# without trig or heron's...

• Feb 24th 2006, 04:06 PM
nancytbee
without trig or heron's...
This isnt a hw problem but just something my teacher posted up:

how could i find the height of a triangle with side lengths of fifteen, fourteen, and thirteen without using trig or herons formula?

thanks
• Feb 24th 2006, 05:23 PM
ticbol
Quote:

Originally Posted by nancytbee
This isnt a hw problem but just something my teacher posted up:

how could i find the height of a triangle with side lengths of fifteen, fourteen, and thirteen without using trig or herons formula?

thanks

There is this popular formula to get the area of any triangle:
Area = (1/2)ab*sinC ---------(1)
where
---a and b are two sides of the triangle
---angle C is the angle included by the sides a and b.

So if you know the 3 sides of the triangle, you just find an interior angle and you get the area.
How to find an interior angle when all 3 sides are known? Why, use the Law of Cosines:
c^2 = a^2 +b^2 -2ac*cosC -------(2)

So, here goes:
Say we get the interior angle between the 15 and 14 sides of the triangle,
13^2 = 15^2 +14^2 -2(15)(14)cosC
2(15)(14)cosC = 15^2 +14^2 -13^2
cosC = [15^2 +14^2 -13^2]/[2(15)(14)] = 0.6
C = arccos(0.6) = 53.13 degrees

Then, we find the area of the triangle,
A = (1/2)(15)(14)sin(53.13deg) = 84 sq.units --------answer.

-------------------------------------------------
Ooppss, no trig formula? Sorry, I missed that.

Then, let us use algebra only.

Here is one way.

Imagine the 13-14-15 triangle be resting on its horizontal side 15.
If we can find the altitude from the horizontal side 15 to the intersection of the sides 14 and 13, then solved the problem.

Draw the altitude. Call it h.
So the side 15 is divided into 2 line segments, into say u and v.
Two right triangles were formed, with these:

Under the side 14,
---hypotenuse = 14
---vertical leg = h
---horizontal leg = u

Under the side 13,
---hypotenuse = 13
---vertical leg = h also
---horizontal leg = v

Then,
In the given triangle, u +v = 15 ----(1)
In the 1st right triangle, by Pythagorean theorem, u^2 +h^2 = 14^2 ----(2)
In the 2nd right triangle, by Pythagorean theorem, v^2 +h^2 = 13^2 ----(3)

There, 3 unknowns in 3 independent equations. Solvable.

(2) minus (3),
u^2 -v^2 = 14^2 -13^2
(u+v)(u-v) = 196 -169
(u+v)(u-v) = 27
Substitute (1) into that,
(15)(u-v) = 27
u-v = 27/15 = 9/5 = 1.8 ----(4)

(1) plus (4),
u +v = 15 ----(1), plus,
u -v = 1.8 ---(4)
--------------
2u = 15 +1.8 = 16.8
u = 16.8/2 = 8.4 -----***
Plug that into (2),
(8.4)^2 +h^2 = 14^2
h^2 = 14^2 -(8.4)^2 = 125.44
h = sqrt(125.44) = 11.2

Therefore, area of given triangle = (1/2)(15)(11.2) = 84 sq.units.
• Feb 24th 2006, 06:58 PM
Rich B.
Hi:

If we lie the longest side on the x-axis of the coordinate plane, with the left-most vertex at the origin, then the other two vertices are at (15,0) and (x,y), where y is the desired height. By the distance formula, x^2 +y^2 = 13^2 and (x-15)^2 + y^2 = 14^2. Subtracting eq#1 from #2 gives (x-15)^2 - x^2 = 14^2 - 13^2 = (14+13)(14-13) = 27. Thus, -30x + 15^2 = 27.

I am confident that you can determine the value of y from here without breaking into a sweat. I hope this helps.

Regards,

Rich B.