At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Think of it this way. The recatngle had an original length, L, and an original width, W. The new rectangle has a new length that is (2/3)L and a new width that is (2/3)W. You follow so far?
Now let's find the area of the original rectangle and the area of the new rectangle. The area of a rectangle is length times width. The original rectangle: Aold = LW The new rectangle: Anew = (2/3)L * (2/3)W = (2/3)*(2/3)*LW = (4/9)LW
Did you choose 4/9 randomly?
Now you see that the new area is (4/9)LW and the old area is LW. That means the new area is 4/9 times the area of the old rectangle.
No, 4/9 = 2/3 * 2/3
Remember the problem states that the new rectangle has a length and width that are 2/3 what they used to be. So the new length is (2/3)L, and the new width is (2/3)W. When you multiply the new length and the new width to get the new area, you are multiplying (2/3)L * (2/3)W which is the same as (2/3)*(2/3)*LW. (2/3) * (2/3) = 4/9, so the new area is (4/9)LW
OOOOOOH. Ok, I believe I understand it now - thank you.
Then when you compare the new area of (4/9)LW with the original area of LW, you see that the new area is 4/9 of the old area.