Consider a rectangle ABCD having 5, 6, 7, 9 p | vedclass

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(D) Let the number of points on sides $AB, BC, CD, DA$ be $n_1=5, n_2=6, n_3=7, n_4=9$ respectively.$\alpha$ is the number of triangles formed by choo

Vedclass · Accepted Answer

$717$

Vedclass · Answer

(D) Let the number of points on sides $AB, BC, CD, DA$ be $n_1=5, n_2=6, n_3=7, n_4=9$ respectively.$\alpha$ is the number of triangles formed by choosing $3$ points from different sides.To form a triangle,we select $3$ sides out of $4$ and then $1$ point from each selected side.$\alpha = (n_1 n_2 n_3) + (n_1 n_2 n_4) + (n_1 n_3 n_4) + (n_2 n_3 n_4)$$\alpha = (5 \cdot 6 \cdot 7) + (5 \cdot 6 \cdot 9) + (5 \cdot 7 \cdot 9) + (6 \cdot 7 \cdot 9)$$\alpha = 210 + 270 + 315 + 378 = 1173$$\beta$ is the number of quadrilaterals formed by choosing $4$ points from different sides.To form a quadrilateral,we select $1$ point from each of the $4$ sides.$\beta = n_1 \cdot n_2 \cdot n_3 \cdot n_4$$\beta = 5 \cdot 6 \cdot 7 \cdot 9 = 1890$Therefore,$(\beta-\alpha) = 1890 - 1173 = 717$.

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