Geometry Puzzle – Challenge 73

Challenge:

The perimeter of a square is 13 cm and the perimeter of an equilateral triangle is 7 cm. What is the difference of a side of the square and a side of a triangle?

A- \(\frac{7}{13}\) cm

B- \(\frac{11}{12}\) cm

C- \(1\frac{1}{4}\) cm

D- \(2\frac{1}{4}\) cm

E- \(\frac{13}{7}\) cm

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The perimeter of a square is 13cm. Therefore, one side of the square is \(\frac{13}{4}\) cm.
The perimeter of the equilateral triangle is 7cm. Therefore, one side of the triangle is \(\frac{7}{3}\) cm.
The difference of a side of the square and a side of a triangle is:
\(\frac{13}{4} – \frac{7}{3} = \frac{39-28}{12} = \frac{11}{12}\) cm

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Frequently Asked Questions

How do I find one side of a square from its perimeter?

Divide by 4. A square has 4 equal sides, so side = perimeter / 4. For perimeter 13: side = 13/4 = 3.25 cm.

How do I find one side of an equilateral triangle from its perimeter?

Divide by 3. An equilateral triangle has 3 equal sides, so side = perimeter / 3. For perimeter 7: side = 7/3 cm.

How do I subtract 13/4 and 7/3?

Find a common denominator. LCM(4,3) = 12. So 13/4 = 39/12, 7/3 = 28/12, and 13/4 – 7/3 = 11/12.

Why use a common denominator?

Fractions can only be added or subtracted directly when they share the same denominator. Different denominators represent different size pieces.

Can I check by converting to decimals?

Yes. 13/4 = 3.25, 7/3 = 2.333…, difference = 0.9166… = 11/12.

How do I simplify 11/12?

11 and 12 share no common factor besides 1, so 11/12 is already in simplest form.

Why does the square have a larger side here?

Because 13/4 = 3.25 is larger than 7/3 ≈ 2.33. The square’s perimeter (13) is closer to 4 times the triangle’s perimeter (7), so each square side is larger than each triangle side.

What if both perimeters were equal?

Equal perimeters give different sides because the shapes have different side counts. With perimeter P: square side = P/4, triangle side = P/3. The triangle’s sides are longer.

Where do shapes with fractional sides appear?

Sewing patterns, packaging design, architectural blueprints, and any field that requires precise measurements smaller than a whole unit.

What grade can solve this?

Middle school (Grade 5-7), once students are comfortable with fraction subtraction and basic perimeter formulas.

Related Lessons You May Like

• How to find the area of rectangles
• How to find the perimeter of rectangles
• How to find the area of triangles
• How to solve systems of equations
• How to solve multi-step word problems

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.