Perimeter of triangle:20
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Check the picture below.
let's check where the quadratic and x = 6 meet, bear in mind we're integrating by "y", so
![\bf -------------------------------\\\\ \displaystyle \int\limits_{0}^{\frac{\pi }{2}}\ [6-6sin(y)]dy\implies \left. \cfrac{}{} 3y^2+6cos(y) \right]_{0}^{\frac{\pi }{2}} \\\\\\ \left[ \cfrac{3\pi 2}{4}+0 \right]-[0+6]\implies \cfrac{3\pi^2}{4}-6\approx 1.4022033](https://tex.z-dn.net/?f=%5Cbf%20-------------------------------%5C%5C%5C%5C%0A%5Cdisplaystyle%20%5Cint%5Climits_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%20%7D%7B2%7D%7D%5C%20%5B6-6sin%28y%29%5Ddy%5Cimplies%20%5Cleft.%20%5Ccfrac%7B%7D%7B%7D%203y%5E2%2B6cos%28y%29%20%5Cright%5D_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%20%7D%7B2%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cleft%5B%20%5Ccfrac%7B3%5Cpi%202%7D%7B4%7D%2B0%20%5Cright%5D-%5B0%2B6%5D%5Cimplies%20%5Ccfrac%7B3%5Cpi%5E2%7D%7B4%7D-6%5Capprox%201.4022033)
let's check where the quadratic and x = 6 meet, bear in mind we're integrating by "y", so