Difference between revisions of "2016 AMC 12B Problems/Problem 9"
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==Problem== | ==Problem== | ||
| − | Carl decided to fence in his rectangular garden. He bought <math>20</math> fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly <math>4</math> yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of | + | Carl decided to fence in his rectangular garden. He bought <math>20</math> fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly <math>4</math> yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl's garden? |
<math>\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512</math> | <math>\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512</math> | ||
==Solution== | ==Solution== | ||
| − | |||
| − | To start, use algebra to determine the number of posts on each side. You have (the long sides count for <math>2</math> because there are twice as many) <math>6x = 20 + 4</math> (each corner is double counted so you must add <math>4</math>) Making the shorter end have <math>4</math>, and the longer end have <math>8</math>. <math>((8-1) | + | To start, use algebra to determine the number of posts on each side. You have (the long sides count for <math>2</math> because there are twice as many) <math>6x = 20 + 4</math> (each corner is double counted so you must add <math>4</math>) Making the shorter end have <math>4</math>, and the longer end have <math>8</math>. <math>((8-1) \cdot 4) \cdot ((4-1) \cdot 4) = 28 \cdot 12 = 336</math>. Therefore, the answer is <math>\boxed{\textbf{(B)}\ 336}</math> |
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| + | ~Albert471 | ||
==Solution 2== | ==Solution 2== | ||
Latest revision as of 14:31, 5 July 2024
Contents
Problem
Carl decided to fence in his rectangular garden. He bought 
fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly 
yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl's garden?
Solution
To start, use algebra to determine the number of posts on each side. You have (the long sides count for 
because there are twice as many) 
(each corner is double counted so you must add ) Making the shorter end have , and the longer end have 
. 
. Therefore, the answer is 
~Albert471
Solution 2
To start, allocate 4 posts from the total 20 to be the corners. Now 16 posts are left. Pairing each long and short side together, you have 8 posts for a segment of long and short (For rectangle ABCD, AB = CD, AC = BD, we are taking what is originally AB + BC + CD + DA = 16, and making it AB + AC = 8). From 8, you need two numbers that add up too 8, that when you add 2 to both of them are at a 1:2 ratio. Brute force says these numbers are 6 and 2. 6 + 2 = 12, and (6+2) = 2(2+2). Then it is trivial to calculate area, as the long side has 8 posts, short side has 4, the total length of the long side is 28, and the short side 12. 28 x 12 = 336.
~Shadow-18
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
