Question 160319
<pre><font size = 4 color = "indigo"><b>
Draw the rectangle and its diagonal, putting
the length {{{15}}} on the diagonal:

{{{drawing(400,400,-1,13,-4,10,

rectangle(0,0,12,9), triangle(0,0,12,0,12,9),
locate(5,5,15)

 )}}}

Label the width {{{w}}}.

Since the length is 3 cm longer than the width,

label the length {{{w+3}}}

{{{drawing(400,400,-1,13,-4,10,

rectangle(0,0,12,9), triangle(0,0,12,0,12,9),
locate(5,5,15), locate(6,0,w+3),locate(12.3,5,w)

 )}}}

Erase the upper left half of the rectangle:

{{{drawing(400,400,-1,13,-4,10,
 triangle(0,0,12,0,12,9),
locate(5,5,15), locate(6,0,w+3),locate(12.3,5,w)
 )}}}

That leaves a right triangle, with legs 

{{{matrix(1,3,w+3,",",w)}}} and hypotenuse {{{15}}}.

So we use the Pythagorean theorem:

{{{a^2+b^2=c^2}}} with {{{matrix(1,5,a=(w+3),",",b=w,",",c=15)}}}

{{{(w+3)^2+w^2=15^2}}}

{{{(w+3)(w+3) + w^2 = 225}}}

{{{(w^2+3w+3w+9)+w^2=225}}}

{{{(w^2+6w+9)+w^2=225}}}

{{{w^2+6w+9+w^2=225}}}

{{{2w^2+6w+9=225}}}

{{{2w^2+6w-216=0}}}

Divide through by 2:

{{{2w^2/2+6w/2-216/2=0/2}}}

{{{w^2+3w-108=0}}}

Think of two integers which
have product -108 and combine
to give +3:

They are -9 and +12, so the
left sides factors as:

{{{(w-9)(w+12)=0}}}

Use the zero-factor principle

{{{matrix(2,3,w-9=0,",",w+12=0,w=9,",",w=-12)}}}

We choose the positive value for the width,

Therefore the width = 9cm
and the length = w+3 = 9+3 = 12cm

Edwin</pre>