My right angled triangle has sides measuring n, n^2, n^3.
Evaluate its area, perimeter and centre of gravity.
If it's assumed that the side that's n^3 is the hypotenuse:
Wolfram Alpha solves n^2 + n^4 = n^6 as
n = sqrt((1+sqrt(5)/2)) =~ 1.45534669022535
The area = n * n^2 / 2 = n^3 / 2 = ((2^(1/2)*5^(1/2))/2 + 1)^(3/2)/2 =~ 2.07341888950732
The perimeter would be n + n^2 + n^3 = 10^(1/2)/2 + 2^(1/2)*(10^(1/2) + 2)^(1/2) + (5^(1/2)*(10^(1/2) + 2)^(1/2))/2 + 1 =~ 8.33456891272915
I'm not up on my calculations of center of gravity.
For other choices of hypotenuse, giving sides less than 1, only side of length n would work as the hypotenuse, as
solving n^4 + n^6 = n^2 gives
n = sqrt(1/2 (sqrt(5) - 1)) =~ 0.786151377757423
which can be continued from there.
Edited on January 15, 2024, 6:26 pm
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Posted by Charlie
on 2024-01-15 18:24:12 |
Partial answer
