# Size image by area

Sometimes I have to insert many relatively small different images, with different aspect ratio, which widths must be adjusted one by one in order to look approximately of the same size.

In such cases it would be useful to be able to size the image by area, that is choose the area and let LaTeX set image width or height so that width x height = area (keeping aspect ratio).

The ideal would be a new command built upon \includegraphics able to accept an area parameter (square root of area would be even better, because in order to get a sense of the size of the area you usually end up in calculating its square root). Such a command, given for example two images img_a (having aspect ratio W:H=2:1) and img_b (having aspect ratio W:H=2:3), with the following code

\includegraphicsbyarea[area=18]{img_a} % suppose area in square centimeters
\includegraphicsbyarea[area=18]{img_b}
\includegraphicsbyarea[area=32]{img_a}


would insert:

• img_a with width of 6 cm and height of 3 cm (6x3=18, 6:3=2:1)
• img_b with width of 3.464 cm and height of 5.196 cm (3.464x5.196=18, 3.464:5.196=2:3)
• img_a with width of 8 cm and height of 4 cm (8x4=32, 8:4=2:1)

Would it be possible?

Thanks in advance for any clue.

• Do your images have the same area before you include them? – user36296 Aug 19 '18 at 11:17
• @samcarter No, of course. – mmj Aug 19 '18 at 14:14

Here is an approach not redefining \includegraphics. However compared to scale option, I cut a branch because I don't have details of graphicx.sty in head, so possibly there would be a way to delegate to driver the final rescaling, which I am losing here. Ping @DavidCarlisle.

(I use xintexpr but xfp of course would do it as well as in Joseph's answer; also, up to some more cumbersome notations one could use only the macros of xintfrac, giving a tiny speed-up as expression parsing is skipped).

\documentclass{article}
\usepackage{graphicx}
\usepackage{xintexpr}
\makeatletter
\define@key{Gin}{sqrtofarea}{%
\def\Gin@req@sizes{%
\edef\Gin@scalex{\xinttheiexpr[5]% round fixed point to 5
% fractional digits
\dimexpr#1\relax/
sqrt(\Gin@nat@height*\Gin@nat@width)
\relax}%
\let\Gin@scaley\Gin@exclamation
\Gin@req@height\Gin@scalex\Gin@nat@height
\Gin@req@width\Gin@scalex\Gin@nat@width
}%
\@tempswatrue}
\makeatother
\begin{document}

\includegraphics[sqrtofarea=2cm]{example-image-a}

\includegraphics[sqrtofarea=3cm]{example-image-a}

\includegraphics[sqrtofarea=4cm]{example-image-a}

\includegraphics[sqrtofarea=5cm]{example-image-a}

\newbox\mybox
Equality only expected up to 5 digits of precision due to intrinsic
limitations of graphicx computations.

\setbox\mybox\hbox{\includegraphics[sqrtofarea=2cm]{example-image-a}}%

\xinttheiiexpr\ht\mybox*\wd\mybox\relax
?=
\xinttheiiexpr\dimexpr2cm\relax*\dimexpr2cm\relax\relax\ (4cm$^2$)

\setbox\mybox\hbox{\includegraphics[sqrtofarea=5cm]{example-image-a}}%

\xinttheiiexpr\ht\mybox*\wd\mybox\relax
?=
\xinttheiiexpr\dimexpr5cm\relax*\dimexpr5cm\relax\relax\ (25cm$^2$)
\end{document}


The sentence at bottom of image must be amended: in TeX all dimensions are integer multiples of 1sp. When we set the area square root as key, we automatically limit the achievable precision of the area. For example 5cm internally in TeX gives 9323399sp, hence a square equal to 86925768913201 as in image above. The previous square is 86925750266404 and the next one is 86925787560000, so they diverge in the 7th digit already and we can never overcome that possible imprecision when comparing a square with a produce height times width. Above we observe discrepancy already in 5th digit so the sentence is probably not completely wrong, but I felt I needed to add this mathematical precision.

