Question

If a convex polygon has $35$ diagonals, then the number of triangles in which exactly one side is common with that of polygon is

• A. $35$
• B. $60$
• C. $120$
• D. $150$

Answer 1

The correct option is B $60$

If polygon has $n$ sides, then number of
diagonals is
${}^n{C}_2-n=35$
$\Rightarrow \frac{n(n-1)}{2}-n=35$
Solving we get $n=10.$

Suppose one of the common sides of the triangle and polygon is $A_1{A}_2.$ Then third vertex cannot be $A_3$ or $A_{10}$ for exactly one side of triangle and polygon to be common. Thus, for the third vertex of triangle only six vertices are left. Therefore there are six triangles in which side $A_1{A}_2$ is common with that of polygon.
Similarly for each of the sides of polygon i.e., $A_2{A}_3,A_3{A}_4\cdots ,A_9{A}_{10},A_{10}{A}_1$ there are six triangles possible.

$\therefore$ Total number of triangles with exactly one side common with that of polygon is
$=10\times 6=60$