Question
In a polygon no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon be 70 then the number of diagonals of the polygon is
A.
20
B.
28
C.
8
D.
None of these
Answer :
20
Solution :
A selection of four vertices of the polygon gives an interior intersection.
∴ the number of sides $$= n$$
$$\eqalign{ & \Rightarrow \,{\,^n}{C_4} = 70 \cr & \Rightarrow \,\,n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right) = 24 \times 70 = 8 \times 7 \times 6 \times 5 \cr & \therefore \,\,n = 8 \cr} $$
∴ the number of diagonals $$ = {\,^8}{C_2} - 8.$$
A selection of four vertices of the polygon gives an interior intersection.
∴ the number of sides $$= n$$
$$\eqalign{ & \Rightarrow \,{\,^n}{C_4} = 70 \cr & \Rightarrow \,\,n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right) = 24 \times 70 = 8 \times 7 \times 6 \times 5 \cr & \therefore \,\,n = 8 \cr} $$
∴ the number of diagonals $$ = {\,^8}{C_2} - 8.$$