Here is same with xfp (basically copied from Joseph's way of using it):

\documentclass{article}
\usepackage{graphicx}
\usepackage{xfp}
\makeatletter
\define@key{Gin}{sqrtofarea}{%
\def\Gin@req@sizes{%
\edef\Gin@scalex{\fpeval{#1/sqrt(\Gin@nat@height*\Gin@nat@width)}}%
\let\Gin@scaley\Gin@exclamation
\Gin@req@height\Gin@scalex\Gin@nat@height
\Gin@req@width\Gin@scalex\Gin@nat@width
}%
\@tempswatrue}
\makeatother
\begin{document}

\includegraphics[sqrtofarea=2cm]{example-image-a}

\includegraphics[sqrtofarea=3cm]{example-image-a}

\includegraphics[sqrtofarea=4cm]{example-image-a}

\includegraphics[sqrtofarea=5cm]{example-image-a}

\end{document}


• no wrapping of #1 in \dimexpr #1\relax needed here; xintexpr could easily be extended to recognize cm, in, pt, etc ... units so that e.g. 2cm is understood automatically, but the problem is that it would then do an exact conversion to a fractional number of sp units, whereas TeX process is more complex than simply using a proportionality factor and proceeds with rounding and truncating in various directions at various stages; so using exact conversion factor means not doing same operations as TeX itself. For this reason, no such units are defined yet in xintexpr and user must go via \dimexpr #1\relax; the xintexpr parser will apply \number to this, triggering TeX's own way to convert dimensional units.

• I am not expert in xfp so I don't know if sometimes such computation could result in a scientific notation which would break TeX later; in the xintexpr solution I applied a transformation to fixed point value with 5 fractional digits. I don't know how one does that in xfp and whether it could be needed here in some cases. In the MWE above it works fine.

• It seems that I wrote too early about "optimal" solutions. Your solution as far as I understand adds the sqrtofarea option (nice name, by the way) to standard includegraphics (which is better than adding a new command). I also tested the sqrtofarea option together with the standard angle option of includegraphics ans it works, whereas if I add the same angle option when using the command includegraphicsbyarea of other solutions I get an error. – mmj Aug 23 '18 at 10:29
• I think @DavidCarlisle should be pinged about correct way to add this option to \includegraphics. Good to hear it works with angle option. I added some mathematical comment at bottom of answer. – user4686 Aug 23 '18 at 10:30
• I think that sqrtofarea should definitively become a standard option of includegraphics. Scaling by area is the best way of sizing many images different in size and aspect ratio so that they look approximately of the same size. – mmj Aug 23 '18 at 10:47
• unfortunately even with e-TeX's \numexpr, \dimexpr, it requires some work to avoid arithmetic overflows when implementing computations at low-level. Already graphicx/graphics has some extra code for obtaining ratio of two lengths which works well enough in 99.99% of real life situations, and is cleverly compact, but sqrtofarea requires extra implementation effort as it seems at least square root must be handled. Naturally graphicx will never load package like xint and all its unneeded macros. Perhaps in future it will have xfp automatically, then of course, sqrtofarea becomes feasible. – user4686 Aug 23 '18 at 10:55
• Maybe sqrt(\Gin@nat@height)*sqrt(\Gin@nat@width) would help in avoiding overflow? (I barely understand what you mean, sorry if I suggested something stupid). – mmj Aug 23 '18 at 11:03

Scaling to an area is an unusual request: normally one knows the target height or width. The following takes the image and scales such that the area is that given in the optional argument, interpreted in square centimetres:

\documentclass{article}
\usepackage{graphicx}
\usepackage{xfp}
\makeatletter
\define@key{Garea}{area}{\def\Garea@area{#1}}
\define@key{Garea}{areaunit}{\def\Garea@unit{#1}}
\define@key{Gin}{area}{} % So we can pass through easily
\define@key{Gin}{areaunit}{}
\newcommand*{\Garea@area}{}
\newcommand*{\Garea@unit}{cm}
\newcommand{\includegraphicsbyarea}[2][]{%
\setkeys{Garea}{#1}%
\ifx\Garea@area\@empty
\gdef\Garea@scale{scale = 1}%
\else
\begingroup
\setbox0=\hbox{\includegraphics[#1]{#2}}%
\xdef\Garea@scale{scale =
\fpeval{sqrt((\Garea@area * 1 \Garea@unit * 1 \Garea@unit)
/(\the\ht0 * \the\wd0))}}%
\endgroup
\fi
\expandafter\includegraphics\expandafter[\Garea@scale,#1]{#2}%
}
\makeatother
\begin{document}
\includegraphicsbyarea[area=100]{example-image-a}
\includegraphicsbyarea[area=100,areaunit=pt]{example-image-a}
\end{document}


It still applies any options to the graphic: I've ordered such that the area scale is first, but one might want to put it last.

