Asked by jeff
One leg of a right triangle has a length of 15 m. The other sides have lengths that are consecutive integers. Find the number of meters in the perimeter.
Answers
Answered by
Damon
225 + x^2= (x+1)^2
225 + x^2 = x^2 + 2 x + 1
2 x = 224
x = 112
x+1 = 113
perimeter = 15 + 112 + 113
225 + x^2 = x^2 + 2 x + 1
2 x = 224
x = 112
x+1 = 113
perimeter = 15 + 112 + 113
Answered by
Bosnian
The other sides :
x and x + 1
where x + 1 is the hypotenuse
So Pythagorean theorem:
( x + 1 )² = x² + 15²
x² + 2 x + 1 = x² + 225
Both x² get cancelled
2 x + 1 = 225
2 x = 225 - 1 = 224
x = 224 / 2 = 112 m
Hypotenuse:
x + 1 = 112 + 1 = 113 m
Perimeter = 15 +112 + 113 = 240 m
Checking the solution:
112² + 15² = 12544 + 225 = 12769 = 113²
x and x + 1
where x + 1 is the hypotenuse
So Pythagorean theorem:
( x + 1 )² = x² + 15²
x² + 2 x + 1 = x² + 225
Both x² get cancelled
2 x + 1 = 225
2 x = 225 - 1 = 224
x = 224 / 2 = 112 m
Hypotenuse:
x + 1 = 112 + 1 = 113 m
Perimeter = 15 +112 + 113 = 240 m
Checking the solution:
112² + 15² = 12544 + 225 = 12769 = 113²
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