Consider the convex polygon which has $35$ diagonals. Then number of triangles in which exactly two sides are common with that of polygon is

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Solution

Given convex polygon with sides $^{'} n^{'}$ which has $35$ diagonals.
Number of diagonals
$= \frac{n (n - 1)}{2} - n = 35$
$\Rightarrow n^{2} - 3 n - 70 = 0$
$\Rightarrow (n - 10) (n + 7) = 0$
$∴ n = 10$
For number of triangles in which exactly two sides are common with that of polygon we have to select three consecutive vertices of the polygon i.e. $A_{1} A_{2} A_{3}, A_{2} A_{3} A_{4}, A_{3} A_{4} A_{5} \dots A_{10} A_{1} A_{2}$ which can be done by $n$ ways.

Hence number of triangles in which exactly two sides are common with that of polygon $= 10$

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\begin{matrix} C o l u m n 1 & C o l u m n 2 a. Number of triangles joining the vertices of the polygon & p. 210 b. Number of points of intersections of diagonal which lies inside the polygon & q. 120 c. Number of triangle in which exactly one side is common with that of polygon & r. 10 d. Number of triangles in which exactly two sides are common with that of polygon & s. 60 \end{matrix}

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