2018 AMC 12B Problem 11

Below is the professionally curated solution for Problem 11 of the 2018 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12B solutions, or check the answer key.

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Concepts:special right trianglearea

Difficulty rating: 1760

11.

A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point AA in the figure on the right. The box has base length ww and height h.h. What is the area of the sheet of wrapping paper?

2(w+h)22(w+h)^2

(w+h)22\dfrac{(w+h)^2}{2}

2w2+4wh2w^2+4wh

2w22w^2

w2hw^2h

Solution:

Following a fold from a corner of the paper to the center of the box top, the distance from a corner of the sheet to its center is w2+h+w2=w+h. \dfrac{w}{2}+h+\dfrac{w}{2}=w+h.

That segment is a leg of a 4545-4545-9090 triangle whose hypotenuse is a full side of the square sheet, so the side length is 2(w+h).\sqrt2\,(w+h).

The area of the sheet is (2(w+h))2=2(w+h)2.\left(\sqrt2\,(w+h)\right)^2=2(w+h)^2.

Thus, the correct answer is A.

Problem 11 in Other Years