the last word problem ?, I promise!!!

[Kstrikeoutgirl]

The problem is:

Originally the dimensins of a rectangle were 20 cm by 23 cm. When both dimensins were decreased by the same amount, the area of the rectangle decreased by 120 cm squared. Find the dimensions of the new rectangle.




Thanks for helping!!!

 

[soroban]

Hello, Kstrikeoutgirl!

Originally the dimensions of a rectangle were 23 cm by 20 cm.
When both dimensions were decreased by the same amount,
the area of the rectangle decreased by 120 cm².
Find the dimensions of the new rectangle.

We already know the following:
. . The original length was 23 cm.
. . The original width was 20 cm.
. . The original area was: $\displaystyle 23\,\times\,20\:=\;460$ cm²

Let x be the amount of decrease in the length and width.
. . The new length is: $\displaystyle 23\,-\,x$ cm.
. . The new width is: $\displaystyle 20\,-\,x$ cm.
. . The new area is: $\displaystyle (23\,-\,x)(20\,-\,x)$ cm².

The new area is 120 cm² less than the original area: $\displaystyle 460\,-\,120\:=\:340$ cm².

There is our equation: $\displaystyle \;(23\,-\,x)(20\,-\,x)\:=\:340$

. . which simplifies to: $\displaystyle \;x^2\,-\,43x\,+\,120\:=\:0$

. . and factors: $\displaystyle \;(x\,-\,3)(x\,-\,40)\:=\:0$

. . and has roots: $\displaystyle \;x\,=\,3$ and $\displaystyle 40$

Since the original rectangle was 23-by-20, we can't reduce the dimensions by 40 cm,
. . so: $\displaystyle \;x\,=\,3$ cm


Don't forget to answer the question (the dimensions).

The new length is: $\displaystyle \,23\,-\,3\:=\:20$ cm.
The new width is: $\displaystyle \;20\,-\,3\:=\:17$ cm.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

<u>Check</u>

The new rectangle has an area of: $\displaystyle \;20\,\times\,17\:=\:340$ cm².

This is 120 cm² less than the original area (460) . . . check!

 

[galactus]

$\displaystyle (20)(23)=460$

The new rectangle will have area 120 less than that, $\displaystyle 340cm^{2}$

Since they're decreased by the same amount:

$\displaystyle (20-x)(23-x)=340$

Solve the quadratic:

$\displaystyle x^{2}-43x+120=0$


Too fast, Soroban.