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

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(B) Total number of points $= 3 + 4 + 5 = 12$. Number of ways to select $3$ points out of $12$ is $^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2

Vedclass · Accepted Answer

$205$

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(B) Total number of points $= 3 + 4 + 5 = 12$. Number of ways to select $3$ points out of $12$ is $^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$. However,points on the same side are collinear and cannot form a triangle. Number of triangles formed by $3$ collinear points on side $AB = ^3C_3 = 1$. Number of triangles formed by $3$ collinear points on side $BC = ^4C_3 = 4$. Number of triangles formed by $3$ collinear points on side $CA = ^5C_3 = 10$. Required number of triangles $= 220 - (1 + 4 + 10) = 220 - 15 = 205$.

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 is:

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