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

#### Circles

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**.##### Extension

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 10cm

^{2}and looks like:##### Extension

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 \(\frac{1}{4}\)

##### Extension

What is the difference between the shaded and the hatched regions in this dodecagon?

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:

ReplyDeletehttp://seekecho.blogspot.fr/2014/05/the-proof-see.html