Difference between revisions of "2025 AMC 8 Problems/Problem 8"

Latest revision as of 18:42, 9 March 2025

Problem

Isaiah cuts open a cardboard cube along some of its edges to form the flat shape shown on the right, which has an area of $18$ square centimeters. What is the volume of the cube in cubic centimeters?

$\textbf{(A)}~3\sqrt{3} \qquad \textbf{(B)}~6 \qquad \textbf{(C)}~9 \qquad \textbf{(D)}~6\sqrt{3} \qquad \textbf{(E)}~9\sqrt{3}$

Solution 1

Each of the $6$ faces of the cube have equal area, so the area of each face is equal to $\frac{18}{6} = 3$ , making the side length $\sqrt3$ . From this, we can see that the volume of the cube is $\sqrt{3}^3 = \boxed{\textbf{(A)}~3\sqrt{3}}$

~Sigmacuber

Solution 2

Let the side length of the cube shown be equal to $s$ centimeters. The surface area of this cube is equal to the area of the net of the cube, which is equal to $18$ square centimeters. The surface area of this cube is also $6s^2$ square centimeters, so we have $6s^2 = 18 \implies s^2 = \frac{18}{6} = 3 \implies s = \sqrt{3}.$ However, we aren't done. We have found that the side length of the cube is $\sqrt{3} cm$ , but the question asks for the volume of the cube, which is equal to $s^3 = \sqrt{3}^3 = \boxed{\textbf{(A)}~3\sqrt{3}}$ cubic centimeters.

~aoum

Video Solution 1 by Cool Math Problems

https://youtu.be/BRnILzqVwHk?si=h0-bM3iwNbCG0cm-&t=253

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution 3

~hsnacademy

Video Solution 4 by Thinking Feet

https://youtu.be/PKMpTS6b988

Video Solution(Quick, fast, easy!)

https://youtu.be/fdG7EDW_7xk

~MC

@@ Line 1: / Line 1: @@
-==Problem==
+== Problem ==
-Isaiah cuts open a cardboard cube along some of its edges to form the flat shape shown on the right, which has an area of 18 square centimeters. What is the volume of the cube in cubic centimeters?
-<math>\textbf{(A)}~3\sqrt{3}\qquad\textbf{(B)}~6\qquad\textbf{(C)}~9\qquad\textbf{(D)}~6\sqrt{3}\qquad\textbf{(E)}~9\sqrt{3}</math>
+Isaiah cuts open a cardboard cube along some of its edges to form the flat shape shown on the right, which has an area of <math>18</math> square centimeters. What is the volume of the cube in cubic centimeters?
-==Solution==
+[[File:Amc8_2025_prob8.PNG]]
-Observe that since the six squares on the net are congruent, each of them has area <math>\frac{18}{6}=3</math>. Hence, the side length of the cube is <math>\sqrt{3}</math>. Its volume is then <math>(\sqrt{3})^3=\boxed{\textbf{(A)}~3\sqrt{3}}</math>. ~cxsmi
+<math>\textbf{(A)}~3\sqrt{3} \qquad \textbf{(B)}~6 \qquad \textbf{(C)}~9 \qquad \textbf{(D)}~6\sqrt{3} \qquad \textbf{(E)}~9\sqrt{3}</math>
+== Solution 1 ==
+Each of the <math>6</math> faces of the cube have equal area, so the area of each face is equal to <math>\frac{18}{6} = 3</math>, making the side length <math>\sqrt3</math>. From this, we can see that the volume of the cube is <math>\sqrt{3}^3 = \boxed{\textbf{(A)}~3\sqrt{3}}</math>
+~Sigmacuber
+== Solution 2 ==
+Let the side length of the cube shown be equal to <math>s</math> centimeters. The surface area of this cube is equal to the area of the net of the cube, which is equal to <math>18</math> square centimeters. The surface area of this cube is also <math>6s^2</math> square centimeters, so we have
+<cmath>6s^2 = 18 \implies s^2 = \frac{18}{6} = 3 \implies s = \sqrt{3}.</cmath>
+However, we aren't done. We have found that the side length of the cube is <math>\sqrt{3} cm</math>, but the question asks for the volume of the cube, which is equal to <math>s^3 = \sqrt{3}^3 = \boxed{\textbf{(A)}~3\sqrt{3}}</math> cubic centimeters.
+~[[User:aoum|aoum]]
+== Video Solution 1 by Cool Math Problems ==
+https://youtu.be/BRnILzqVwHk?si=h0-bM3iwNbCG0cm-&t=253
+== Video Solution 2 by SpreadTheMathLove ==
+https://www.youtube.com/watch?v=jTTcscvcQmI
+== Video Solution 3 ==
+[//youtu.be/VP7g-s8akMY?si=UuALQxA6xGUGW8hN&t=577 ~hsnacademy]
+== Video Solution 4 by Thinking Feet ==
+https://youtu.be/PKMpTS6b988
+==Video Solution(Quick, fast, easy!)==
+https://youtu.be/fdG7EDW_7xk
+~MC
+== See Also ==
+{{AMC8 box|year=2025|num-b=7|num-a=9}}
+{{MAA Notice}}
+[[Category:Introductory Geometry Problems]]

2025 AMC 8 (Problems • Answer Key • Resources)
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