An alternative interface uses the square root of the area of the image: conveniently this has units we can work with directly

\documentclass{article}
\usepackage{graphicx}
\usepackage{xfp}
\makeatletter
\define@key{Garea}{sqrtarea}{\def\Garea@sqrtarea{#1}}
\define@key{Gin}{sqrtarea}{}
\newcommand*{\Garea@sqrtarea}{}
\newcommand{\includegraphicsbyarea}[2][]{%
\setkeys{Garea}{#1}%
\ifx\Garea@diag\@empty
\gdef\Garea@scale{scale = 1}%
\else
\begingroup
\setbox0=\hbox{\includegraphics[#1]{#2}}%
\xdef\Garea@scale{scale =
\fpeval{\Garea@sqrtarea /(sqrt(\the\ht0 * \the\wd0))}}%
\endgroup
\fi
\expandafter\includegraphics\expandafter[\Garea@scale,#1]{#2}%
}
\makeatother
\begin{document}
\includegraphicsbyarea[sqrtarea = 5cm]{example-image-a}
\includegraphicsbyarea[sqrtarea = 10cm]{example-image-a}
\end{document}


(See the answer by mmj for an alternative name for the key.)

• Your answer is clean and works well, but I would like to choose the unit of measure of the area. To obtain that it might be helpful using as parameter the square root of area instead of area. I think that using the square root of area is better, because to get a sense of the area you usually end up calculating the square root (I updated my question to underline that). – mmj Aug 20 '18 at 22:05
• I've added a follow up with an improved (IMHO) version of your solution. If you agree, please consider replacing your actual solution with that one, and I will delete the follow up, just for clarity towards future readers. – mmj Aug 22 '18 at 12:23
• @mmj I've added a version using the diagonal as the scaling determinant: that has a square-root relationship to area, and has units, and has a clear name :) – Joseph Wright Aug 22 '18 at 12:43
• @mmj You'd be better to post your answer code as an answer, rather than edit it into the question. The site structure works best that way: as it stands, its a bit confusing. – Joseph Wright Aug 22 '18 at 12:44
• Diagonal is a nice name and of course varies linearly with the width and height of an image, but it does not vary linearly when the aspect ratio changes, and the command is supposed to deal with images having different aspect ratios. What you call diagonal is actually the same parameter I used to modify your code, it is the square root of the area. Calling it diagonal is misleading the user because he/she might think that a 16x1 image and a 4x4 image will have the same diagonal, whereas they will have the same area (16) but very different diagonals, about 16 and 5.66 respectively. – mmj Aug 22 '18 at 14:29

# Update 3

This is the last revision, I promise.

The only change is the use of pgfmath.sty in place of fp-eval.sty which, in turn, required a little special handling to avoid dimension too large errors. As well, the use of pgfmath.sty allowed for considerable simplification. The new output is identical to the old.

%===8><-----%

I was unhappy with the way that I left the \scaletoarea macro. It has been altered to be, hopefully, a little more user-friendly. I have made it possible to use the options to \includegraphics for all of the macros. Take a look at the examples.

%===8><---%

I take it that you want all of the graphics to have the same area regardless of the aspect ratio. The following code does that. There are two approaches outlined here.

The first scales subsequent graphics based on the size of the first graphic Use \fpic for the first graphic; using \spic for the subsequent graphics will size them to have the same area as the first graphic. Use the options to \fpic (the same as those to \includegraphics) to size the first graphic accordingly.

Second, if you know the target area, you can use \scaletoarea{<unitless area>}{<linear unit of area>}{<name of graphic>}. For example, if you were to scale a graphic, foo, so that it is 5 square inches in area, you would write: \scaletoarea{5}{in}{foo}. Any of the units recognized by TeX can be used.

