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  • The sides AB, BC, and CA of a triangle ABC whose interior points are 6, 7, and 8. The number of triangles that may be formed with these points as vertices is  <div clas

The sides AB, BC, and CA of a triangle ABC whose interior points are 6, 7, and 8. The number of triangles that may be formed with these points as vertices is

 

Option: 1

1345


Option: 2

1215


Option: 3

1302


Option: 4

1219


Answers (1)

best_answer

Given that,

The interior points of a triangle ABC are 6, 7, and 8.

To construct a triangle we need 3 points.

The total number of ways to form a triangle is,

6 + 7  + 8 = 21

Thus, the total number of ways to construct the triangle using the 3 points is given by,

\mathrm{\begin{aligned} &{ }^{21} C_3=\frac{21 !}{3 ! 18 !}\\ &{ }^{21} C_3=\frac{21 \times 20 \times 19}{3 \times 2}\\ &{ }^{21} C_3=1330 \end{aligned}}

The points on the sides AB, BC, and CA are all collinear.

The number of ways for the sides AB is given by,

\mathrm{\begin{aligned} &{ }^6 C_3=\frac{6 !}{3 ! 3 !}\\ &{ }^6 C_3=\frac{6 \times 5 \times 4}{3 \times 2}\\ &{ }^6 C_3=20 \end{aligned}}

The number of sides for the side BC is given by,

\mathrm{\begin{aligned} { }^7 C_3 & =\frac{7 !}{3 ! 4 !} \\ { }^7 C_3 & =\frac{7 \times 6 \times 5}{3 \times 2} \\ { }^7 C_3 & =35 \end{aligned}}

The number of sides for the side CA is given by,

\mathrm{\begin{aligned} &{ }^8 C_3=\frac{8 !}{3 ! 5 !}\\ &{ }^8 C_3=\frac{8 \times 7 \times 6}{3 \times 2}\\ &{ }^8 C_3=56 \end{aligned}}

The total number of triangles is given by,

\mathrm{\begin{aligned} & N={ }^{21} C_3-\left({ }^6 C_3+{ }^7 C_3+{ }^8 C_3\right) \\ & N=1330-(20+35+56) \\ & N=1219 \end{aligned}}

Therefore, the total number of triangles formed is 1219.

 

Posted by

shivangi.shekhar

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