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A New Method to Calculate The Area of
A Triangle from Its Three Sides
The old way to calculate the area of a triangle from its three sides a, b and c is:
First find s = ½ (a + b + c), the area of the triangle = [s(s-a)(s-b)(s-c)]1/2
a, b ,c > 0 and s(s-a)(s-b)(s-c) > 0.
A new method to calculate a triangle area is by placing a triangle in an X-Y plane
as shown below: Assign a set of coordinates to each point:
A(X1,Y1), B(X2,Y2), C(X3,Y3)
Y
C(X3,Y3)
b
A(X1,Y1) a
X
c
B(X2,Y2)
Three will be six variables X1, X2, X3, Y1, Y2, Y3
and three equations (Distance Theorem):
(X2-X1)2
+ (Y2-Y1)2 = c2
(X3-X1)2 + (Y3-Y1)2
= b2
(X3-X2)2
+ (Y3-Y2)2 = a2
With three equations and six variables, there will be three undetermined
variables and infinite sets of solutions.
In order to find a fixed solution, we need to eliminate three variables.
Why not put point A at the origin and point B at any point in the X-axis
(other than the origin), three variables are eliminated automatically.
A(X1,Y1) =A(0,0)
B(X2,Y2)
=B(X2,0)
Y
C(X3,Y3)
b a
X
A(0,0) c B(X2,0)
The three equations become:
(X2-0)
2 + (0-0)2 = c2 (I)
(X3-0) 2 + (Y3-0)2 = b2 (II)
(X3-X2) 2 + (Y3-0)2
= a2 (III)
From
(I) X22 = c2
X2 = ± c
Now, only two variables X3 and Y3 are left and three equations are reduced to two:
(X3-0) 2 + (Y3-0)2
= b2 (I)
(X3-X2) 2 + (Y3-0)2
= a2 (II)
After expansions, two equations become:
X32 + Y32
= b2
(I)
X32- 2X3X2 + X22
+ Y32 = a2 (II)
Type (a): When X2 = +c
Y
C(X3,Y3)
b a
A(0,0) c B(+c,0) X
C(X3,Y3)
(When X3 >0, ÐCAB is an acute angle)
Y
C(X3,Y3)
b a
A(0,0) c
B(+c,0) X
C(X3,Y3)
(When X3 =0, ÐCAB is a right angle)
Y
C(X3,Y3)
b a
A(0,0) c B(+c,0) X
C(X3,Y3)
(When X3 <0, ÐCAB is an obtuse angle)
X32 + Y32
= b2 (I)
X32- 2cX3 + c2 + Y32 = a2 (II)
From (I), substitute X32 + Y32 = b2 into (II)
b2 - 2cX3 + c2 = a2
X3 = (b2 + c2 - a2)/2c
Substitute the value of X3 into (I)
Y3 = ± ( b2 - X32)1/2
b2 – X32
must be greater than 0, or the triangle doesn’t exist
∣Y3∣ is also the height of the triangle with base AB
The area of Δ ABC = ½ c∣Y3∣, when Y3 is not equal to 0
Type (b): When X2 = - c
Y
C(X3,Y3)
a
b
B(-c,0) c
A(0,0) X
C(X3,Y3)
(When X3 < 0, ÐCAB is an acute angle)
Y
C(X3,Y3)
a b
B(-c,0)
c A(0,0) X
C(X3,Y3)
(When X3 =0, ÐCAB is a right angle)
Y
C(X3,Y3)
a b
B(-c,0) c A(0,0) X
C(X3,Y3)
(When X3 >0, ÐCAB is an obtuse angle)
X32 + Y32
= b2 (I)
X32 + 2cX3 + c2 + Y32 = a2 (II)
From (I), substitute X32 + Y32 = b2 into (II)
b2 + 2cX3 + c2 = a2
X3 = -(b2 + c2 - a2)/2c
Substitute the value of X3 into (I)
Y3 = ± ( b2 - X32)1/2
b2 – X32
must be greater than 0, or the triangle doesn’t exist
∣Y3∣ is also the height of the triangle with base AB
The area of Δ ABC = ½ c∣Y3∣, when Y3 is not equal to 0
By : Tienchien Fu Any comments or questions? E-mail: Tienchien@aol.com