Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively. Let α be the number of triangles having these points from different sides as vertices and β be the number of quadrilaterals having these points from different sides as vertices. Then (β - α) is equal to :

Solution

Maths sir · 3 steps

One point from each chosen side: just multiply the counts, and no three such points can ever be collinear.

1. Why multiplying works. A straight line meets the boundary of a rectangle in at most 2 points unless it runs along a side. So points taken from different sides are never collinear, and every choice of one point per side gives a genuine figure.
2. Quadrilaterals. A quadrilateral needs one point from each of the 4 sides, so β = 5 × 7 × 6 × 9 = 1890.

   Sides used for the triangle	Number of triangles
   AB, CD, BC	5 × 7 × 6 = 210
   AB, CD, DA	5 × 7 × 9 = 315
   AB, BC, DA	5 × 6 × 9 = 270
   CD, BC, DA	7 × 6 × 9 = 378
   Total α	210 + 315 + 270 + 378 = 1173

3. So. Pick 3 of the 4 sides in C(4, 3) = 4 ways and add the products to get α = 1173. Then β - α = 1890 - 1173 = 717, which is option (a).

The idea to keep. When every vertex must come from a different group, first choose which groups to use, then multiply the group sizes. Check for collinearity before trusting the count.

Answer from the NTA final answer key. Solution by PrepWiser.