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Difference between revisions of "2012 AMC 8 Problems/Problem 6"

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==Problem==
 
==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 8 inches high and 10 inches wide. What is the area of the border, in square inches?
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A painting with dimensions 10 inches by 14 inches is placed in a picture frame (of constant width), increasing its area to 221 square inches. How many inches is the width of the picture frame?
  
 
<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>

Revision as of 12:01, 15 January 2019

Problem

A painting with dimensions 10 inches by 14 inches is placed in a picture frame (of constant width), increasing its area to 221 square inches. How many inches is the width of the picture frame?

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

Solution

First, we start with a sketch. It's always a good idea to start with a picture, although not as detailed as this one. (Tip: On the actual AMC 8 test, you should NOT color your picture in, even if it is beautiful)

2012amc86.png

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 $168-80 = \boxed{\textbf{(E)}\ 88}$.

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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