Geometry Problem

[MathStudent1999]

8. A triangle whose verticies (0,0), (5,12) & (8,6) is seperated into two parts by a vertical line so that each part has the same area. What us the cordinates of the intersection of the deviding line and the x-axis? Express your answer in simplest radical form.

Can someone give me a few hint and tips on this question?

 

[tkhunny]

Chart the points.
Draw the triangle.
Play around with vertical lines in various places and see what you get.
What sorts of secondary shapes are produced when yudraw these various vertical lines?
Develope an exact solution from your exploration.

 

[MathStudent1999]

Chart the points.
Draw the triangle.
Play around with vertical lines in various places and see what you get.
What sorts of secondary shapes are produced when yudraw these various vertical lines?
Develope an exact solution from your exploration.

I got : [3 SQRT(5)]/2. Can someone verify this is correct?

 

[pappus]

8. A triangle whose verticies (0,0), (5,12) & (8,6) is seperated into two parts by a vertical line so that each part has the same area. What us the cordinates of the intersection of the deviding line and the x-axis? Express your answer in simplest radical form.

Can someone give me a few hint and tips on this question?

Here is how would have done this problem:

1. Draw a sketch.

2. The total area of triangle ABC is 33 units. Thus you are looking for a triangle whose area is 16.5 units.

3. The triangle in question (with the thick lined sides) can be obtained as difference of 2 right triangles (blue - red):

The area of the blue triangle is calculated by:

$\displaystyle a_{\Delta blue}=\frac12 \cdot x \cdot \underbrace{\frac{12}5 x}_{why?}=\frac65 x^2$

The area of the red triangle is calculated by:

$\displaystyle a_{\Delta red}=\frac12 \cdot x \cdot \underbrace{\frac34 x}_{why?} = \frac38 x^2$

4. Solve for x

$\displaystyle \frac65 x^2 - \frac38 x^2=\frac{33}2$

EDIT: Sorry Dennis, I didn't intend to copy your solution (Sometimes an extra pair of eyes could be helpful)

 

[MathStudent1999]

You're amazing, young man! That's not quite it, but it is an existing length:
the vertical length from bottom side of the given triangle to the x axis;
answer is 2SQRT(5), so I'm sure you just got mixed up.

                 B(5,12)
  
              E(u,e)
 

                         C(8,6)

              D(u,d)



A(0,0)   u    U  W       V

That's the "picture": given triangle ABC, vertical line ED (extended to U on x axis)
bisecting the area, BW and CV both verticals (used to create similar right triangles...

Your 3SQRT(5) / 2 is length of DU. You need length of AU, right?

I'll let you try again. By the way, length of EU = 24SQRT(5) / 5

I attacked it this way: area_right_triangle_AEU - area_right_triangle_ADU = 33/2.

Your right! I just got the wrong line