Difference between revisions of "2017 AMC 8 Problems/Problem 11"

m (Problem 11)
(6 intermediate revisions by 4 users not shown)
Line 1: Line 1:
==Problem 11==
+
==Problem==
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor? what does diagonals mean
+
A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
  
  
Line 8: Line 8:
 
Since the number of tiles lying on both diagonals is <math>37</math>, counting one tile twice, there are <math>37=2x-1\implies x=19</math> tiles on each side. Hence, our answer is <math>19^2=361=\boxed{\textbf{(C)}\ 361}</math>.
 
Since the number of tiles lying on both diagonals is <math>37</math>, counting one tile twice, there are <math>37=2x-1\implies x=19</math> tiles on each side. Hence, our answer is <math>19^2=361=\boxed{\textbf{(C)}\ 361}</math>.
  
==See Also==
+
==Video Solution==
 +
Associated video: https://youtu.be/QCWOZwYVJMg
 +
 
 +
==See Also:==
 
{{AMC8 box|year=2017|num-b=10|num-a=12}}
 
{{AMC8 box|year=2017|num-b=10|num-a=12}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:01, 18 January 2021

Problem

A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?


$\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369$

Solution

Since the number of tiles lying on both diagonals is $37$, counting one tile twice, there are $37=2x-1\implies x=19$ tiles on each side. Hence, our answer is $19^2=361=\boxed{\textbf{(C)}\ 361}$.

Video Solution

Associated video: https://youtu.be/QCWOZwYVJMg

See Also:

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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