Difference between revisions of "2025 AMC 8 Problems/Problem 8"
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==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? | 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? | ||
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| + | [[File:Amc8_2025_prob8.PNG]] | ||
<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> | <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== | ==Solution== | ||
| − | + | Since the figure is a cube, each of the six sides are equal, making the area of one of the faces 18/6 = 3, which makes the side length √3. Therefore, the volume of the cube is (√3)^3 = (A): 3√3 | |
==Vide Solution 1 by SpreadTheMathLove== | ==Vide Solution 1 by SpreadTheMathLove== | ||
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==Video Solution by Thinking Feet== | ==Video Solution by Thinking Feet== | ||
https://youtu.be/PKMpTS6b988 | https://youtu.be/PKMpTS6b988 | ||
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| + | ==See Also== | ||
| + | {{AMC8 box|year=2025|num-b=7|num-a=9}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:17, 31 January 2025
Contents
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?
Solution
Since the figure is a cube, each of the six sides are equal, making the area of one of the faces 18/6 = 3, which makes the side length √3. Therefore, the volume of the cube is (√3)^3 = (A): 3√3
Vide Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution by Thinking Feet
See Also
| 2025 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
