This problem solving activity has a measurement focus.

Does a square peg **fit better** into a round hole or does a round peg fit better into a square hole?

- Draw a diagram to represent a physical situation for a general case.
- Apply Pythagoras’ theorem to a general situation.
- Devise and use problem solving strategies to explore situations mathematically.

Students may be familiar with the phrase "you can't fit a square peg in a round hole".

This problem is a straightforward application of the formulae for the areas of a circle and a square.

### The Problem

Does a square peg fit better into a round hole or does a round peg fit better into a square hole?

### Teaching Sequence

- Get students to draw both shapes.
- Talk about what a "better fit" might mean. Suggest a relation with percentage of area left over.
- From their visual perceptions ask them to vote.

Square peg in round hole. Circle peg in square hole. Both the same - Students may need to work from a specific radius measurement, for example, 5 cm, if they have difficulty setting up the general case.
- Focus questions that can help students get started include:
*How can we set this up?**What information do we know?**How would we label the diagrams?**What mathematical knowledge/formulae could we apply to this situation?**How will we compare the cases?* - Encourage students to clearly justify their reasoning by writing a concluding statement to explain their answer.
- Shares answers.
- Reflect back on the original votes. Discuss the difference between visual factors that might make the square look like a better fit.

### Solution

Begin by agreeing on a definition for "fits better". This can be connected to the "waste space" after the peg is fitted. This "waste space" is best expressed as a percentage of the larger shape. This is because the answer is then a ratio that is independent of the size of the objects.

Apply this approach to each situation.

**Square peg in round hole.**

If the radius of the circle is r, then by Pythagoras’ Theorem, the side of the square is √(r^{2} + r^{2}) = √ (2r^{2}) = r √2.

Hence the area of the square is r √2 x r √2 = √ (2r^{2}) x √ (2r^{2}) = 2r^{2}.

The area of the circle is πr^{2}.

Students could look at the ratio area square/area circle = 2r^{2}/πr^{2} = 2/π.

As a percentage this is 200/π which is approximately 63.66%.

**Round peg in square hole**

Alternatively they may then subtract the square area from the circle area to find the wasted area.

2 (π - 2). The percentage wasted is π - 2/ π x 100 = 36.33%

The area of the circle is πr^{2}.

From the diagram, the side length of the square is 2r.

Hence the area of the square is (2r)^{2} = 4r^{2}.

Students could find the ratio area circle/area square = πr^{2}/4r^{2} = p/4. As a percentage this is 100π/4 = 78.54%.

The circle covers more of the square than the square does the circle. It’s therefore better to be a round peg in a square hole than a square peg in a round hole.

Alternatively they may then subtract the area of the circle from the square area to find the wasted area.

4 r^{2 }- πr^{2 }= r^{2 }(4 - π). The percentage wasted is then 4 - π/4 = 21.46%.