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# Find the base of the new rectangle

## Tags: Mathematik

19:06 Uhr, 21.11.2022

Tag!
I need your help to see if this problem is correct and if not what I did wrong.

''The perimeter of a rectangle is $44m\phantom{\rule{0.12em}{0ex}}$. If the base of this rectangle were 3 meters longer and its height 4 meters less, the area of the new rectangle would be $30$ square meters less than the area of the first rectangle. Find the base of the new rectangle''.

I kind of lean in this direction
data given
$P\phantom{\rule{0.12em}{0ex}}=2\left(b\phantom{\rule{0.12em}{0ex}}\right)+2\left(h\phantom{\rule{0.12em}{0ex}}\right)$
Perimeter=44 $m\phantom{\rule{0.12em}{0ex}}$
so if $P\phantom{\rule{0.12em}{0ex}}=44,$ then,
$b\phantom{\rule{0.12em}{0ex}}\cdot h\phantom{\rule{0.12em}{0ex}}=22$

$2\left(a\phantom{\rule{0.12em}{0ex}}\right)+2\left(b\phantom{\rule{0.12em}{0ex}}\right)=44$

$a\phantom{\rule{0.12em}{0ex}}+b\phantom{\rule{0.12em}{0ex}}=22$

$3a\phantom{\rule{0.12em}{0ex}}+3b\phantom{\rule{0.12em}{0ex}}=66$

If the base of this rectangle were 3 meters longer $=\left(b\phantom{\rule{0.12em}{0ex}}+3\right)$
and its height 4 meters less $=\left(h\phantom{\rule{0.12em}{0ex}}-4\right)$
then,
the area of the new rectangle would be $30$ square meters less than the area of the first rectangle.
then I set it up like this,

(b×h)−(b+3(h−4) $=30{m\phantom{\rule{0.12em}{0ex}}}^{2}$

bh−(bh−4b+3h−12) $=30m\phantom{\rule{0.12em}{0ex}}2$

bh−bh+4b−3h+12=30m^2

4b−3h+12=30m^2

4b−3h=30〖−12〗
4b−3h=18

I went back to re-take my first equation and add my last one to it

$3h\phantom{\rule{0.12em}{0ex}}+3b\phantom{\rule{0.12em}{0ex}}=66$
−3h+4b=18

$7b\phantom{\rule{0.12em}{0ex}}=84$

$b\phantom{\rule{0.12em}{0ex}}=12$

the new base of the new triangle is $12m\phantom{\rule{0.12em}{0ex}}$

is this okay?

Vielen Dank.

Für alle, die mir helfen möchten (automatisch von OnlineMathe generiert):
"Ich möchte die Lösung in Zusammenarbeit mit anderen erstellen."

19:28 Uhr, 21.11.2022

$2\cdot \left(a\phantom{\rule{0.12em}{0ex}}+b\phantom{\rule{0.12em}{0ex}}\right)=44$
$a\phantom{\rule{0.12em}{0ex}}+b\phantom{\rule{0.12em}{0ex}}=22$
$a\phantom{\rule{0.12em}{0ex}}=22-b\phantom{\rule{0.12em}{0ex}}$

$A\phantom{\rule{0.12em}{0ex}}=a\phantom{\rule{0.12em}{0ex}}\cdot b\phantom{\rule{0.12em}{0ex}}=\left(22-b\phantom{\rule{0.12em}{0ex}}\right)\cdot b\phantom{\rule{0.12em}{0ex}}$

$\left(22-b\phantom{\rule{0.12em}{0ex}}+3\right)\cdot \left(b\phantom{\rule{0.12em}{0ex}}-4\right)=\left(22-b\phantom{\rule{0.12em}{0ex}}\right)\cdot b\phantom{\rule{0.12em}{0ex}}-30$

$b\phantom{\rule{0.12em}{0ex}}=..$.

$a\phantom{\rule{0.12em}{0ex}}=..$.

21:12 Uhr, 21.11.2022

I am sorry walbus, there is a typo in my first post which may change things
$i\phantom{\rule{0.12em}{0ex}}$ wrote $h\phantom{\rule{0.12em}{0ex}}\cdot b\phantom{\rule{0.12em}{0ex}}=22$ when it should have been $h\phantom{\rule{0.12em}{0ex}}+b\phantom{\rule{0.12em}{0ex}}=22$

dose that change your setup, or not?
because I simplified your setup and got $b\phantom{\rule{0.12em}{0ex}}=10,$ but that is not good,

22:17 Uhr, 21.11.2022

You got confused with $b\phantom{\rule{0.12em}{0ex}},h\phantom{\rule{0.12em}{0ex}}$ vs. $a\phantom{\rule{0.12em}{0ex}},b\phantom{\rule{0.12em}{0ex}}!$ You yourself used both different set of names which sure is something to avoid!

It once again shows that it does not make much sense to use one-letter-names without explaining what is meant.

You got base length $b\phantom{\rule{0.12em}{0ex}}=12m\phantom{\rule{0.12em}{0ex}}$ which is the correct answer.

In walbus' answer $b\phantom{\rule{0.12em}{0ex}}$ is the heigth and a is the base. So his $b\phantom{\rule{0.12em}{0ex}}=10$ is correct, too. because of $a\phantom{\rule{0.12em}{0ex}}=22-b\phantom{\rule{0.12em}{0ex}}=12$ he gets the same result as yours.

Actually I don't understand his answer. You asked "is this okay?" and the answer should simply have been "Yes!".
Of course apart from the typo you mentioned yourself and the confusing mix $\left(b\phantom{\rule{0.12em}{0ex}};h\phantom{\rule{0.12em}{0ex}}\right)$ vs. $\left(a\phantom{\rule{0.12em}{0ex}};b\phantom{\rule{0.12em}{0ex}}\right)$.
And even though the question definitely was "Find the base of the new rectangle", his approach would in first case get you the height which he has chosen to name "b".

01:04 Uhr, 22.11.2022

Yes, ROmans. you're right. I should have stuck to $b\phantom{\rule{0.12em}{0ex}}$ for base and $h\phantom{\rule{0.12em}{0ex}}$ for height.
Walbus $b\phantom{\rule{0.12em}{0ex}}$ for height threw me off completely. Honestly, don't know why he chose $b\phantom{\rule{0.12em}{0ex}}$ because in German height is Höhe , right?. So, the logical variable of choice would have been $h\phantom{\rule{0.12em}{0ex}}$.
But I am thankful to him for his answer and interest. Thanks to you to my friend!

03:40 Uhr, 22.11.2022

Its sure not unusual to name the sides of a rectangle a and $b\phantom{\rule{0.12em}{0ex}}$. And if you draw a rectangle it would be quite natural to name the horizontal length a and vertical $b\phantom{\rule{0.12em}{0ex}}$ as walbus had done.

But as your text is talking about a "base" and a "height", it would have been necessary to clearly state which variable name is supposed to indicate the "base" and which the "height" (and yes, in this case using $b\phantom{\rule{0.12em}{0ex}}$ and $h\phantom{\rule{0.12em}{0ex}}$ as names is the better choice).
This is a recurring problem, even here in the forum, that answers (usually well-intentioned) are given that are often relatively meaningless because there is no explanation of what the variables and terms presented are supposed to represent.

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