Q90 / 180 · COMEDK 2023 (Morning Shift) · Mathematics
Three vertices are chosen randomly from the eight vertices of a regular 8-sided polygon. The probability that they form the vertices of an isosceles triangle, is
- Correct answerA$\dfrac{3}{7}$
- B$\dfrac{4}{7}$
- C$\dfrac{5}{7}$
- D$\dfrac{2}{7}$
Solution
Total ways to choose 3 vertices from 8: $\binom{8}{3} = 56$
For isosceles triangles, fix any vertex as the apex. For vertex 1, the symmetric pairs are $(2,8), (3,7), (4,6)$, giving 3 isosceles triangles per apex.
Since 8 is not divisible by 3, no equilateral triangles exist, so there is no overcounting.
Total isosceles triangles $= 8 \times 3 = 24$
$P(\text{Isosceles}) = \dfrac{24}{56} = \dfrac{3}{7}$
For isosceles triangles, fix any vertex as the apex. For vertex 1, the symmetric pairs are $(2,8), (3,7), (4,6)$, giving 3 isosceles triangles per apex.
Since 8 is not divisible by 3, no equilateral triangles exist, so there is no overcounting.
Total isosceles triangles $= 8 \times 3 = 24$
$P(\text{Isosceles}) = \dfrac{24}{56} = \dfrac{3}{7}$
Hence, the correct answer is option: 'A'.