# Thread: Finding area of triangle

1. ## Finding area of triangle

I have a problem...I am given a triangle, with different heights, $h_{1}$, $h_{2}$, and $h_{3}$, green, blue, and red respectively.

With these, I am to find the area of the triangle, but I'm not sure how if I'm not given a base at all. Can anyone help, please? Thank you!

2. use Heros formula

3. No, that doesn't work, I tried. Heron's formula involves using lengths of each side (a, b, c). My teacher said he would like a solution with heights alone, which is why I'm stumped.

4. Originally Posted by nathan02079
No, that doesn't work, I tried. Heron's formula involves using lengths of each side (a, b, c). My teacher said he would like a solution with heights alone, which is why I'm stumped.
This looks interesting.
a,b,c are the known perpendicular lengths
x,y,z are the UNKNOWN base or sides.

a*x/2 = b*y/2 = c*z/2

After a little thought:
x1+x2 = x , base to a
y1+y2 = y , base to b
z1+z2 = z , base to c

$a^2 + (x1)^2 = (z1+z2)^2$
$a^2 + (x2)^2 = (y1+y2)^2$

$b^2 + (y1)^2 = (x1+x2)^2$
$b^2 + (y2)^2 = (z1+z2)^2$

$c^2 + (z1)^2 = (y1+y2)^2$
$c^2 + (z2)^2 = (x1+x2)^2$

Six equations & six unknowns.
I'll be back.

5. Just to make sure I'm understanding it correctly...

are you saying that the a, b, and c (the known perpendicular lengths) in your equation is equivelant to $h_{1}, h_{2}$, and $h_{3}$?

6. Let us say that the area of triangle is $\Delta$

and $h_1,h_2,h_3$ be the heights
and $a,b,c$ be the sides of the triangle respectively.

Then,
$\Delta=\frac{1}{2}ah_{1}=\frac{1}{2}bh_{2}=\frac{1 }{2}ch_{3}$

which gives
$a=\frac{2\Delta}{h_{1}}$

$b=\frac{2\Delta}{h_{2}}$

$c=\frac{2\Delta}{h_{3}}$

and hence, semi-perimeter,s
$s=\frac{a+b+c}{2}$

$s=\Delta(\frac{1}{h_{1}}+\frac{1}{h_{2}}+\frac{1}{ h_{1}})$

Now, substitute these in Heros formula

7. Originally Posted by nathan02079
Just to make sure I'm understanding it correctly...

are you saying that the a, b, and c (the known perpendicular lengths) in your equation is equivelant to $h_{1}, h_{2}$, and $h_{3}$?
YES.
----------------------------------------------------------

After a waste of some scrap paper (and some time), the fog surrounding me evaporated. Someone has already solved this.

Check here:
http://mathforum.org/library/drmath/view/61893.html

Synopsis:

I = $\left ( \frac {1}{h_1} + \frac {1}{h_2} + \frac{1}{h_3} \right )$

I1 = $\left( \frac{-1}{h_1} + \frac{1}{h_2} + \frac{1}{h_3} \right )$

I2 = $\left ( \frac{1}{h_1} + \frac{-1}{h_2} + \frac{1}{h_3} \right )$

I3 = $\left ( \frac{1}{h_1} + \frac{1}{h_2} + \frac{-1}{h_3} \right )$

AREA_of_TRIANGLE = $\frac{1}{ \sqrt{ I \times I_1 \times I_2 \times I_3} }$

[reason for separation]
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