Answer:
5 5/6
Step-by-step explanation:
49,98,147,196,245,294,343,392,441,490,539,588,637,686,735,784,833,882,931,980,1029,1078,1127,1176,1225,1274,1323,1372,1421,1470,1519,1568,1617,1666,1715,1764,1813,1862,1911,1960,2009,2058,2107,2156,2205,2254,2303,2352,2401,
-1,3 because your moving the quadrant counter clockwise into the 2nd quadrant
Check the picture below.
so notice, the sides AB and AC you can pretty much count them off the grid.
now, to get the CB side.
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) C&({{ -5}}\quad ,&{{ 1}})\quad % (c,d) B&({{ 3}}\quad ,&{{ -5}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ CB=\sqrt{[3-(-5)]^2+[-5-1]^2}\implies CB=\sqrt{(3+5)^2+(-5-1)^2} \\\\\\ CB=\sqrt{8^2+(-6)^2}\implies CB=\sqrt{100}\implies CB=10](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0AC%26%28%7B%7B%20-5%7D%7D%5Cquad%20%2C%26%7B%7B%201%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0AB%26%28%7B%7B%203%7D%7D%5Cquad%20%2C%26%7B%7B%20-5%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ACB%3D%5Csqrt%7B%5B3-%28-5%29%5D%5E2%2B%5B-5-1%5D%5E2%7D%5Cimplies%20CB%3D%5Csqrt%7B%283%2B5%29%5E2%2B%28-5-1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ACB%3D%5Csqrt%7B8%5E2%2B%28-6%29%5E2%7D%5Cimplies%20CB%3D%5Csqrt%7B100%7D%5Cimplies%20CB%3D10)
sum all three sides up, and that's the perimeter of the triangle.