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Mathematics 15 Online
OpenStudy (anonymous):

See attachment

OpenStudy (anonymous):

OpenStudy (anonymous):

@ghazi

OpenStudy (mathstudent55):

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?

OpenStudy (anonymous):

yes.

OpenStudy (mathstudent55):

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

OpenStudy (anonymous):

Ok.

OpenStudy (anonymous):

Did you choose 4/9 randomly?

OpenStudy (mathstudent55):

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.

OpenStudy (mathstudent55):

No, 4/9 = 2/3 * 2/3

OpenStudy (anonymous):

Oh,ok

OpenStudy (mathstudent55):

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

OpenStudy (anonymous):

OOOOOOH. Ok, I believe I understand it now - thank you.

OpenStudy (mathstudent55):

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.

OpenStudy (mathstudent55):

you're welcome

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