For a right triangle with two legs of the right triangle a and b, draw three circles with radius r and one ellipse as shown in the diagram (the major axis of the ellipse is parallel to the hypotenuse of the right triangle). Find a relationship between a, b, and r

How can I make some triangles in a bigger triangle knowing its perimeters like this picture?

This is my file
 

restart;
F := proc(ee, LL) Typesetting:-mrow(InertForm:-Typeset(InertForm:-Display(eval(eval(InertForm:-MakeInert(factor(ee)), [`%*` = `*`]) = InertForm:-MakeInert(map(sort, algsubs(a*x = InertForm:-MakeInert(a*x), ee), order = plex(b))), `=`~([a, b], LL)), inert = false)), Typesetting:-mo("="), InertForm:-Typeset(eval(ee, `=`~([a, b], LL)))); end proc;
p := a^3*x^3 + 3*a^2*b*x^2*y + 3*a*b^2*x*y^2 + b^3*y^3;
L := [[2, 3], [1, 2], [1/3, -sqrt(2)]];
ans := F~(p, L);
print~(ans);

I get 

I want to the order of terms of the polynomial, e.g, (2 x + 3y)^3 like this

(2 x + 3 y)^3 = (2 x)^3 + 3 (2 x)^2 (3 y) + 3 (2 x) (3 y)^2 + (3 y )^ 3 = 8*x^3 + 36*x^2*y + 54*x*y^2 + 27*y^3. 

How can I get that? 

I have this file
 

restart:
F := proc(ee,LL)
  uses InertForm, Typesetting;
  mrow(Typeset(Display(eval(eval(MakeInert(factor(ee)),[`%*`=`*`])
                                 =MakeInert(subs(b=MakeInert(b*y)/y,
                        a=MakeInert(a*x)/x,p)),[a,b]=~LL),
                       inert=false)),
       mo("="),Typeset(eval(ee,[a,b]=~LL)))
end proc:
p := (a*x)^2 - 2*a*x*b*y + (b*y)^2:
L := [[sqrt(2),3],[2,5],[3,12],[1/3,5/7]];
ans := F~(p, L):
print~(ans):

How can I put the results like this
 

\documentclass[12pt,a4paper]{article}
\usepackage[left=2cm, right=2cm, top=2cm, bottom=2cm]{geometry}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{enumitem}
\theoremstyle{definition}
\newtheorem{ex}{Exercise}
\begin{document}
\begin{ex}
\[(\sqrt{2} x-3 y)^2=(\sqrt{2} \cdot x)^2-2 \cdot(\sqrt{2} \cdot x) \cdot(3 \cdot y)+(3 \cdot y)^2=2 x^2-6 \sqrt{2} x y+9 y^2.\]
\end{ex}

\begin{ex}
\[(2 x-5 y)^2=(2 \cdot x)^2-2 \cdot(2 \cdot x) \cdot(5 \cdot y)+(5 \cdot y)^2=4 x^2-20 x y+25 y^2. \]
\end{ex}

\begin{ex}
\[(3 x-12 y)^2=(3 \cdot x)^2-2 \cdot(3 \cdot x) \cdot(12 \cdot y)+(12 \cdot y)^2=9 x^2-72 x y+144 y^2. \]
\end{ex}

\begin{ex}
\[\left(\frac{x}{3}-\frac{5 y}{7}\right)^2=\left(\frac{1}{3} \cdot x\right)^2-2 \cdot\left(\frac{1}{3} \cdot x\right) \cdot\left(\frac{5}{7} \cdot y\right)+\left(\frac{5}{7} \cdot y\right)^2=\frac{1}{9} x^2-\frac{10}{21} x y+\frac{25}{49} y^2. \]
\end{ex}
 
\end{document} 

How can I find Mean, Median, Quartiles, Variance, StandardDeviation of data in this table

I use Mathamatica and get the result

Clear["Global`*"]
boundaries = Range[0, 10, 5/2];
frequencies = {18, 11, 13, 6};
binMeans = Mean /@ Partition[boundaries, 2, 1];
weighted = WeightedData[binMeans, frequencies];
weightedHist = HistogramDistribution[weighted, {5/2}];
Through[{Mean, Median, Quartiles, Variance, StandardDeviation}[
  weightedHist]]