The area of the triangle formed by the lines joining the vertex of the parabola $x^2=20 \mathrm{y}$ to the end of its Latus rectum is

For the parabola $x^2 = 20y$, comparison with the standard form $x^2 = 4ay$ yields $4a = 20$, so $a = 5$. The vertex is at the origin: $V = (0, 0)$.

The latus rectum lies along $y = a = 5$. Substituting into the equation gives $x^2 = 100$, so $x = \pm 10$. The endpoints are $L_1 = (-10, 5)$ and $L_2 = (10, 5)$.

The triangle formed by these points has base length 20 and height 5. Its area is $\frac{1}{2} \times 20 \times 5 = 50$.