Áp dụng định lí Pytago vào `ΔABD`, ta có:

`AD^2 = AB^2 + BD^2`.

`=> BD^2 = AD^2 - AB^2`.

`=> BD = \sqrt{AD^2 - AB^2} = \sqrt{y^2 - x^2}`.

$\\$

$\color{#1fdb91}{\texttt{C}}\color{#0aaec9}{\texttt{+}}\color{#e9418e}{\texttt{+}}$

#include <iostream>
#include <iomanip>
#include <cmath>

#define ld long double

using namespace std;

ld x, y, z;

int main() {
  cin >> x >> y;
  z = sqrt(y * y - x * x);
  cout << setprecision(2) << fixed;
  cout << x * z << endl << (x+z) * 2;
}

$\\$
\begin{array}{c|c|c}\color{#ffd710}{\texttt{\{}} \color{#8655d6}{\texttt{\{}}\ \ \color{#8cdcda}{\text{Daoanhviet96}}\ \ \color{#8655d6}{\texttt{\}}} \color{#ffd710}{\texttt{\}}}\end{array}

Để tính diện tích hình chữ nhật ta cần tính `BD`

Ta có: `BD^2=y^2-x^2` (ĐL Py ta go)

`=> BD=sqrt(y^2-x^2)`

$\\$

#include <bits/stdc++.h>

using namespace std;

int main()
{
    double x,y;
    cin>>x>>y;
    double BD=sqrt(y*y-x*x);
    double dt=BD*x;
    double cv=(BD+x)*2;
    cout<<fixed<<setprecision(2)<<dt<<endl<<cv;
}