# Thread: Area of Triangle AHF in Rectangle ABCD

1. ## Area of Triangle AHF in Rectangle ABCD

Hello everyone!

This is a really tricky problem for me; I have no clue where to start. If anyone could spare some advice, I'd appreciate it.

In rectangle ABCD points F and G lie on segment AB so that AF = FG = GB and E is the midpoint of segment DC. Also, segment AC intersects segment EF at H and segment EG at J. The area of rectangle ABCD is 70 square units. What is the area of triangle AHF?

2. ## Re: Area of Triangle AHF in Rectangle ABCD

Let w = the width of the rectangle

Let h = the height of the rectangle

w * h = 70.

The formula for the area of a triangle is what?

What is the height of triangle EFG?

What is the base of triangle EFG?

So what is its area in terms of w and h?

Now can you answer the question?

A good place to start any problem is to name your variables, in this case the height and width of the rectangle, and to write down what you already know numerically about those variables.

3. ## Re: Area of Triangle AHF in Rectangle ABCD

I've done what I could with what you suggested, but I'm confused about how triangle EFG can be applied to triangle AHF.

w = width of rectangle
h = height of rectangle
w * h = 70
height of triangle EFG = 70/w
width of triangle EFG = 70/3h

area of triangle EFG = 1/2(height)(width) = 1/2(70/w)(70/3h) = 4900/6wh = 2450/3wh

I'm confused about how all this would apply to triangle AHF. Is there a formula, postulate, rule, or something I'm supposed to use?

4. ## Re: Area of Triangle AHF in Rectangle ABCD

Hi,
This will work if you know about equations of lines.

Assign coordinates to your rectangle with A = (0,0) and C = (w,h). You then know the coordinates of the remaining points. Find the equation of the line through A, C and the equation of the line through E and F. Now solve these two simultaneously to find the coordinates of H. You can now write down the area of triangle AHF in terms of w, h. This turns out to be in terms of wh. Done.

5. ## Re: Area of Triangle AHF in Rectangle ABCD

You are correct. I answered way too quickly.

Now the simple way to go is to slap down a coordinate system with A at (0, 0), B at (0, h), C at (w, h), and D at (w, 0).

E is at (w/2, h), F is at (w/3, 0), and G is at (2w/3, 0).

So the line AC is represented by the equation $\dfrac{y - 0}{x - 0} = \dfrac{0 - h}{0 - w} \implies -wy = -hx \implies y = \dfrac{h}{w} * x.$

The line EF is represented by the equation $\dfrac{y - 0}{x - \frac{1}{3}w} = \dfrac{0 - h}{\frac{1}{3}w - \frac{1}{2}w} \implies y\left(- \dfrac{w}{6}\right)= - hx + \dfrac{hw}{3} \implies y = \left(\dfrac{6h}{w} * x\right) - 2h.$

Those two lines intersect at H where $\dfrac{hx}{w} = \dfrac{6hx}{w} - 2h \implies \dfrac{5hx}{w} = 2h \implies x = \dfrac{2w}{5} \implies y = what?$

Now can you figure out what is the area of AHF?

6. ## Re: Area of Triangle AHF in Rectangle ABCD

Thanks, I think I got the answer!

line AC: y = hx/w
line EF: y = 6hx/w - 2h
wh = 70
H coordinates: x = 2w/5; y = hx/w --> y = h/w(2w/5) = 2h/5
Area AHF = 1/2(base)(height) = 1/2(2h/5)(w/3) = 2wh/30 = wh/15 = 70/15 = 14/3

Is that what I should have gotten?

7. ## Re: Area of Triangle AHF in Rectangle ABCD

Looks good to me