Cutting an Isosceles Triangle in Half

[Denyven]

Homework Statement

Take an Isosceles triangle with a height of 10 and draw a line parallel to the base. How far from the top of the triangle does the drawn line have to be for the top half's area to equal the bottom half's area.

Homework Equations

The area of a triangle A=\frac{bh}{2}
The area of a trapezoid A=\frac{h(b_{1}+b_{2})}{2}

The Attempt at a Solution

Nothing Major, I know the top half of the triangle (the one created by the line drawn through) is similar to whole triangle, but that's about it.

Thanks in Advance

 

[Hurkyl]

Turn it into an algebra problem. Solve the algebra problem.

 

[Denyven]

How do I do it if there are so many variables involved?

 

[Hurkyl]

The same way you always do. If you're solving for x, then an equation with a 2 in it is no different than an equation with a c in it.

Maybe you'll decide that you want to eliminate a variable... but you know how to do that too, right?


By the way, what variables are you using, and what are they defined to be?

 

[Denyven]

I'm currently using x for the various heights, with the top Isosceles triangle heights being x and the bottom trapezoid being 10-x, since the height of the total large triangle that contains everything is ten. But that's what I have so far, what should I do about the bases of the various figures?

 

[Denyven]

To help visualize this problem please observe this attached picture.

 

[Mark44]

The top triangle has all the same angles as the larger triangle, so it (top triangle) is similar to the large triangle. That means that the ratio of the base of the small triangle to that of the large triangle is equal to the ratio of the height of the small triangle to that of the large triangle. That should get you another equation.

 

[Denyven]

Thanks, but I don't know the base of the large triangle, how can I figure that out?

 

[Hurkyl]

I'm currently using x for the various heights, with the top Isosceles triangle heights being x and the bottom trapezoid being 10-x, since the height of the total large triangle that contains everything is ten.

I assume what you meant was "I'm currently using x for the height of the smaller triangle".

But that's what I have so far, what should I do about the bases of the various figures?

If you can't figure out how to express it in terms of what you've already named, then name it with a new variable and keep going.

 

[Mark44]

You can't, but you can get an equation that involves it and the base of the smaller triangle.

 

[um0123]

i've been working on this with Denyven and i am also stumped.

Since the base of the smaller triangle is the same as the top base of the trapazooid and the areas must be the same i have gotten this:

let the height of the smaller triangle be 10-x and the heght of the trapazoid be x.

\frac{1}{2}B_1(10-x) = \frac{1}{2}(B_1 + B_2)(x)

10B_1 - xB_1 = xB_1 + xB_2

from there i don't know where to go

 

[Hurkyl]

Humor me a bit -- write out what you got by translating the geometry problem into an algebra problem. Do not solve the problem, do not partially solve the problem. Just write down the algebra problem you need to solve.

 

[um0123]

<br /> \frac{1}{2}B_1(10-x) = \frac{1}{2}(B_1 + B_2)(x)<br />

that is what i got when i translated the geometry into algebra, two formulas for area that must be equal to each other

 

[Mentallic]

<br /> \frac{1}{2}B_1(10-x) = \frac{1}{2}(B_1 + B_2)(x)<br />

that is what i got when i translated the geometry into algebra, two formulas for area that must be equal to each other

Ok good that seems right. But now you have to give an expression for B₁ and B₂ in terms of something useful, such as the height of the triangle (10, and x). You might want to involve another variable that can be used to relate the base lengths - think trigonometry.


Interestingly enough, the answer is totally independent of how "squishy" the isosceles triangle is, the height x is always a set height above the base of the triangle.

 

[Hurkyl]

<br /> \frac{1}{2}B_1(10-x) = \frac{1}{2}(B_1 + B_2)(x)<br />

that is what i got when i translated the geometry into algebra, two formulas for area that must be equal to each other

I'm pretty sure that isn't just a translation of the geometry -- you've done quite a bit of solving too. (Also, that's not an algebra problem, that's an equation)

First off, you should have no problem solving that equation for x.

Secondly, does your arithmetic problem capture all relevant geometric information?

The reason I asked you to just translate without doing any solving is to make it much more obvious whether or not the algebraic problem reflects all relevant geometric information.

You don't have to go through that exercise in full detail, but you really should check to make sure you haven't lost any relevant geometric information when you translated into algebra...