Problem 6 · 2012 AMC 8
Easy
Geometry & Measurement
outer-minus-inner
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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Answer: E — 88 square inches.
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Hint 1 of 2
The border is an awkward picture-frame shape. Don't measure it directly — it's a big rectangle (frame + photo) with the photo punched out of the middle.
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Hint 2 of 2
This is complementary area: outer area − inner area. The only trap is the width — a 2-inch border adds 2 on the left and 2 on the right, so each dimension grows by 4, not 2.
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Approach: outer rectangle minus the photo (complementary area)
- The border is everything in the outer rectangle except the photo, so its area = outer area − photo area. That sidesteps adding up four separate strips.
- The 2-inch border hits both sides, so each dimension gains 2 + 2 = 4: the outer rectangle is 12 × 14 = 168.
- Subtract the photo: 8 × 10 = 80, so border = 168 − 80 = 88 sq in.
- Watch the doubling — a border/margin of width w always adds 2w to each side length, the classic slip in frame and walkway problems.
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