How many non-right triangles can be created? Math220DaalenProject (004 XFile C:/Users/daale/AppData/Local/Microsoft/Windows/INetCache/Content.Outlook/3DTXZJYI/Math220Da.. *PDFwith Microsoft Edge, finish setting up your browser.Not syncingComplete setupMaybe later+のA Read aloud 7 DrawFall 2020 Math 220-100 Discrete Mathematics ProjectHighlight vErase |品日Academic Integrity Remainder:Please read Tu sculum University the Academic Misconduct and Plagiarism Policy, which can be found on the Canvaspage of this class under Tu sculum University Institution al Policies, before attempting this project.This is an individual assignm ent. Collaboration is not permitted. You can refer to lecture notes or the textbookof this class but you are not allowed to use third-party resources to supplement your work.You should not look up solutions to these problems and derive your solutions by copying these.Project Instructions: You can find the detailed instructions and tutorials regarding this project on the Canvaspage of the cdass. Please read the instructions of this project before attempting this assignment. The due of thisassignment is 12/1/2020, 11:59 and no late work will be accepted.Question : Con sider all the triangles you can create using the points shown below as vertices. Note, we are notallowing degen er ate triangles (ones with all three vertices on the same line). How many non-right triangles can youcreat e? Please explain your thinking and your answer in great detail as st ated in the guidelines of this project.11:06②へ疊@ 小 ENG11/30/20日thing

From the given set of points to make triangles (non-degenerate): All the three vertices cannot be co-linear. So, select any two co-linear points and other vertex which is not co-linear to these selected two. This can be done in two ways for the given set of points. case (i): Two points from vertical co-linear points and third from horizontal set of points. Total number of vertical co-linear points are 5 Selecting any 2 points from these 5 points is done in C25 number of ways. Selecting third vertex from 6 horizontal set of points is done in C16 number of ways. Then the number of triangles formed in this case is: C25×C16=10×6=60 case (ii): Two vertices from the horizontal co-linear points and third vertex from vertical set of points. Total number of horizontal points are 6 Selecting any two points from these 6 points is done in C26 number of ways. Selecting third vertex from 5 vertical points is done in C15 number of ways. Then the number of triangles formed in this case is: C26×C15=15×5=75 Therefore, total number of triangles that can be formed using given set of points is: 60+75=135.For the triangle to be right angled, the vertex which is common with vertical and horizontal set of points must be one of the three vertices of the triangle. Then the other two vertices are one from each set of horizontal and vertical points. Leaving the fixed corner vertex, remaining points in horizontal set is 6 and 4 in vertical set. Now, Selecting one point from 6 horizontal set is done in C16 number of ways. Selecting one point from 4 vertical set is done in C14 number of ways. Then, the number of right triangles formed is: C16×C14=6×4=24 So, number of right angled triangles that can be formed using given set of points is 24. The number of non-right triangles is thus the difference between total number of triangles and number of right angled triangles. That is: 135-24=111 Therefore, number of non-right angled triangles that can be formed is 111.