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"

(→‎Solution)
m (→‎Solution)
 
(7 intermediate revisions by 7 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>
  
−
===Solution===
+
==Solution==
−
As per the tringle inequality, the sum of the length of any 2 sides is greater than the largest side. So, let <math>x</math> be the third side and <math>19</math> be the largest side. <math>5+x > 19</math>. This gives the smallest value of <math>x</math> as <math>15</math>. The perimeter would be <math>5 + 15 + 19 = 39</math> The smallest side which would be greater than the perimeter would be <math>\box{43}</math>.
+
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.
 +
 
 +
 
 +
==Solution 2==
 +
<math>s</math>, fulfills <math>s<5+19=24</math>, meaning that s<24. The greatest value of s can be 23. 23+5+19 is 47. 48 is our nearest answer choice so its <math>\boxed{\textbf{(D)}\ 48}</math>.-TheNerdWhoIsNerdy.
 +
 
 +
==Video Solution (HOW TO THINK CRITICALLY!!!)==
 +
https://youtu.be/rxXHhn5PB9w
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://youtu.be/zUiKAoX2D_Q
 +
 
 +
~savannahsolver
 +
 
 +
==See Also==
 +
 
 +
{{AMC8 box|year=2015|num-b=7|num-a=9}}
 +
{{MAA Notice}}

Latest revision as of 19:50, 15 September 2024

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.


Solution 2

$s$, fulfills $s<5+19=24$, meaning that s<24. The greatest value of s can be 23. 23+5+19 is 47. 48 is our nearest answer choice so its $\boxed{\textbf{(D)}\ 48}$.-TheNerdWhoIsNerdy.

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/rxXHhn5PB9w

~Education, the Study of Everything

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=227292"