Geometry Puzzle – Challenge 75
A cube has surface area 3a and volume 2a. Find the total edge length (a cube has 12 edges). Set 6x^2 = 3a and x^3 = 2a, solve to get x = 4, and total edge length = 12 times 4 = 48 feet.
Key takeaways:
- Cube: 6 faces (each x^2), 12 edges (each x), volume x^3.
- Surface area equation: 6x^2 = 3a, so a = 2x^2.
- Volume equation: x^3 = 2a = 2(2x^2) = 4x^2.
- Solve x^3 = 4x^2: divide by x^2, x = 4.
- Total edge length = 12 times 4 = 48.
Here you find a math puzzle to develop your child’s math skills and geometry facts in a fun way.
Challenge:
If the total surface area of a cube is 3a square feet and the volume of the cube is 2a cubic feet, then what is the total length, in feet, of all the cube’s edges?
A- 12
B- 24
C- 48
D- 64
E- 96
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The correct answer is C.
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A cube has 12 edges. Let x be the edge of the cube. So:
Surface area of the cube = 6 (one side)\(^2 = 6x^2 =3a\)
\(x^2 = \frac{1}{2} a → x = \sqrt{\frac{1}{2 }a} = (\frac{1}{2} a)^{\frac{1}{2}}\)
Volume of the cube = (one side)\(^3 = x^3 =2a → x = \sqrt[3]{2a} = (2a)^{\frac{1}{3}}\)
Solve for a for both equations:
\((\frac{1}{2} a)^{\frac{1}{2}} = (2a)^{\frac{1}{3}}\)
Both sides to the power of 6:
\((\frac{1}{2} a)^3 = (2a)^2 → \frac{1}{8} a^3 = 4a^2
→ \frac{1}{8} a = 4 → a = 32\)
Therefore, the volume of the cube is 64 cubic feet and one side is 4.
The total length of all the cube’s edges is 48 feet. (4 × 12 = 48)The Best Books to Ace Algebra
Frequently Asked Questions
How do you solve math puzzles step by step?
Translate each clue into an equation, combine them, and check the result. For this cube, the surface area clue 6x² = 3a gives a = 2x². The volume clue x³ = 2a then becomes x³ = 4x², so x = 4 feet. A cube has 12 edges, giving a total edge length of 48 feet. Check: surface area 96 = 3 × 32 and volume 64 = 2 × 32.
Is mathematics a puzzle?
Mathematics is broader than puzzles, but it often works like one: a few clues fit together to give one answer. In this cube puzzle, two clues share an unknown a. Surface area is 3a and volume is 2a. Neither clue alone tells you the edge, but together they force x = 4, which is what makes the problem satisfying to solve.
What are challenging math puzzles for 7th graders?
Good challenges for seventh graders use familiar formulas in an unfamiliar way. This cube puzzle is a strong example because students already know surface area 6x² and volume x³, but here they must link them through a shared value a. Solving x³ = 4x² and remembering that a cube has 12 edges leads to a total length of 48 feet.
Are there printable 7th grade math puzzles?
Yes, many 7th grade math puzzles are available to print, and geometry puzzles like this cube challenge work well on paper. Encourage your student to draw the cube and label one edge x before writing any equations. A quick variation to try next: if the surface area is 6a and the volume is a, the edge is 1 and the total edge length is 12.
Are there challenging math puzzles in PDF?
Yes, challenging math puzzle collections are easy to find as PDFs, and this one is worth printing with its answer covered. Have your student solve it two ways. The quick way substitutes a = 2x² into x³ = 2a. The longer way finds a = 32 first, which gives a volume of 64 cubic feet and an edge of 4 feet. Both lead to 48.
Related Lessons You May Like
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If your student enjoys puzzles like this, Geometry for Beginners works the same relationships inside a full curriculum. Pre-Algebra for Beginners covers the algebraic foundations gently.
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