Using the Heron's formula will determine the area and height for each triangle
"\\quad A = \\sqrt{s(s-a)(s-b)(s-c)} \\quad \\text{, where } s = \\cfrac{a+b+c}2 \\\\ \\\\\n\\quad h_b = \\cfrac{2A}b"
(41, 18, 41) A = 360, h = 40
(41, 39, 50) A = 780, h = 40
(41, 51, 58) A = 1020, h = 40
(41, 84, 85) A = 1680, h = 40
(41, 105, 104) A = 2100, h = 40
(41, 207, 202) A = 4140, h = 40
(41, 408, 401) A = 8160, h = 40
(50, 60, 50) A = 1200, h = 40
(50, 72, 58) A = 1440, h = 40
(50, 105, 85) A = 2100, h = 40
(50, 126, 104) A = 2520, h = 40
(50, 228, 202) A = 4560, h = 40
(50, 429, 401) A = 8580, h = 40
(58, 84, 58) A = 1680, h = 40
(58, 117, 85) A = 2340, h = 40
(58, 138, 104) A = 2760, h = 40
(58, 240, 202) A = 4800, h = 40
(58, 441, 401) A = 8820, h = 40
(85, 150, 85) A = 3000, h = 40
(85, 171, 104) A = 3420, h = 40
(85, 273, 202) A = 5460, h = 40
(85, 474, 401) A = 9480, h = 40
(104, 192, 104) A = 3840, h = 40
(104, 294, 202) A = 5880, h = 40
(104, 495, 401) A = 9900, h = 40
(202, 396, 202) A = 7920, h = 40
(202, 597, 401) A = 11940, h = 40
(401, 798, 401) A = 15960, h = 40
Comment: First I find all right triangles with integer side lengths having one leg of length 40. Then combine it with each other.
<code js>
{
const a=[];
// Search for right-angled triangles with integer side lengths having one leg of length 40
for(let c, b=1; b<1e4; b++) {
c=Math.sqrt(1600 + b*b); // hypotenuse
if(Number.isInteger(c)) a.push([b,c])
}
// Combine finded triangles
for(let i=0, L=a.length; i<L; i++)
for(let j=i; j<L; j++)
console.log('%d, %d, %d',
a[i][1], // hypotenuse of one triangle
a[i][0]+a[j][0], // Sum of legs
a[j][1] // hypotenuse of other triangle
);
}
</code>
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