Question 77304
Let a=shorter leg, b=longer leg, c=hypotenuse

So we have

{{{c=3a+4}}} "The hypotenuse of a right triangle is four feet longer than three times the shorter leg"

{{{b=c-1}}} "The longer leg is one foot less than the hypotenuse"

Now solve for a:

{{{c=3a+4}}}

{{{c-4=3a}}}

{{{a=(1/3)c-4/3}}}

Plug {{{(1/3)c-4/3}}} into a and plug in {{{c-1}}} into b of the Pythagorean theorem

{{{((1/3)c-4/3)^2+(c-1)^2=c^2}}}

{{{((1/9)c^2-(8/9)c+16/9)+(c^2-2c+1)=c^2}}} foil the terms on the left side

{{{((10/9)c^2-(26/9)c+25/9)=c^2}}} Combine like terms

{{{((1/9)c^2-(26/9)c+25/9)=0}}} Subtract {{{c^2}}} from both sides

Now use the quadratic formula to solve for c (note:this solver uses decimals instead of fractions):

*[invoke quadratic "c", 1/9, -26/9, 25/9 ]

So the hypotenuse is:

c=1 or c=25

Use this to find a
Let c=1 
{{{1=3a+4}}}
{{{a=-1}}}
Since this length is negative, we must disregard the hypotenuse length of 1

Let c=25
{{{25=3a+4}}}
{{{a=7}}}

Now use c=25 to find b

{{{b=25-1}}}
{{{b=24}}}

So our dimensions are:
a=7,b=24,c=25

<p>
Check:
{{{a^2+b^2=c^2}}} Plug in a=7, b=24, c=25
{{{7^2+24^2=25^2}}}
{{{49+576=625}}}
{{{625=625}}} works

{{{c=3a+4}}} plug in a=7 and c=25
{{{25=3(7)+4}}}
{{{25=25}}} works

{{{b=c-1}}}
{{{24=25-1}}}
{{{24=24}}} works