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My algebra is so rusty, this should be an easy one!

The area of a rectangle is

$$A = W \times H$$

If you're given $A$ (say $150$) and a ratio of $W:H$ (say $3:2$) how can you calculate $W$ and $H$?

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4 Answers 4

up vote 4 down vote accepted

Right so you have $W:H$ is $3:2$, which means $$150 = 3x \times 2x$$ So this is just a simple equation which you can solve for. That is $6x^{2}=150$ which says $x^{2} =25$ and hence $x =\sqrt{25}$ that is $x =5 $. So $W = 3 \times 5 =15 $ and $H= 2 \times 5 =10$ and you can see that they have ratio $3 : 2$.

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X=5 (then you get 3x * 5x = 15x10 = 150) - but you're showing X = 5 srt 2 - is that correct? – Ryan Nov 18 '10 at 23:42
@Ryan: Edited now. Sorry made a egregious mistake – anonymous Nov 18 '10 at 23:50
Cheers - skool is coming back to me after many years ;) Got stuck with simplifying 3x X 2x => 3 X x X 2 X x => 6x X x => 6x^2. Also found this*2x&sol‌​vf=AUTO – Ryan Nov 19 '10 at 0:02

HINT $\rm\displaystyle\ \ \frac{W}H = \frac{3}2\ \Rightarrow\ 2\:W = 3\:H\:,\ $ so $\rm\ 300 = 2\:W\:H = 3\:H\:H\ \Rightarrow\ H = \ldots $

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Bill - your hint has flummoxed me! ;) – Ryan Nov 19 '10 at 0:04
@Ryan: What don't you follow? – Bill Dubuque Nov 19 '10 at 1:42

I was looking for an answer for the same question. It's just that the other posts are kind of too complicated for me. I see it this way:

If you have given a width $a$ and a height $b$ and you know the $\textsf{area}$. You want to stretch each side with a fixed ratio of $x$ until you get the final size of the area. So you stretch $a$ with $x$ and $b$ with $x$ equals the $\textsf{area}$.

$$\begin{align*} (a\cdot x)\cdot(b\cdot x)&=\textsf{area}\\\\ (a\cdot b)\cdot x^2&=\textsf{area}\\\\ x^2&=\frac{\textsf{area}}{a\cdot b}\\\\ x&=\sqrt{\frac{\textsf{area}}{a\cdot b}} \end{align*}$$ $x$ is the ratio you are stretching. So you can multiply $x$ by $a$ and $b$ to get the new size of $a$ or $b$.

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  1. $ratio = \frac{x}{y}$
  2. $area = x \cdot y$


$x = y \cdot ratio$

Using substitution:

$area = (y\cdot ratio) \cdot y$

$area = y^{2} \cdot ratio$

$\frac{area}{ratio} = y^{2}$

$\sqrt{\frac{area}{ratio}} = y$

Now that you have y:

$x = \frac{area}{y}$

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