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# Chocolate Cake

##### Age 11 to 14 Challenge Level:

The recipe uses a tin of radius 10cm and depth 7.5cm. This has a volume of

$$\pi\times10^2\times7.5 cm^3 = 2400 cm^3$$

We don't know the depth of the 23cm round tin.  If its depth is also 7.5cm, then its volume is given by:

$$\pi\times11.5^2\times7.5 cm^3 = 3100 cm^3$$

which would be fine.

The limit for the depth of this tin could be found by trial and error, or you could rearrange the formula for the volume of a cylindrical tin to find the height which gives a volume of 2400 cm3.

$$\pi\times11.5^2\times h cm^3 = 2400 cm^3$$ $$h = \frac{2400}{\pi\times11.5^2} cm = 5.8 cm$$

So depending on the depth of the 23cm round tin, all could be well and Toby gets his cake!

The volume of the square tin is $15^2\times6cm^3=1350cm^3$, which isn't large enough.  For a large enough square tin of the same depth, we need:

$$l^2\times 6 cm^3 = 2400 cm^3$$
$$l = \sqrt{\frac{2400}{6}} cm = 20 cm$$

where l is the length of the side of the tin.

If you think of the diagonal of the square tin as being equivalent to the diameter of a round tin, the length of the diagonal, d, is:

$$d^2 = 2\times l^2$$
$$d = \sqrt{2\times20^2} = 28.3 cm$$

so quite a bit longer than the diameter of the 20cm round tin!