2018 AMC 12B Problem 11

Question

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

A

in the figure on the right. The box has base length

w

and height

h.

What is the area of the sheet of wrapping paper?

2(w+h)^2

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

2w^2+4wh

2w^2

w^2h

Answer: A

Concepts:special right triangle area

Difficulty rating: 1760

Small hint:

The distance from a corner of the sheet to its center is

\tfrac{w}{2}+h+\tfrac{w}{2}

Big hint:

A side of the square sheet is

\sqrt2

times that corner-to-center distance

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

\dfrac{w}{2}+h+\dfrac{w}{2}=w+h.

That segment is a leg of a

45

-

45

-

90

triangle whose hypotenuse is a full side of the square sheet, so the side length is

\sqrt2\,(w+h).

The area of the sheet is

\left(\sqrt2\,(w+h)\right)^2=2(w+h)^2.

Thus, the correct answer is A.