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Difference between revisions of "2012 AMC 8 Problems/Problem 6"
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<math> \textbf{(A)}\hspace{.05in}36\qquad\textbf{(B)}\hspace{.05in}40\qquad\textbf{(C)}\hspace{.05in}64\qquad\textbf{(D)}\hspace{.05in}72\qquad\textbf{(E)}\hspace{.05in}88 </math>
 
<math> \textbf{(A)}\hspace{.05in}36\qquad\textbf{(B)}\hspace{.05in}40\qquad\textbf{(C)}\hspace{.05in}64\qquad\textbf{(D)}\hspace{.05in}72\qquad\textbf{(E)}\hspace{.05in}88 </math>
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==Solution==
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In order to find the area of the border, we need to subtract the area of the photograph from the area of the photograph and the frame. Right off the bat, we can see that the area of the photograph is <math> 8 \times 10 = 80 </math> square inches. The dimensions of the whole frame (including the photograph) would be <math> 8+2+2 = 12 </math> by <math> 10+2+2 = 14 </math>. Therefore, the area of the photograph and frame would be <math> 12 \times 14 = 168 </math> square inches. Subtracting the area of the photograph from the area of both the frame and photograph, we find the answer to be <math> \boxed{\textbf{(E)}\ 88} </math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=5|num-a=7}}
 
{{AMC8 box|year=2012|num-b=5|num-a=7}}

Revision as of 10:16, 24 November 2012

A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures 8 inches high and 10 inches wide. What is the area of the border, in square inches?

$\textbf{(A)}\hspace{.05in}36\qquad\textbf{(B)}\hspace{.05in}40\qquad\textbf{(C)}\hspace{.05in}64\qquad\textbf{(D)}\hspace{.05in}72\qquad\textbf{(E)}\hspace{.05in}88$

Solution

In order to find the area of the border, we need to subtract the area of the photograph from the area of the photograph and the frame. Right off the bat, we can see that the area of the photograph is $8 \times 10 = 80$ square inches. The dimensions of the whole frame (including the photograph) would be $8+2+2 = 12$ by $10+2+2 = 14$. Therefore, the area of the photograph and frame would be $12 \times 14 = 168$ square inches. Subtracting the area of the photograph from the area of both the frame and photograph, we find the answer to be $\boxed{\textbf{(E)}\ 88}$.

See Also

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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
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