## Area of Triangle

Without using Heron’s formula, find the area of the triangle.

Source: NCTM Mathematics Teacher, January 2006

SOLUTION

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

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

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.
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