Problem 4. A field in the shape of a rectangle, with a length $20 \mathrm{~m}$ longer than its width, is fenced with 3 strands of wire, using a total of $840 \mathrm{~m}$ of wire. What is the perimeter and what are the lengths of the sides of the field?
Solution. First method. Let the length of the field be denoted by $a$, the width by $b$, and the perimeter by $L$. Then, $a=20+b$ and $L=2(a+b)$.

Since the field is enclosed by three rows of wire whose total length is 840 meters, we get $3L=840$, i.e., $L=280 \, \text{m}$. Now, from $L=2(a+b)$, we have $2(a+b)=280$, i.e., $2(20+b+b)=280$, so $20+2b=140$, from which we get $b=60 \, \text{m}$. Finally, $a=20+b=20+60=80 \, \text{m}$.

Second method. The field is enclosed by three rows of wire, so its perimeter is equal to $840:3=280 \, \text{m}$. According to this, the sum of the sides of the field is equal to $280:2=140 \, \text{m}$. The longer side is 20 meters longer, so the length of the shorter side is equal to (140-20):2=60m. Finally, the length of the longer side is equal to $60+20=80 \, \text{m}$.