    \documentclass{article}
\usepackage{graphicx,calc,pgfmath}
\usepackage[margin=0.5in]{geometry}

\newlength\fpicw
\newlength\fpich
\newlength\spicw
\newlength\spich
\newsavebox{\firstpic}
\newsavebox{\nextpics}
\newlength{\targetarea}

%% \fpic[<options to \includegraphics>]{<name of graphic>}
\newcommand{\fpic}[2][]{% First picture, use \includegraphics, and options
\sbox{\firstpic}{\includegraphics[#1]{#2}}%
\settoheight{\fpich}{\usebox{\firstpic}}%
\settowidth{\fpicw}{\usebox{\firstpic}}%
\usebox{\firstpic}%
}

%% \spic[<options to \includegraphics>]{<graphic name>}
\newcommand{\spic}[2][]{% Second and succeeding pictures
\sbox{\nextpics}{\includegraphics[#1]{#2}}%
\settoheight{\spich}{\usebox{\nextpics}}%
\settowidth{\spicw}{\usebox{\nextpics}}%
\pgfmathsetmacro{\scaling}{sqrt((\fpicw/\spicw)*(\fpich/\spich))}%
\scalebox{\scaling}{\usebox{\nextpics}}%
\typeout{**The scaling (#2) = \scaling}% Comment-out if not needed
}

%% \scaletoarea[<options to \includegraphics>]{<unitless area>}{<unit of area (linear)>}{<graphic name>}
\newcommand{\scaletoarea}[4][]{%
\pgfmathsetmacro{\mytmp}{sqrt(#2)}%
\setlength{\targetarea}{\mytmp #3}%
\sbox{\nextpics}{\includegraphics[#1]{#4}}%
\settoheight{\spich}{\usebox{\nextpics}}%
\settowidth{\spicw}{\usebox{\nextpics}}%
\pgfmathsetmacro{\scaling}{sqrt((\targetarea/\spicw)*(\targetarea/\spich))}%
\scalebox{\scaling}{\usebox{\nextpics}}%
\typeout{++The scaling (#4) = \scaling}% Comment-out if not needed
}

%% \scaletoareab[<options to \includegraphics>]{<square root of desired area>}{<name of graphic>}
\newcommand{\scaletoareab}[3][]{%
\setlength{\targetarea}{#2}%
\sbox{\nextpics}{\includegraphics[#1]{#3}}%
\settoheight{\spich}{\usebox{\nextpics}}%
\settowidth{\spicw}{\usebox{\nextpics}}%
\pgfmathsetmacro{\scaling}{sqrt((\targetarea/\spicw)*(\targetarea/\spich))}%
\scalebox{\scaling}{\usebox{\nextpics}}%
\message{++The scaling (#3) = \number\scaling}%
}

\begin{document}

\fpic[width=1.25in]{Peppers}\spic{Pasta}\spic{OldImage}\spic{BethanyDrawing}

\scaletoarea{3}{in}{Peppers}\scaletoarea{12}{cm}{Pasta}\scaletoarea{2}{in}{OldImage}\scaletoarea{61}{pc}{BethanyDrawing}

\scaletoarea[width=0.5in,height=3in,keepaspectratio=false]{3}{in}{Peppers}
\scaletoareab[width=0.5in,height=3in,keepaspectratio=false]{1.732050807568877in}{Peppers}

\end{document}


• My math was wrong for \scaletoarea. The first argument to \scaletoarea should be the square root of the desired area. Made the correction in the code listing. – sgmoye Aug 19 '18 at 18:45
• I have modified my original answer to modify how \scaletoarea works. – sgmoye Aug 20 '18 at 11:58
• I tested the previous version of your scaletoarea command and it worked well, but above all I liked the use of square root of area as parameter instead of area. In fact using the square root of area is better, because to get a sense of the area you have in any case to calculate the square root. And you save a parameter in comparison to the new current version that needs an additional parameter for the unit of measure. Please consider restoring the use of square root of area as parameter for scaletoarea command. – mmj Aug 20 '18 at 21:31
• I'm a little tied up now, but I will get to it later today or tomorrow. Thanks. – sgmoye Aug 20 '18 at 22:28
• @mmj Check now. The macro (last one in the code) is \scaletoareab[<options to \includegraphics>]{<square root of desired area>}{<name of graphic>}. Included example and output. – sgmoye Aug 21 '18 at 0:41

Here is an MCVE that fulfills your requirements provided th. I have also added commented the code and added a full explanation with regards to the logic of the program.

