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

#### Ellipses

The area of an ellipse is

*πab*where*a*and*b*are the distances from the centre of the ellipse to the closest and furthest points on the ellipse.In the first ellipse,

*a*=5cm and*b*=4cm, so the area is 20*π*cm^{2}. In the second ellipse,*a*=5cm and*b*=3cm, so the area is 15*π*cm^{2}. Hence,**the first ellipse has the larger area**.##### Extension

How far apart should the pins be placed to give the ellipse with the largest area?

#### Triangle Numbers

T

T

= (

_{n}=^{1}/_{2}*n*(*n*+1), so:T

_{n}+T_{n+1}=^{1}/_{2}*n*(*n*+1) +^{1}/_{2}(*n*+1)(*n*+2)= (

*n*+1)^{2}So, we are looking for

*n*such that (*n*+1)^{2}+(*n*+3)^{2}=(*n*+5)^{2}. This is true when**(6***n*=5^{2}+8^{2}=10^{2}).##### Extension

Find

*n*such that T_{n}+T_{n+1}+T_{n+1}+T_{n+2}=T_{n+2}+T_{n+3}.#### Ticking Clock

The second hand will always be pointing at one of the 60 graduations. If the minute and hour hand are 120° away from the second hand they must also be pointing at one of the graduations. The minute hand will only be pointing at a graduation at zero seconds past the minute, so the second hand must be pointing at 0. Therefore the hand are either pointing at: hour: 4, minute: 8, second: 0; or hour: 8, minute: 4, second: 0. Neither of these are real times, so

**it is not possible**.##### Extension

If the second hand moves continuously instead of moving every secone, will there be a time when the hands of the clock are all 120° apart?

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