2025 AMC 8 Problems/Problem 8: Difference between revisions
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Since the figure is a cube, each of the six sides are equal, making the area of one of the faces <math>\frac{18}{6} = 3</math>, which makes the side length <math>\sqrt3</math>. Therefore, the volume of the cube is <math>\sqrt3^3 = \textbf{(A)}~3 \sqrt3</math> | Since the figure is a cube, each of the six sides are equal, making the area of one of the faces <math>\frac{18}{6} = 3</math>, which makes the side length <math>\sqrt3</math>. Therefore, the volume of the cube is <math>\sqrt3^3 = \textbf{(A)}~3 \sqrt3</math> | ||
~ | ~Sigmacuber | ||
==Vide Solution 1 by SpreadTheMathLove== | ==Vide Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
Revision as of 17:02, 1 February 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?
Solution
Since the figure is a cube, each of the six sides are equal, making the area of one of the faces 
, which makes the side length 
. Therefore, the volume of the cube is 
~Sigmacuber
Vide Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/VP7g-s8akMY?si=UuALQxA6xGUGW8hN&t=577 ~hsnacademy
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
