To solve the problem of finding the number of triangles whose vertices are at the vertices of a regular octagon, but none of whose sides is a side of the octagon, we can follow these steps:
First, calculate the total number of triangles that can be formed by choosing any three vertices from the octagon. A regular octagon has 8 vertices. The number of ways to choose 3 vertices from 8 is given by the combination formula:
\(^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\).
Next, subtract the number of triangles where at least one side is a side of the octagon from the total number of triangles.
To determine these triangles, notice that for a triangle to have a side of the octagon, it must include two adjacent vertices:
There are 8 such pairs of adjacent vertices (one pair for each side of the octagon). Thus, the number of unwanted triangles (where one side is a side of the octagon) is:
\(8 \times 5 = 40\).
Finally, subtract the number of unwanted triangles from the total number possible:
\(56 - 40 = 16\).
Thus, the number of triangles where none of the sides is a side of the octagon is 16.
Therefore, the correct answer is 16.
The number of triangles having no side common with an \( n \)-sided polygon is given by:
\[ \text{no. of triangles having no side common with a } n \text{-sided polygon} = \binom{n}{1} \times \binom{n-4}{2} \div 3 \]
Substitute \( n = 8 \):
\[ = \binom{8}{1} \times \binom{4}{2} \div 3 \]
\[ = 8 \times 6 \div 3 \]
\[ = 16. \]
Thus, the number of such triangles is 16.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,