Comparing Areas

I've found a few misconceptions arise from questions like this.

 

A misconception that might come up is that if the shape doesn't have a known formula, then it doesn't have an area.

 

Staircase Regulations

I'm considering a loft extension and it requires a staircase, so I've done some research. I'm amazed at how much contradicatory information there is online and how many conditions there are in the building regulations for this.

Disclaimer (to avoid legal repurcussions) - The figures I've used in these questions are based on various online sources. They are roughly accurate based on what I've seen, to provide real-life questions for students, but they shouldn't be used as guidance for building staircases.

 

Solutions

\(tan(pitch)=\frac{rise}{tread}\)

\(tan^{-1}(\frac{17}{29})=30.4^{\circ}\)

 \(\frac{14}{tan(28)}=26.3 cm\)

Solutions

A - No. \(22+2(16)=54<55\)

B - Yes

C - No. \(t=\frac{15}{tan(40)}=17.9<24.5\)

D - Yes

E - Yes

F - No. \(\frac{18.5}{tan(35)}=26.4>26\)

G - No. The tread is 28>26

H - No. \(24.5+2(15)=54.5<55\)

Solution

\(tan^{-1}(\frac{2.9}{3.7})=38.1^{\circ}\) which is less than 42, so yes.

Consider bounds.

For the rise:
\(\frac{290}{15}=19.33\) and \(\frac{290}{22}=13.18\)

For the tread:
\(\frac{370}{24.5}=11.02\) and \(\frac{370}{26}=14.23\)

The only integer (as we need an integer number of steps) that satisfies both conditions is 14.

And finally, a Desmos graph to demonstrate the scenario.

Bricks in a Box

Solution

The brick is \(15\times5\times8=600 cm^3\)

The box is \(130\times80\times100=1 040 000 cm^3\)

A common misconception is then \(1 040 000\div 600=1733.33...\) so the box can hold \(1733\) because we have to round down. The problem there is that some bricks might need cutting up to fit in the remaining space.

Here's another question that serves as a prompt to reconsider the first part.

 

One brick doesn't even fit in the box, despite the volumes being \(40 cm^3\) and \(2000 cm^3\) suggesting that \(50\) boxes should fit.

So how do we tackle the first part?

This is an example of a packing problem and they get exceptionally difficult. Even with 11 squares in 2D, this is incredibly complicated, as Walter Trump showed in 1979 with his diagram below showing an optimal packing.

  

Our problem involves 3 dimensions and I'm almost certain I don't have the optimal packing.

We could arrange the bricks in different orientations, keeping faces of the bricks parallel to those of the box. For each orientation we can include a maximum number of rows, columns and layers. Here are those results.

LWH - 8, 16, 12 - 1536
LHW - 8, 10, 20 - 1600
WLH - 26, 5, 12 - 1560
WHL - 26, 10, 6 - 1560
HWL - 16, 16, 6 - 1536
HLW - 16, 5, 20 - 1600

The LHW orientation allows us to fit 8 rows of 15cm into the 130cm length, 10 columns of 8cm into the 80cm width and 20 layers of 5cm into the 100cm height, resulting in 1600 boxes. This is a new lower bound, but I suspect there are ways of stacking bricks diagonally once the box is nearing capacity to optimise the number.

 

 I'm going no further with this. I'll leave that as an exercise for you.

Ultimately, the point of this was to get students thinking about whether their calculations and the mathematical findings apply to the real-life situation. Sometimes we need to stop and look at the bigger picture.

Youtube's Pop-up Survey

While watching some videos on Youtube, this pop-up appeared asking for some feedback on the video recommendations.

Thoughts? Intentionally designed for specific data or data collecting oversight?

 

Pyramid Patterns

Maths is just patterns of patterns... of patterns.

For each of these pyramids, identify the rule and fill in the missing boxes. Make it a race to the top if you want to utilise peer-on-peer competition. Some might have shortcuts to get the top quickly.

Also available as a worksheet and PowerPoint document with answers here.

 

Rotating Cockpits in Vehicles

I stumbled across this video by Aren Khachatryan showing an innovative rotating cockpit idea to reduce the impact of head on collisions through circular redirection.

Which led to the following question for students. It lends itself nicely to modelling restrictions (the human being a particle for example) and Aren goes into much more detail here, showing the maths behind it, providing some super-curricular material for students to explore.

Solution

The initial speed is \(72 kmh^{-1} = 20 ms^{-1}\). Take directly below the centre to be \(\theta=0\).

By the convservation of mechanical energy, \(\frac{1}{2}(100)(20^2) =\frac{1}{2}(100)(v^2) + 100g(0.5 - 0.5cos\theta)\)

Resulting in \(v^2=400+gcos\theta -g\)

Then \(a=\frac{v^2}{r}=\frac{400+gcos\theta -g}{0.5}=800-2g+2gcos\theta\)

This has a maximum value of \(800\) when \(\theta=0\) which is roughly \(80 g_0\).

Note and further reading

I saw Aren Khachatryan's idea and thought it would make an interesting problem to look at. The figures in my solution are of a crude model to make it more accessible to students learning about circular motion. They are not a review of Aren's work. His work is much more in depth, taking into account decelerating centre, multiple bodies and modelling considerations. It makes for an interesting read.

While doing some background digging for this, I also came across a car crash calculator that has some explanations with accessible maths for A-Level students.

Types of Triangles