Calculate width and height of rectangle containing given area and conforming to given ratio.
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 $3 : 2$.
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 $b$.
Assume:
- $ratio = \frac{x}{y}$
- $area = x \cdot y$
So:
$x = y \cdot ratio$
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:
$x = \frac{area}{y}$