Difference between revisions of "2018 AMC 10B Problems/Problem 15"

Revision as of 02:56, 21 September 2021

The following problem is from both the 2018 AMC 12B #11 and 2018 AMC 10B #15, so both problems redirect to this page.

Problem

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?

$[asy] size(270pt); defaultpen(fontsize(10pt)); filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey); dot((-3,3)); label("$A$",(-3,3),NW); draw((1,3)--(-3,-1),dashed+linewidth(.5)); draw((-1,3)--(3,-1),dashed+linewidth(.5)); draw((-1,-3)--(3,1),dashed+linewidth(.5)); draw((1,-3)--(-3,1),dashed+linewidth(.5)); draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5)); draw((0,3)--(0,-3),linetype("2.5 2.5")+linewidth(.5)); draw((3,0)--(-3,0),linetype("2.5 2.5")+linewidth(.5)); label('$w$',(-1,-1),SW); label('$w$',(1,-1),SE); draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle); draw((4.5,0)--(8.5,0)); draw((6.5,2)--(6.5,-2)); label("$A$",(6.5,0),NW); dot((6.5,0)); [/asy]$

$\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h$

Solution 1

Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length of each dotted line is $h$ . The area of the rectangle that is $w$ by $h$ is $wh$ . The combined figure of the two triangles with base $h$ is a square with $h$ as its diagonal. Using the Pythagorean Theorem, each side of this square is $\sqrt{\frac{h^2}{2}}$ . Thus, the area is the side length squared which is $\frac{h^2}{2}$ . Similarly, the combined figure of the two triangles with base $w$ is a square with area $\frac{w^2}{2}$ . Adding all of these together, we get $\frac{w^2}{2} + \frac{h^2}{2} + wh$ . Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting $4\left(\frac{w^2}{2} + \frac{h^2}{2} + wh\right) = 2\left(w^2 + h^2 + 2wh\right) = \boxed{\textbf{(A) } 2(w+h)^2} \qquad$ .

Solution 2

The sheet of paper is made out of the surface area of the box plus the sum of the four triangles. The surface area is $2w^2 + 2wh + 2wh$ which equals $2w^2 + 4wh$ .The four triangles each have a height and a base of $h$ , so they each have an area of $\frac{h^2}{2}$ . There are four of them, so multiplied by four is $2h^2$ . Together, paper's area is $2w^2 + 4wh + 2h^2$ . This can be factored and written as $\boxed{\textbf{(A) } 2(w+h)^2} \qquad$ .

Solution 3

The sheet of paper is made out of $4$ squares. Each square has a side length of $\frac{w}{\sqrt{2}} + \frac{h}{\sqrt{2}}$ , which we get from the Pythagorean Theorem (a $45-45-90$ triangle's legs is the hypotenuse divided by $\sqrt2$ ). Thus, to find the area of the entire paper, we square our side length and multiply by 4. So, $4\cdot\left(\frac{w}{\sqrt{2}} + \frac{h}{\sqrt{2}}\right)^2 \Rightarrow \boxed{\textbf{(A) } 2(w+h)^2} \qquad$ , which is the answer.

Sol by IronicNinja

@@ Line 29: / Line 29: @@
 ==Solution 1==
-Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length of each dotted line is <math>h</math>. The area of the rectangle that is <math>w</math> by <math>h</math> is <math>wh</math>. The combined figure of the two triangles with base <math>h</math> is a square with <math>h</math> as its diagonal. Using the Pythagorean Theorem, each side of this square is <math>\sqrt{\frac{h^2}{2}}</math>. Thus, the area is the side length squared which is <math>\frac{h^2}{2}</math>. Similarly, the combined figure of the two triangles with base <math>w</math> is a square with area <math>\frac{w^2}{2}</math>. Adding all of these together, we get <math>\frac{w^2}{2} + \frac{h^2}{2} + wh</math>. Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting <math>4(\frac{w^2}{2} + \frac{h^2}{2} + wh) = 2(w^2 + h^2 + 2wh) = \boxed{\textbf{(A) } 2(w+h)^2} \qquad</math>.
+Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length of each dotted line is <math>h</math>. The area of the rectangle that is <math>w</math> by <math>h</math> is <math>wh</math>. The combined figure of the two triangles with base <math>h</math> is a square with <math>h</math> as its diagonal. Using the Pythagorean Theorem, each side of this square is <math>\sqrt{\frac{h^2}{2}}</math>. Thus, the area is the side length squared which is <math>\frac{h^2}{2}</math>. Similarly, the combined figure of the two triangles with base <math>w</math> is a square with area <math>\frac{w^2}{2}</math>. Adding all of these together, we get <math>\frac{w^2}{2} + \frac{h^2}{2} + wh</math>. Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting <math>4\left(\frac{w^2}{2} + \frac{h^2}{2} + wh\right) = 2\left(w^2 + h^2 + 2wh\right) = \boxed{\textbf{(A) } 2(w+h)^2} \qquad</math>.
 ==Solution 2==

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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 AMC 12 Problems and Solutions

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