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Difference between revisions of "2012 AMC 8 Problems/Problem 6"
 
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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?
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==Problem==
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A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures <math>8</math> inches high and <math>10</math> inches wide. What is the area of the border, in square inches?
  
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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>
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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>
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==Solution==
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In order to find the area of the frame, we need to subtract the area of the photograph from the area of the photograph and the frame together. The area of the photograph is <math> 8 \times 10 = 80 </math> square inches. The height of the whole frame (including the photograph) would be <math> 8+2+2 = 12</math>, and the width of the whole frame,  <math> 10+2+2 = 14 </math>. Therefore, the area of the whole figure 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>168-80 = \boxed{\textbf{(E)}\ 88} </math>.
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==Video Solution==
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https://youtu.be/P89ysBIaPfs ~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=5|num-a=7}}
 
{{AMC8 box|year=2012|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 10:26, 13 June 2022

Problem

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 frame, we need to subtract the area of the photograph from the area of the photograph and the frame together. The area of the photograph is $8 \times 10 = 80$ square inches. The height of the whole frame (including the photograph) would be $8+2+2 = 12$, and the width of the whole frame, $10+2+2 = 14$. Therefore, the area of the whole figure 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 $168-80 = \boxed{\textbf{(E)}\ 88}$.

Video Solution

https://youtu.be/P89ysBIaPfs ~savannahsolver

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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