, and D lies on line AB, then Line CD is perpendicular to side AB. 

Then the fence that you intend to build necessarily has to be built along line CD, doesn't it?

I think so because one of the assumptions is "a fence is built perpendicular to side AB".

So, the fence will be the same as the height, that is CD, with D on side AB.

 

Now, remember what kind of triangles are divided into congruent triangles by their height?

Yes.  The isosceles triangles. The height of isosceles triangles do some fancy things: they bisect the vertex and are the perpendicular bisector of its opposite side.  So...

In order for the fence to divide equally the area of ΔABC, the triangle has to be isosceles and therefore, the distance from D to A will be:?

 

Remember:

AB=48 m

The fence is segment CD

D lies on side AB, and

Segment CD is the perpendicular bisector of side AB