A convex polygon has 44 diagonals. What is the total number of distinct triangles that can… — JAMB UTME

A convex polygon has $44$ diagonals. What is the total number of distinct triangles that can be formed by joining any three vertices of this polygon?

Cevap: 165

165

The number of diagonals of an

n

-sided convex polygon is given by

\binom{n}{2} - n = \frac{n(n-3)}{2}

. Setting this equal to

44

yields

n(n-3) = 88

, which simplifies to

n^2 - 3n - 88 = 0

. Factoring gives

(n-11)(n+8) = 0

, so

n = 11

. The total number of distinct triangles formed by choosing any 3 vertices from an 11-sided polygon is

\binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165

Set up the equation for the number of diagonals in terms of the number of vertices

n

\frac{n(n-3)}{2} = 44

Choosing any 2 vertices from

n

vertices gives

\binom{n}{2}

total connecting line segments. Subtracting the

n

boundary sides leaves the diagonals.

Solve the quadratic equation for

n

n^2 - 3n - 88 = 0 \implies (n-11)(n+8) = 0 \implies n = 11

A polygon must have a positive integer number of vertices, so

n = 11

Compute the number of distinct triangles using combinations

\binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165

In a convex polygon, no three vertices are collinear, so every unique combination of 3 vertices forms a distinct triangle.

Application of combinations to geometry (polygon diagonals and triangle selection)

Tahmini Süre:2m 0s

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