Question

Two squares have side lengths that are consecutive odd integers. The total area of the squares is 650cm^2. What is their total perimeter?

Answer 1

$the\ consecutive\ odd\ integers\ is:2n+1\ and\ 2n+3\ (n\in\mathbb{N})\\\\Area_1=(2n+1)^2=4n^2+4n+1\\\\Area_2=(2n+3)^2=4n^2+12n+9\\\\Area_1+Area_2=4n^2+4n+1+4n^2+12n+9=8n^2+16n+10\\\\8n^2+16n+10=650\\\\8n^2+16n+10-650=0\\\\8n^2+16n-640=0\ \ /:8$

$n^2+2n-80=0\\\\\Delta=2^2-4\cdot1\cdot(-80)=4+320=324;\ \sqrt\Delta=\sqrt{324}=18\\\\n_1=\frac{-2-18}{2\cdot1} < 0;\ n_2=\frac{-2+18}{2\cdot1}=\frac{16}{2}=8\\\\2n+1=2\cdot8+1=16+1=17\ (m)\\\\2n+3=2\cdot8+3=16+3=19\ (m)\\\\\\the\ total\ perimeter=4\cdot17+4\cdot19=68+76=144\ (m)$