Checking for Consistent Units

If students don't seem to take the "make sure you state the units" or "check that units are consistent" seriously, then these might provide a nice reminder of how important they can be. Here are five examples where inconsistent units caused issues. Search online for more information - some of them have interesting anecdotes to share with students.

 

Interactive Cobweb or Staircase Diagrams

An interactive cobweb or staircase diagram showing the iterative process behind finding a root of an equation.

Load up this Desmos graph, change the formula, initial value and number of iterations using the sliders to match the question you're looking at.

Equal Pieces

Following on from the earlier Cake problem...

Solution

Focus on the triangle with base length \(y\). It has a perpendicular height of \(\frac{1}{2}(x+y)\) therefore an area of \(\frac{y(x+y)}{4}\).

The entire cake has an area of \(\frac{3}{4}(x+y)^2\) so this triangle is a quarter of that.

\(\frac{y(x+y)}{4}=\frac{1}{4} \times \frac{3}{4}(x+y)^2\)

\(y(x+y)=\frac{3}{4}(x+y)^2\)

\(y=3x\)

So the ratio is 1:3.
 

 

Crawl Through a Hole in a Sheet of A4 Paper

One of my favourite ice-breakers that involves thinking outside the box.

 Typically, students see it as a 2-dimensional problem. This video by The Laughing Clock serves as a nice prompt think more broadly.

Below is an image which can be printed out. Cutting along each line forms a "snake" with a line running along it. Carefully opening it up creates a hole larger than the sheet you started with.

With a thinner snake, it starts to look more like lined paper. It also becomes incredibly delicate.

The hole in the paper has a perimeter double the length of the line on the snake (the line in red) and given that we are only limited by the number of lines we draw, we can theoretically make a hole as big as we want. Realistically, the thin paper makes it very easy to tear, but it's interesting to think that we can create a hole in something that's bigger than the original object.

Congruent Cake Slices

Here's a plan (top down view) of a cake. The challenge is to cut the cake into congruent pieces. Two and three pieces are pretty easy and may serve as a nice warm up, but then four becomes more challenging.

Here's a solution. There may be others.

An interesting finding while trying to solve the problem is below. This cuts the cake into four pieces of equal size, but they are not congruent. One of my students did, however, point out that it looks 3 dimensional - four congruent rectangles arranged in an L shape - which I thought was a clever loophole of sorts. 

 

Overlapping Circles

A nice little investigation involving Pythagoras' Theorem and geometry. I like how this one diagram, starting with two overlapping circles, contains the square roots of 1 to 5.

The next question is whether \(\sqrt{6}\) can be found in the diagram without drawing too many additional lines.

A 3:4:5 Triangle from a Square

Solution

Considering we will be using midpoints, I'm going to say the length of side AB is \(2n\).

\(BY^2=CY^2+BC^2=n^2+(2n)^2=5n^2\implies BY=\sqrt{5}n\)

Given that BX=BY (which can be proved in the same way) we can find angle XBA, which I will temporarily call \(\alpha\).

\(sin(\alpha)=\frac{AX}{XB}=\frac{n}{\sqrt{5}n}=\frac{1}{\sqrt{5}}\)

Now, let's use basic trigonometry on the triangle BYZ, with angle ZYB being \(2\alpha\).

\(YZ=BY cos(2\alpha)=\sqrt{5}n (1-2{sin}^2 (\alpha))=\sqrt{5}n (1-\frac{2}{5})=\frac{3\sqrt{5}n}{5}\)

That's two sides of the right-angled triangle, so use Pythagoras' Theorem to find BZ.

\(BZ^2=BY^2-ZY^2=(\sqrt{5}n)^2-({\frac{3\sqrt{5}n}{5}})^2=5n^2-\frac{9n^2}{5}=\frac{16n^2}{5}\)

Therefore \(BZ=\frac{4n}{\sqrt{5}}\) and the ratio YZ:ZB:BY is \(\frac{3n}{\sqrt{5}}:\sqrt{5}n:\frac{4n}{\sqrt{5}}\) which when multiplied through by \(\frac{\sqrt{5}}{n}\) gives \(3:5:4\)

The triangle is therefore a 3:4:5 triangle.