SOLUTION: what is the are of rectangle whose length is twice its width and whose perimeter is equal to that of a square whose area is 1?
Algebra ->
Angles
-> SOLUTION: what is the are of rectangle whose length is twice its width and whose perimeter is equal to that of a square whose area is 1?
Log On
Question 93998: what is the are of rectangle whose length is twice its width and whose perimeter is equal to that of a square whose area is 1? Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! A square who's area is 1 has a side equal to 1. So, it's perimeter is 4.
Let L=Length and W=width
L=2W
P=2*(L+W) Perimeter formula
Substitute 2W for L: P=2*(2W+W)=2(3W)=6W
6W=4
divide both sides by 6: W=4/6=2/3
L=2W=2*2/3=4/3
Area=L*W=(2/3)*(4/3)=8/9
