Question 1125269

let legs be {{{a}}} and {{{b}}}, and hypotenuse {{{c}}}
 
One leg of a right triangle is one inch shorter than the other leg, we have

{{{a=b-1}}}....eq.1



if the hypotenuse is {{{5}}} inches , find the length of the shorter leg using Pythagorean theorem

{{{c^2=a^2+b^2}}}...eq.2

plug in {{{a}}} from eq.1 in eq.2 and {{{c=5}}}


{{{5^2=(b-1)^2+b^2}}}...solve for {{{b}}}

{{{25=b^2-2b+1+b^2}}}

{{{25=2b^2-2b+1}}}

{{{25-1^2=2b^2-2b}}}

{{{0=2b^2-2b-24}}}.............simplify

{{{0=b^2-b-12}}}

use quadratic formula:


{{{b = (-(-1) +- sqrt( (-1)^2-4*1*(-12) ))/(2*1) }}}

{{{b = (1 +- sqrt( 1+48 ))/2 }}}

{{{b = (1 +- sqrt( 49 ))/2 }}}

{{{b = (1 +- 7)/2 }}}

solution: we need only positive solution

{{{b = (1 + 7)/2 }}}

{{{b = 8/2 }}}

{{{b = 4 }}} inches

go to {{{a=b-1}}}....eq.1, substitute {{{b}}}

{{{a=4-1}}}

{{{highlight(a=3)}}} inches ->the length of the shorter leg