AMC8 2015 8, Problem 19

A triangle with vertices as $A=(1,3)$, $B=(5,1)$, and $C=(4,4)$ is plotted on a $6\times5$ grid. What fraction of the grid is covered by the triangle? [asy] draw((1,0)--(1,5),linewidth(.5)); draw((2,0)--(2,5),linewidth(.5)); draw((3,0)--(3,5),linewidth(.5)); draw((4,0)--(4,5),linewidth(.5)); draw((5,0)--(5,5),linewidth(.5)); draw((6,0)--(6,5),linewidth(.5)); draw((0,1)--(6,1),linewidth(.5)); draw((0,2)--(6,2),linewidth(.5)); draw((0,3)--(6,3),linewidth(.5)); draw((0,4)--(6,4),linewidth(.5)); draw((0,5)--(6,5),linewidth(.5));  draw((0,0)--(0,6),EndArrow); draw((0,0)--(7,0),EndArrow); draw((1,3)--(4,4)--(5,1)--cycle); label("$y$",(0,6),W); label("$x$",(7,0),S); label("$A$",(1,3),dir(210)); label("$B$",(5,1),SE); label("$C$",(4,4),dir(100)); [/asy]

Solution: The area of $\triangle ABC$ is equal to half the product of its base and height.  By the Pythagorean Theorem, we find its height is $\sqrt{1^2+2^2}=\sqrt{5}$, and its base is $\sqrt{2^2+4^2}=\sqrt{20}$.  We multiply these and divide by $2$ to find the area of the triangle is $\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5$.  Since the grid has an area of $30$, the fraction of the grid covered by the triangle is $\frac 5{30}=\fbox{\frac{1}{6}}$.

Answer: \frac{1}{6}