Question 162455
Let L = the length of the field and W = the width of the field.
From the problem description, you have:
L = W+10 "...length is 10 meters longer than its width..."
The perimeter of a rectangle is given by:
P = 2(L+W) and this is 100 meters, substituting L = W+10 and P = 100, you get:
100 = 2((W+10)+W) Simplifying this, you get:
100 = 2(2W+10) Divide both sides by 2.
50 = (2W+10) Subtracting 10 from both sides gives you:
40 = 2W Finally, dividing both sides by 2, you'll get:
W = 20 and L = W+10 = 20+10 = 30.
The length of the field is 30 meters and the width of the field is 20 meters.