The correct answer is 16

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side
Hence, \(a = \sqrt{(3−1)^2+(6−2)^2}\)
\(= \sqrt{20}\)
Thereafter, \(\frac{b}{2} = \frac{4}{\sqrt5}\)
\(⇒ b = \frac{8}{\sqrt5}\)
So, the Area ab \(= 2 \sqrt5 . \frac{8}{\sqrt5}\)
= 16
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
Let \( \alpha = \dfrac{-1 + i\sqrt{3}}{2} \) and \( \beta = \dfrac{-1 - i\sqrt{3}}{2} \), where \( i = \sqrt{-1} \). If
\[ (7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}, \] then the value of \( m \) is ___________.