Asked by lloyd
how to find the dimension if the diagonal of the rectangle is 8 meters longer than its shorter side and the area of the rectangle is 60 square meters?
Answers
Answered by
Steve
If the dimensions are x and y, then we have
√(x^2+y^2) = x+8
xy = 60
You can grind out the solution, but I think you can recognize which factors of 60 form a convenient right triangle that works, no?
√(x^2+y^2) = x+8
xy = 60
You can grind out the solution, but I think you can recognize which factors of 60 form a convenient right triangle that works, no?
Answered by
Scott
S * L = 60
S² + L² = (S + 8)²
S² + (60 / S)² = (S + 8)²
S² + (3600 / S²) = (S + 8)²
S⁴ + 3600 = S⁴ + 16 S³ + 64 S²
S³ + 4 S² - 225 = 0
S = 5
S² + L² = (S + 8)²
S² + (60 / S)² = (S + 8)²
S² + (3600 / S²) = (S + 8)²
S⁴ + 3600 = S⁴ + 16 S³ + 64 S²
S³ + 4 S² - 225 = 0
S = 5
Answered by
Scott
Steve is referring to the 5-12-13 Pythagorean triple
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