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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
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The correct answer is:
20
A combination of four vertices is equivalent to one interior point of intersection of diagonals.
\(\therefore\) No. of interior points of intersection
\(\begin{aligned}
& ={ }^{\mathrm{n}} \mathrm{C}_4=70 \\
& \Rightarrow \mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)=5.6 .7 .8 \\
& \therefore \mathrm{n}=8
\end{aligned}\)
So, number of diagonals \(={ }^8 \mathrm{C}_2-8=20\)
\(\therefore\) No. of interior points of intersection
\(\begin{aligned}
& ={ }^{\mathrm{n}} \mathrm{C}_4=70 \\
& \Rightarrow \mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)=5.6 .7 .8 \\
& \therefore \mathrm{n}=8
\end{aligned}\)
So, number of diagonals \(={ }^8 \mathrm{C}_2-8=20\)
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