Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AMC 8 Problems/Problem 19"
(→‎Solution 3)
Line 31: Line 31:
  
 
===Solution 3===
 
===Solution 3===
−
By the shoelace theorem the area of <math>\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12}|=|\dfrac{1}{2}(-10)|=5</math>
+
By the shoelace theorem the area of <math>\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12)|=|\dfrac{1}{2}(-10)|=5</math>
 
This means the fraction of the total area is <math>\boxed{\textbf{(A)}~\dfrac{1}{6}}</math>
 
This means the fraction of the total area is <math>\boxed{\textbf{(A)}~\dfrac{1}{6}}</math>
 +
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2015|num-b=18|num-a=20}}
 
{{AMC8 box|year=2015|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:31, 25 November 2015

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?

$\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}$

[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 1

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 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 $\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}$.

Solution 2

Note angle $\angle ACB$ is right, thus the area is $\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=(10)\times \dfrac{1}{2}=5$ thus the fraction of the total is $\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}$

Solution 3

By the shoelace theorem the area of $\triangle ABC=|\dfrac{1}{2}(15+4+4-1-20-12)|=|\dfrac{1}{2}(-10)|=5$ This means the fraction of the total area is $\boxed{\textbf{(A)}~\dfrac{1}{6}}$

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_19&oldid=73160"