Problem: Find area of triangle with sides 13, 14 and 15.


                                         Sketch of Triangle with unknowns h and y assigned

Use:       Pythagorean theorem: C2 = A2 + B2

First right triangle:

h2 + y2 = 132

h2 + y2 - y2 = 132 - y2 Subtract y2 from both sides

Gives:            h2 = 132 - y2
 
 

Second right triangle:

h2 + (15 - y) 2 = 142

h2 + (15 - y) 2 - (15 - y) 2 = 142 - (15 - y)2                     Subtract (15 - y) 2 from both sides

Gives:           h2 = 142 - (15 - y) 2

Now substitute h2 from first equation gives:

132 - y2 = 142 - (15 - y) 2       Expand (15 - y) 2

          15 - y
Times 15 - y
        - 15y + y2                     Multiply 15 - y by - y               Notice -y times -y is +y2
225  - 15y ____                   Multiply 15 - y by 15
225 - 30y + y2

132 - y2 = 142 – (225 - 30y + y2)                      Remove Parens

132 - y2 = 142 – 225 + 30y - y2                           Add y2 to both sides i.e. delete - y2 from both sides

132 = 142 – 225 + 30y                                           Multiply squared terms i.e. 132 and 142

169 = 196 – 225 + 30y                                           Move constants to same side of equation

169 + 225 - 196 = +30y                                         Add constants

198 = 30y                                                               Divide by 30

y = 6.6