Question 494417

The hypotenuse of a right triangle has length 20cm. 
<pre>
a² + b² = 20²
a² + b² = 400
</pre>
The sum of the lengths of the other two sides is 28cm.
<pre>
a + b = 28

So we have this system of equations:

<font face = "symbol">ì</font>a² + b² = 400
<font face = "symbol">í</font>
<font face = "symbol">î</font>a + b = 28

Solve the second equation for one of the letters, say b

             a + b = 28
                 b = 28 - a

Substitute (20 - a) for x in the first equation:

              a² + b² = 400
       a² + (28 - a)² = 400
a² + (28 - a)(28 - a) = 400
  a² + 784 - 56a + a² = 400
      2a² - 56a + 784 = 400 

Subtract 400 from both sides to get 0 on the right

      2a² - 56a + 384 = 0

Divie every term by 2

       a² - 28a + 192 = 0

Factor as

     (a - 16)(a - 12) = 0
     
  a - 16 = 0         a - 12 = 0
       a = 16             a = 12 
       b = 28 - a         b = 28 - a
       b = 28 - 16        b = 28 - 12
       b = 12             b = 16

It looks as though there are two solutions,
but there is really only one, as it is only
a matter of which side is chosen to have length
represented by the letter "a" and which is to 
have length represented by letter "b".

Regardless, the other two sides are 12 and 16 

Edwin</pre>