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?

$2(w + h)^2$

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

$2w^2 + 4wh$

$2w^2$

$w^2 h$

Answer: A

Concepts:paper folding Pythagorean Theorem square (geometry)

Difficulty rating: 1730

Solution:

Let the sheet have side $s.$ The base sits as a square of side $w$ turned $45^\circ,$ so the center is $\tfrac{w}{2}$ from each base edge. A corner of the sheet lies $\tfrac{s}{\sqrt2}$ from the center. Folding that corner up to the top center traces a straight line: $\tfrac{w}{2}$ out to the base edge, then $h$ up the side, then $\tfrac{w}{2}$ across the top. So $\tfrac{s}{\sqrt2} = \tfrac{w}{2} + h + \tfrac{w}{2} = w + h.$ Then $s = \sqrt2\,(w + h),$ and the area is $s^2 = 2(w + h)^2.$ Thus, A is the correct answer.

2018 AMC 10B Problem 15

Answer: A

Difficulty rating: 1730

Solution:

Problem 15 in Other Years