Area of Triangle

Without using Heron’s formula, find the area of the triangle.
image
Source: NCTM Mathematics Teacher, January 2006

SOLUTION
image
Applying the Law of Cosines
a^2=b^2+c^2-2bc\,\textrm{cos} A
14^2=13^2+15^2-2\times 13\times 15\,\textrm{cos} A
196=169+225-390\,\textrm{cos} A
390\,\textrm{cos} A=198
\textrm{cos} A=198/390
\textrm{sin}^2A+\textrm{cos}^2A=1
\textrm{sin}^2A=1-\textrm{cos}^2A
=1-(198/390)^2
=.7422485207
\textrm{sin}A=\sqrt{.7422485207}
=.8615384615
Area of triangle
(1/2)bc\,\textrm{sin} A=(1/2)13\times 15\times (.8615384615)
=84

Answer: 84 square units

Alternative solution 1
image
Most textbooks state the Law of Sines as
a/\textrm{sin} A=b/\textrm{sin} B=c/\textrm{sin} C
We have a more definitive Law of Sines
a/\textrm{sin} A=b/\textrm{sin} B=c/\textrm{sin} C = diameter of circumcircle
Let’s apply it to a and \textrm{sin} A
a/\textrm{sin} A = diameter
14/.8615384615 = diameter
diameter =16.25
radius r=16.25/2=8.125
In right triangle OMA
h^2+MA^2=r^2
h^2+(13/2)^2=(8.125)^2
h^2=23.76562501
h=\sqrt{23.76562501}=4.875
Area of triangle OAC
(1/2)bh=(1/2)13\times 4.875=31.6875
We calculate similarly the areas of the triangle OAB and OBC
Area of triangle OAB
(1/2)15\times 3.125=23.4375
Area of triangle OBC
(1/2)14\times 4.125=28.875
Area of triangle ABC
31.6875+23.4375+28.875=84

Alternative solution 2
image
In right triangle AHC
x^2+y^2=13^2\qquad\qquad (1)
In right triangle AHB
x^2+z^2=15^2\qquad\qquad (2)
Comparing the value of x^2 from Eq. (1) and Eq. (2)
13^2-y^2=15^2-z^2
z^2-y^2=15^2-13^2
(z+y)(z-y)=(15+13)(15-13)
14(z-y)=56
z-y=4
Adding the two equations
z-y=4
z+y=14
—————-
2z=18
z=9
Substituting the value of z in Eq. (2)
x^2+9^2=15^2
x^2=144
x=12
Area of triangle
(1/2)14\times 12=84

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , , , , , . Bookmark the permalink.

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