Question

The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the sides of the octagon is

• A. 24
• B. 52
• C. 48
• D. 16

Answer 1

The correct option is D 16

Number of all possible triangles
= Number of selections of 3 points from 8 vertices
$={}^8{C}_3=56$

Number of triangle with one side common with octagon = 8 $\times$ 4 = 32
(Consider side $A_1{A}_2$. Since two points $A_3,A_8$ are adjacent, 3rd point should be chosen from remaining 4 points).
Number of triangles having two sides common with octagon: All such triangles have three consecutive vertices, viz., $A_1{A}_2{A}_3,A_2{A}_3{A}_4.\cdots A_8{A}_1{A}_2$.
Number of such triangles = 8
$\therefore$ Number of triangles with no side common
= 56 - 32 - 8 = 16.