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Question

Consider a convex polygon which has 35 diagonals. Then match the following columns:
​​​​​​Column 1Column 2a. Number of triangles joining the vertices of thepolygon p. 210b. Number of points of intersections of diagonal which lies inside the polygon q. 120c. Number of triangle in which exactly one side is common with that of polygon r. 10d. Number of triangles in which exactly two sides are common with that of polygon s. 60

A
aq; bp; cs; dr.
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B
as; br; cq; dp.
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C
aq; bs; cp; dr.
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D
as; bq; cr; dp.
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Solution

The correct option is A aq; bp; cs; dr.
a. If polygon has n sides, then number of diagonals is nC2n=35 (given). Solving we get n=10. Thus, there are 10 vertices, from which 10C3(=120) triangles can be formed.

b. Four vertices can be selected in 10C4(=210) ways. Using these four vertices two diagonals can be formed, which has exactly one point of intersection lying inside the polygon.

c. Suppose one of the sides of the triangle is A1A2. Then the third vertex cannot be A3 or A10
Thus, for third vertex, six vertices are left. There are six triangles in which side A1A2 is common with that of the polygon.
Similarly, for each of the sides A2A3, A3A4, ..., A9A10, there are six triangles. Then total number of triangles is 60.

d. Triangles A1A2A3, A2A3A4, ...,A8A9A10 have two sides common with that of polygon. Hence, there are 10 such triangles.

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