Sample output:

Now the explanation:

The code!:

\documentclass[12pt]{article}
\usepackage{graphicx, color}
\usepackage[margin=0.5in]{geometry}
\usepackage{calculator}         % Very important package. Provides
% basic mathematical operations like
% dividing, rounding, truncating, and
% finding the GCD (greatest common divisor).
\usepackage{caption, amsmath, amssymb}

\makeatletter
\newcommand*{\getlength}[1]{\strip@pt#1}
\makeatother

% For the area, and ratio (width:height) = (\xRatio:\yRatio)
\newlength{\areaGiven}
\newlength{\xRatio}
\newlength{\yRatio}

% Accepts 4 argments: area, width ratio, height ratio, and name of image (respectively).
\setlength\areaGiven{#1 cm}
\setlength\xRatio{#2 cm}
\setlength\yRatio{#3 cm}

\noindent Given information~~~~~~~~~~~: \textbf{area} = #1cm,
\textbf{ratio} $(width:height) = #2:#3$.

% Convert the unit from pt to cm. Note, 1pt = 0.0352778cm.
% If you really care about exact numbers, 1.0000008pt = 0.035277806cm.
% If inches is the unit to focus on, then change 0.0352778 to 0.0138889.
\MULTIPLY{\getlength{\areaGiven}}{0.0352778}{\areaInCm}
\MULTIPLY{\getlength{\xRatio}}{0.0352778}{\xRatioInCm}
\MULTIPLY{\getlength{\yRatio}}{0.0352778}{\yRatioInCm}

\noindent Converting from pt to cm : \textbf{area} = \areaInCm cm,
\textbf{ratio} $(width:height) = \xRatioInCm cm:\yRatioInCm cm$.

% Keep in mind that there will be rounding errors. The answer
% we will obtain is a decimal answer. Since pixels are atomic, we need to
% truncate (or round) the decimal values to secure a whole number.
% I used the TRUNCATE command, you can comment the next three line and
% uncomment the ROUND commands if you want to round.
\TRUNCATE[0]{\areaInCm}{\areaInCm}
\TRUNCATE[0]{\xRatioInCm}{\xRatioInCm}
\TRUNCATE[0]{\yRatioInCm}{\yRatioInCm}
% OR use round instead. It is better for images with areas 100+.
% \ROUND[0]{\areaInCm}{\areaInCm}
% \ROUND[0]{\xRatioInCm}{\xRatioInCm}
% \ROUND[0]{\yRatioInCm}{\yRatioInCm}

\noindent After truncating~~~~~~~~~~~~~: \textbf{area} = \areaInCm,
\textbf{ratio} $(width:height) = \xRatioInCm cm:\yRatioInCm cm$.

% The following are the steps of find the result mathematically.
% This is explained in details at the end of the docmument.
\MULTIPLY{\xRatioInCm}{\yRatioInCm}{\k}   % 1) finds (a*b) in ak*bk = area.
\DIVIDE{\areaInCm}{\k}{\k}                % 2) finds (area/(a*b)) in k^2 = (area/(a*b)).
\SQUAREROOT{\k}{\kFinal}                  % 3) finds of k in k = sqrt((area/(a*b))).
\MULTIPLY{\xRatioInCm}{\kFinal}{\x}       % 4) multiplies the x-ratio by k.
\MULTIPLY{\yRatioInCm}{\kFinal}{\y}       % 5) multiplies the y-ratio by k.

\noindent \textbf{Final dimensions are}~~~:
\textbf{width} = \x cm, \textbf{height} = \y cm.

\begin{center}
\includegraphics[width = \x cm, height = \y cm]{#4} % 6*3 cm
\captionof{figure}{
\textbf{area} = #1cm, \textbf{ratio}
$(width:height) = #2:#3$ $\longrightarrow$
\textbf{width} = \x cm, \textbf{height} = \y cm.
}
\end{center}
}

\begin{document}
\noindent
\textcolor{red}{\textbf{Example A:}}\newline
\adjustImage{18}{2}{1}{example-image-a} % area = 18cm, ratio (width:height) = 2:1
\textcolor{red}{\textbf{Example B:}}\newline
\adjustImage{18}{2}{3}{example-image-b} % area = 18cm, ratio (width:height) = 2:3
\textcolor{red}{\textbf{Example C:}}\newline
\adjustImage{32}{2}{1}{example-image-c} % area = 32cm, ratio (width:height) = 2:1
% Maximum area is 575 units!
% \adjustImage{575}{2}{1}{TestImage.png} % area = 575cm, ratio (width:height) = 2:1

