# rectangle: L decr 2.4cm W incr 30%, result: area 4% bigger. find original L and W

• Sep 7th 2012, 10:48 PM
rectangle: L decr 2.4cm W incr 30%, result: area 4% bigger. find original L and W
Hi.

So I'm looking for help with this geometry problem. It is:

the L of the rectangle is decreased by 2.4cm . The W of the rectangle is increased by 30%. As the result, the area of the rectangle grew 4%. Find the original L and W.

First i did an equation (L - 2,4cm) x 1,3W = 1.04LW and from there i got that the L is 12. now i don't know what to do on as every equation or whatever i do with the information
I keep ending up with things like W = W or i get the answer of L again aka I'm stuck. Please help.

Thank you.
• Sep 7th 2012, 10:59 PM
Prove It
Re: rectangle: L decr 2.4cm W incr 30%, result: area 4% bigger. find original L and W
\displaystyle \begin{align*} 1.3 W \left( L - 2.4 \right) &= 1.04 LW \\ 1.3LW - 3.12W &= 1.04LW \\ 0.26LW - 3.12W &= 0 \\ W \left( 0.26 L - 3.12 \right) &= 0 \end{align*}

So we have \displaystyle \begin{align*} W = 0 \end{align*} (impossible) or \displaystyle \begin{align*} 0.26L - 3.12 = 0 \end{align*}. Solve for L.

Without being given any extra information, like what the original or resulting areas or perimeters actually were, you can't solve for W.
• Sep 7th 2012, 11:26 PM
Re: rectangle: L decr 2.4cm W incr 30%, result: area 4% bigger. find original L and W
you did a mistake there on the action in line 2-3. you deducted 0.04 from 1.3, not 1.04, so it should result in 0.26L - 3,12 = 0, but that doesnt matter. the problem is taken from 2009 test from my uni which if done >75% grants you "done" or w/e in "basic math". i'm gonna have to do it too, but i hope it's their mistake then. it should be done with pre-uni knowledge.
• Sep 7th 2012, 11:33 PM