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 8"
(25 intermediate revisions by 14 users not shown)
Line 1: Line 1:
 +
==Problem==
 +
 
What is the smallest whole number larger than the perimeter of any triangle with a side of length <math> 5</math> and a side of length <math>19</math>?
 
What is the smallest whole number larger than the perimeter of any triangle with a side of length <math> 5</math> and a side of length <math>19</math>?
  
 
<math>\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57</math>
 
<math>\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57</math>
−
The area of <math>\triangle ABC</math> is equal to half the product of its base and height.  By the Pythagorean Theorem, we find its height is <math>\sqrt{1^2+2^2}=\sqrt{5}</math>, and its base is <math>\sqrt{2^2+4^2}=\sqrt{20}</math>.  We multiply these and divide by 2 to find the of the triangle is <math>\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5</math>.  Since the grid has an area of <math>30</math>, the fraction of the grid covered by the triangle is <math>\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}</math>.
+
 
 +
==Solution==
 +
We know from the triangle inequality that the last side, <math>s</math>, fulfills <math>s<5+19=24</math>. Adding <math>5+19</math> to both sides of the inequality, we get <math>s+5+19<48</math>, and because <math>s+5+19</math> is the perimeter of our triangle, <math>\boxed{\textbf{(D)}\ 48}</math> is our answer.
 +
 
 +
==Video Solution==
 +
https://youtu.be/zUiKAoX2D_Q
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 10:33, 30 March 2022

Problem

What is the smallest whole number larger than the perimeter of any triangle with a side of length $5$ and a side of length $19$?

$\textbf{(A) }24\qquad\textbf{(B) }29\qquad\textbf{(C) }43\qquad\textbf{(D) }48\qquad \textbf{(E) }57$

Solution

We know from the triangle inequality that the last side, $s$, fulfills $s<5+19=24$. Adding $5+19$ to both sides of the inequality, we get $s+5+19<48$, and because $s+5+19$ is the perimeter of our triangle, $\boxed{\textbf{(D)}\ 48}$ is our answer.

Video Solution

https://youtu.be/zUiKAoX2D_Q

~savannahsolver

See Also

2015 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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_8&oldid=173062"