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Monday, 12 May 2014

Sunday Afternoon Maths XII Answers & Extensions

This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.


Let \(4x\) be the side length of the square. This means that the radius of the red circle is \(2x\) and the radius of a blue circle is \(x\). Therefore the area of the red circle is \(4\pi x^2\).
The area of one of the blue squares is \(\pi x^2\) so the blue area is \(4\pi x^2\). Therefore the two areas are the same.
Is the red or blue area larger?

Largest Triangle

As our shape is a triangle, the 4cm and 5cm sides must be adjacent. Call the angle between them be \(\theta\).
The area of the triangle is \(\frac{1}{2}\times 4\times 5 \times \sin{\theta}\) or \(10\sin{\theta}\). This has a maximum value when \(\theta=90^\circ\), so the largest triangle has and area of 10cm2 and looks like:
What is the largest area triangle with a perimeter of 12cm?

Unit Octagon

Name the regions as follows:
\(E\) is a 1×1 square. Placed together, \(A\), \(C\), \(G\) and \(I\) also make a 1×1 square. \(B\) is equal to \(H\) and \(D\) is equal to \(F\).
Therefore \(B+E+F=A+C+D+G+H+I\). Therefore The hatched region is \(C\) larger than the shaded region. The area of \(C\) (and therefore the difference) is \(\frac{1}{4}\)
What is the difference between the shaded and the hatched regions in this dodecagon?

1 comment:

  1. Those red and blue circles - in a way we should know it instantly, without pi or x... because the area of any shape scales up by 9 when its length goes up by 3. Just been thinking about this in relation to Pythagoras theorem: