Question

An 8cm long wire is to be bent into rectangle , can a rectangle with diagonal 2cm be made from it?​

Answer 1

Answer:

2(a+b)=8

=> a+b=4

=>a=(4-b)

therefore,

a^2 +b^2= 2

=>(4-b)^2 + b^2=2

=> {16+b^2-2.(4).(b)}+b^2=2

=>16+2b^2-8b=2

=>2b^2-8b+14=0

=>b^2-4b+7=0

=>b= {(root over 11) -2} or {-(root over 11) -2}

which is not possible....because it cannot be greater than 4...