\newpage
\newgeometry{margin=1in}

% Here is the explanation of the logic.
\noindent\textbf{Explaining the logic with an example:}\newline
The logic is a simple mathematical manipulation.\newline
First, we need our three variables,
the total area we will call $area$ and the ratio $x : y$.\newline\newline
Let $area = 18$ and the ratio $x : y = 2 : 1$.\newline
We know that $x \times y = area$ or in other words, $2x \times y = 18$.\newline
For simplicity, let's only use one variable, $k$. Substitute $k$ \newline
in our previous equation for both $x$ and $y$, $2k \times k = 18$.\newline
Solve for $k$:
\begin{align*}
2k \times k &= 18\\
2k^{2}      &= 18\\
k^{2}       &= 9\\
k           &= 3
\end{align*}
Now, we multiply $k = 3$ in our original ratio $x : y = 2 : 1$.\newline
Hence, our desired ratio is :
\begin{align*}
x : y &= 2(k) : 1(k)\\
x : y &= 2(3) : 1(3)\\
x : y &=    6 : 3\\
\end{align*}\bigskip

% Formal definition.
\noindent\textbf{Formal definition:}\newline
Let $area$ represent the total area of the image and the ratio $x : y$ respectively
represent the $width : height$ of the image.\newline
Find a value $k$ such that $area = kx \times ky$:
\begin{align*}
area &= (k)x \times (k)y\\
area &= k^{2}\times (xy)\\  % Step 1 (in the preamble) stores x*y.
\frac{area}{xy} &= k^{2}\\  % Step 2 (in the preamble) stores area/(x*y)
\sqrt{\frac{area}{xy}} &= k\\% Step 3 (in the preamble) stores sqrt((area/(x*y)))
\end{align*}
Now, multiply $k$ by the our ratio given, $x : y$. % Steps 4 and 5 (in the preamble).
$kx : ky$
Which yields to $area = kx \times ky$. \hfill $\blacksquare$
\end{document}

• I don't understand your answer, maybe my question wasn't clear enough, please see the update. – mmj Aug 19 '18 at 14:46
• I have updated the answer, check it out. – M. Al Jumaily Aug 20 '18 at 5:17
• You say that my desire is not achievable, but if you know an area A = W x H and an aspect ratio R = W / H, you can definitively find unique values for width W and height H, which is what I'm asking. – mmj Aug 20 '18 at 22:06
• Did you read all the post? If you elimanate the real values then sure, you can find a ratio assuming that W and H are not at the lowest ratio form, like W=2 and H=8... which you can have a reduced ratio of 1:4.... Now, if your ratio is 4:5, you cannot reduce it further because it's in the lowest form already... if you want you can enter the real values and divide it by 2 to obtain 2:2.5 but then you entered the real values where there are infinite solutions. You can divide by 2 or 3 or 4 or even 10000000^10000.... all are valid decimal ratios. – M. Al Jumaily Aug 20 '18 at 22:06
• I think there is a misunderstanding. The aspect ratio of an image is always a rational number. You cannot "reduce" the ratio because width x height must be equal to a given and arbitrary number, the area. My previous comment explain the simple math behind the question. – mmj Aug 20 '18 at 22:17
\documentclass[12pt]{article}
\usepackage{graphicx}
\begin{document}

\begin{figure}% only needed if you want captions
\includegraphics[width=\linewidth]{example-image-a}

\bigskip
\makebox[\linewidth]{% only needed if the textwidth is smaller than the images
\includegraphics[width=6cm,height=3cm,keepaspectratio=false]{example-image-b}
\includegraphics[width=3.464cm,height=5.196cm,keepaspectratio=false]{example-image-b}
\includegraphics[width=8cm,height=4cm,keepaspectratio=false]{example-image-b}%
}
\end{figure}

\end{document}


• I don't understand your answer, maybe my question wasn't clear enough, please see the update. – mmj Aug 19 '18 at 14:45
• then use [width=...,height=...,keepaspectratio=false]{image}. see edited answer – user2478 Aug 19 '18 at 17:23
• I don't want to change the aspect ratio of images, I want a LaTeX command which use the aspect ratio R = W / H of image AND an area parameter A chosen by me, to calculate the width W and height H of image so that W x H = A and W / H = R. – mmj Aug 20 '18 at 22:01