On the sides AB, BC, CA of a Delta ABC,3, 4, | vedclass

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(B) The total number of points chosen on the sides is $3 + 4 + 5 = 12$.To form a triangle,we need to select $3$ points out of these $12$ points.The to

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$205$

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(B) The total number of points chosen on the sides is $3 + 4 + 5 = 12$.To form a triangle,we need to select $3$ points out of these $12$ points.The total number of ways to select $3$ points from $12$ is $^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$.However,points lying on the same side cannot form a triangle. We must subtract the cases where all $3$ points are chosen from the same side.Number of ways to choose $3$ points from $3$ points on side $AB$ is $^3C_3 = 1$.Number of ways to choose $3$ points from $4$ points on side $BC$ is $^4C_3 = 4$.Number of ways to choose $3$ points from $5$ points on side $CA$ is $^5C_3 = 10$.Total collinear cases to subtract $= 1 + 4 + 10 = 15$.Therefore,the number of triangles $= 220 - 15 = 205$.

On the sides $AB, BC, CA$ of a $\Delta ABC$,$3, 4, 5$ distinct points (excluding vertices $A, B, C$) are respectively chosen. The number of triangles that can be constructed using these chosen points as vertices are:

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