Question 662313


{{{x cm}}},

{{{(x+1)cm}}} and 

{{{(x+2)cm}}}.........this is hypotenuse


{{{((x+2)cm)^2= x^2+((x+1)cm)^2}}}

{{{((x+2)cm)^2= x^2+(x^2+2x+1)cm^2}}}

{{{((x+2)cm)^2= (2x^2+2x+1)cm^2}}}

{{{(x^2+4x+4)cross(cm^2)= (2x^2+2x+1)cross(cm^2)}}}

{{{x^2+4x+4= 2x^2+2x+1}}}

{{{0= 2x^2-x^2-4x-4+2x+1}}}

{{{0= x^2-2x-3}}}

{{{x^2-2x-3=0}}}...replace {{{-2x}}} with {{{-3x+x}}}

{{{x^2-3x+x-3=0}}}...group

{{{(x^2-3x)+(x-3)=0}}}

{{{x(x-3)+(x-3)=0}}}

{{{(x+1)(x-3)=0}}}

solutions:

if {{{x+1=0}}}...=>...{{{x=-1}}}...we can't take negative value for a sside of a triangle

if {{{x-3=0}}}...=>...{{{x=3}}}...this is our solution:


so, the sides are

{{{x=3 cm}}},

{{{(x+1)=4cm}}} and 

{{{(x+2)=5cm}}}