Geometry Puzzle – Challenge 76

An equilateral triangle has perimeter 2x and area x (numerically). Each side is (2x)/3. Use the equilateral area formula A = s^2 sqrt(3)/4, set equal to x, and solve for x. Substitute back to get side length = 2 sqrt(3) meters.

Key takeaways:

  • Perimeter of equilateral triangle = 3 times side, so side s = (2x)/3.
  • Area of equilateral triangle = s^2 sqrt(3) / 4.
  • Set s^2 sqrt(3)/4 = x: substitute s = 2x/3 to get (4x^2/9)(sqrt(3)/4) = x.
  • Simplify: x^2 sqrt(3)/9 = x, so x = 9/sqrt(3) = 3 sqrt(3).
  • Side = 2x/3 = 2(3 sqrt(3))/3 = 2 sqrt(3) meters.

This is a perfect math challenge for those who enjoy solving complicated mathematics and critical thinking challenges. Let’s challenge your brain!

Geometry Puzzle – Challenge 76

Challenge:

If the perimeter of an equilateral triangle is 2x meters and its area is x square meters, then what is the length of one side of the triangle in meters?

A- \(\sqrt{3}\)

B- \(\frac{\sqrt{3}}{2}\)

C- \(2\sqrt{3}\)

D- \(\frac{2\sqrt{3}}{3}\)

E- 3

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The correct answer is C.

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The perimeter of the equilateral triangle is 2x meters. So, one side is \(\frac{2}{3}x \) meters.
The area of an equilateral triangle \(= \frac{s^2 \sqrt{3}}{4}\) (s is one side of the triangle)
The perimeter of the triangle is twice its area. So:
\(2x = 2 (\frac{s^2 \sqrt{3}}{4}) → 2x = (\frac{s^2 \sqrt{3}}{2})\)
Replace the s with \(\frac{2}{3}x\). Then:
\(2x = \frac{(\frac{2}{3} x)^2 \sqrt{3}}{2} = \frac{\frac{4}{9} x^2 \sqrt{3}}{2 }→ 4x = \frac{4}{9} x^2 \sqrt{3} → 4 = \frac{4}{9} x\sqrt{3} → 9 = x\sqrt{3}→
\frac{9}{\sqrt{3} }= x → \frac{9}{\sqrt{3} } × \frac{\sqrt{3}}{\sqrt{3} } = x → x = 3\sqrt{3}\)
Then, one side of the triangle is: \(\frac{2}{3}x =\frac{ 2}{3}(3\sqrt{3}) = 2\sqrt{3}\)

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Frequently Asked Questions

How do you solve math shape puzzles?

Start by writing the formulas for the shape, then turn the puzzle’s clues into an equation. For this equilateral triangle, the perimeter 2x means each side is 2x/3, and the area formula is s²√3/4. Setting the area equal to x gives (4x²/9)(√3/4) = x, so x = 9/√3 = 3√3, and each side is 2√3 meters.

How do you solve math puzzles step by step?

Name the unknown, write what each clue means, solve, then check. Here: side = 2x/3 from the perimeter, area = s²√3/4 from the equilateral formula, and area = x from the puzzle. Solve to get x = 3√3 and side = 2√3. Check it: the perimeter is 6√3 and the area is 12√3/4 = 3√3, so the perimeter is exactly twice the area.

What are challenging math puzzles for middle school?

Challenging middle school math puzzles combine two or more ideas, such as perimeter and area, in one short problem. This equilateral triangle puzzle is a good stretch for strong seventh and eighth graders. The perimeter is 2x and the area is x, and finding the side means working with square roots and simplifying 9/√3 to 3√3.

Are there 7th grade math puzzles in PDF with answers?

Yes, you can find printable 7th grade math puzzle sets with answer keys, and this triangle challenge prints neatly with its worked solution. For seventh graders still new to square roots, try a friendlier version first: if the perimeter is 3x and the area is x, each side is x, so x²√3/4 = x and x = 4/√3 = 4√3/3.

Related Lessons You May Like

If your student enjoys these puzzles, Geometry for Beginners works the same relationships inside a full curriculum. Pre-Algebra for Beginners covers the algebra foundations.

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