Question

The shortest side of a fight triangle measures20 m. The lengths of the other two sides are censecutive odd integers. Find the length of the two aides

Answer 1

a² + b² = c²
20² + n² = (n+2)²
400+ n² = (n+2)(n+2)
400 + n²=n² + 2n + 2n +4
400 + n² = n² +4n +4
400 + n²-n² = n²-n² + 4n+4
400 = 4n + 4
400-4 = 4n +4-4
396 = 4n
396/4 = 4n/4
n= 99

20²+ 99² = (99+2)²
400 + 9801 =10201
10201=10201