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?
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.
Does a square peg fit better into a round hole or does a round peg fit better into a square hole?
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.
If the radius of the circle is r, then by Pythagoras’ Theorem, the side of the square is √(r2 + r2) = √ (2r2) = r √2.
Hence the area of the square is r √2 x r √2 = √ (2r2) x √ (2r2) = 2r2.
The area of the circle is πr2.
Students could look at the ratio area square/area circle = 2r2/πr2 = 2/π.
As a percentage this is 200/π which is approximately 63.66%.
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 πr2.
From the diagram, the side length of the square is 2r.
Hence the area of the square is (2r)2 = 4r2.
Students could find the ratio area circle/area square = πr2/4r2 = 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 r2 - πr2 = r2 (4 - π). The percentage wasted is then 4 - π/4 = 21.46%.
Printed from https://nzmaths.co.nz/resource/square-pegs-round-holes at 9:02pm on the 26th April 2024