Question 949469

let  the hypotenuse be {{{c}}},the longer leg {{{a}}} and the shorter leg {{{b}}}

we know that in a right triangle {{{c^2=a^2+b^2}}}.....eq.1

given:
in a right triangle the length of the longer leg {{{a}}} is {{{2cm}}} more than twice the length of the shorter leg {{{b}}}, then we have

 {{{a=2b+2cm}}}

the length of the hypotenuse {{{c}}}is {{{8cm}}} more than the length of the shorter leg {{{b}}}, then we have
{{{c=b+8cm}}} 

go to {{{c^2=a^2+b^2}}}.....eq.1, substitute {{{a}}} and {{{c}}}

{{{(b+8cm)^2=(2b+2cm)^2+b^2}}}

{{{cross(b^2)+16b*cm+64cm^2=4b^2+8b*cm+4cm^2+cross(b^2)}}}

{{{16b*cm+64cm^2=4b^2+8b*cm+4cm^2}}}

{{{0=4b^2-16b*cm+8b*cm+4cm^2-64cm^2}}}

{{{4b^2-8b*cm-60cm^2=0}}}....simplify, divide by {{{4}}}

{{{b^2-2b*cm-15cm^2=0}}}...factor completely, replace {{{-2b*cm}}} with {{{-5b*cm+3b*cm}}}

{{{b^2-5b*cm+3b*cm-15cm^2=0}}}...group

{{{(b^2-5b*cm)+(3b*cm-15cm^2)=0}}}

{{{b(b-5cm)+3cm(b-5cm)=0}}}

{{{(b-5cm)(b+3cm) = 0}}}

solutions:

if 
{{{(b-5cm) = 0}}}=>{{{b=5cm}}}
if 
{{{(b+3cm) = 0}}}=>{{{b=-3cm}}}..-> disregard negative value because we are looking for the length

so, our solution is {{{highlight(b=5cm)}}}

now find {{{a}}} and {{{c}}}

{{{a=2b+2cm}}}=>{{{a=2*5cm+2cm}}}=> {{{highlight(a=12cm)}}}

{{{c=b+8cm}}}=>{{{c=5cm+8cm}}}=>{{{highlight(c=13cm)}}}

check if Pythagoras theorem works:

{{{c^2=a^2+b^2}}}
{{{(13cm)^2=(12cm)^2+(5cm)^2}}}
{{{169cm^2=144cm^2+25cm^2}}}
{{{169cm^2=169cm^2}}}............true; so, these are sides of right angle triangle