Question

Consider a convex polygon which has $35$ diagonals. Then match the following columns:
​​​​​​$\begin{array}{cc}Column1& Column2\\ \text{a. Number of triangles joining the vertices of the}& \\ \text{polygon}& \text{p.}210\\ \text{b. Number of points of intersections of diagonal}& \\ \text{which lies inside the polygon}& \text{q.}120\\ \text{c. Number of triangle in which exactly one side}& \\ \text{is common with that of polygon}& \text{r.}10\\ \text{d. Number of triangles in which exactly two sides}& \\ \text{are common with that of polygon}& \text{s.}60\end{array}$

• A. $a\to q;b\to p;c\to s;d\to r.$
• B. $a\to s;b\to r;c\to q;d\to p.$
• C. $a\to q;b\to s;c\to p;d\to r.$
• D. $a\to s;b\to q;c\to r;d\to p.$

Answer 1

The correct option is A $a\to q;b\to p;c\to s;d\to r.$

$a.$ If polygon has n sides, then number of diagonals is ${}^n{C}_2-n=35$ (given). Solving we get $n=10$. Thus, there are $10$ vertices, from which ${}^{10}{C}_3(=120)$ triangles can be formed.

$b.$ Four vertices can be selected in ${}^{10}{C}_4(=210)$ 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 $A_1{A}_2$. Then the third vertex cannot be $A_3$ or $A_{10}$
Thus, for third vertex, six vertices are left. There are six triangles in which side $A_1{A}_2$ is common with that of the polygon.
Similarly, for each of the sides $A_2{A}_3,A_3{A}_4,...,A_9{A}_{10},$ there are six triangles. Then total number of triangles is 60.

$d.$ Triangles $A_1{A}_2{A}_3,A_2{A}_3{A}_4,...,A_8{A}_9{A}_{10}$ have two sides common with that of polygon. Hence, there are $10$ such triangles.