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. 

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.

Analytical vs Numerical

Here are two questions that look similar, but require different methods for solving.

The first, a problem involving areas, perimeters and some basic trigonometry. 

 The second involving equations and Trial and Improvement.

Solutions

Problem 1

The perimeter is \(30\), which means the area is also \(30\). The perpendicular height is the area divided by base width, \(30 \div 10 = 3\). The acute angle is \(x\) where \(sin(x) = \frac{3}{4}\) so \(x=\sin^{-1}(\frac{3}{4})=36.87^{\circ}\) resulting in the obtuse angle \(180-36.87=143.13^{\circ}\)

Problem 2

The perpendicular height of the parallelogram is \(xsin(x)\) so the area is \(10xsin(x)\). The perimeter is \(2x+20\) so we can form the equation \[2x+20=10xsin(x)\]Divide through by 2 and rearrange to get \[5xsin(x)-x=10\] One of the \(x\)'s is trapped inside the sine function and the other two are not. We don't have an analytical solution for this and have to resort to numerical methods. Through some trial and improvement, you can find the angle to be \(18.09^{\circ}\)

Recurring Decimals Link Up

From what I've seen, recurring decimal questions seem to focus on two skills:

• convert between decimal and fractional form
• compare their size with other numbers

But why do we not explore them in greater detail? Recurring decimals are numbers - they can be used anywhere that numbers appear!

This worksheet has 10 questions which include recurring decimals in other topics. I've found that the additional time working with recurring decimals helps students to appreciate that they are just numbers and gain more confidence with the topic. By working on harder questions, students begin to see the "standard exam questions" as being relatively simple in comparison to these cross-topic questions.

Multiply the lengths to find the volume\[3.\dot{5}\times2.25\times1.2=3\frac{5}{9}\times\frac{9}{4}\times\frac{6}{5}=\frac{32}{9}\times\frac{9}{4}\times\frac{6}{5}=\frac{48}{5}\]Convert the density to an improper fraction\[8.7=8\frac{7}{9}=\frac{79}{9}\]Multiply the volume and density to find the mass\[\frac{48}{5}\times\frac{79}{9}=\frac{3792}{45}\]We can convert that the a recurring decimal to give us a mass in grams.\[3792\div45=82.4\dot{6}\]

Circle Theorems - Link Up

What can Circle Theorems be merged with to form AO3 questions? This worksheet has 19 questions which merge different topics, such as Probability, Sequences and Bounds with circle theorems. There's also written solutions in case you want let students self-assess.

From experience of using this resource, it's great for developing problem solving skills, recalling knowledge from previously studied content and it's a useful diagnostic tool for teachers as it highlights what topics students find most challenging.

 Here's a question not found on the worksheet, combining Circle Theorems with angles in polygons.