Area and volume

Area and volume > Enlargements

1234Enlargements

Solutions to the exercises

Exercise 1

The enlargement factor in length is $20$ , so that the factor for the surface must be $20^{2} = 400$ and that for the volume $20^{3} = 8000$ .
The surface of the sculpture will be $1400 \cdot 400 = 560000$ cm² which is $56$ m³.
The volume of the sculpture will be $3000 \cdot 8000 = 24000000$ cm³ which is $24$ m³.

Exercise 2

$\sqrt[3]{5} \approx 1,71$ times.

If the metal sheet is equally thick then you just increase the surface, so the enlargement factor would be ${(\sqrt[3]{5})}^{2} \approx 2,92$ .
If the metal sheet was also getting thicker then the volume increases, so the enlargement factor would be $5$ .

Exercise 3

$\frac{1}{10}$ of your height.

$\frac{1}{1000}$ of your weight.

$\frac{1}{100}$ times less.

The energy requirement of humans varies greatly between individuals. There is also a difference between men and women. Assuming you eat about $500$ gram of food per day and that you weigh $80$ kg. This would mean that you eat the equivalent of $0,625$ % of your body weight every day.
Your Lilliputian twin would then have to eat $\frac{1}{100} \cdot 500 = 5$ grams per day.

See the answer to the previous question. The Lilliputian weighs $\frac{1}{1000} \cdot 80 = 0,08$ kg, or $80$ grams. He needs to eat the equivalent of $6,25$ % of his body weight every day, so by comparison that is $10$ times as much as you!

Because the energy requirement is proportional to the surface area of the body, and the enlargement factor for surface is $l^{2}$ .

The food requirement is proportional to $l^{2}$ and the weight is proportional to $l^{3}$ , so the food requirement per kg weight is proportional to $l^{(\frac{2}{3})}$

Exercise 4

The part of the cone that sticks out of the cube is $\frac{1}{4}$ part of the whole cone.
The enlargement factor of the the height of the cone that sticks out compared to the height $h$ of the whole cone is therefore $\sqrt[3]{0,25} \approx 0,63$ .
Since the cube has edges of $6$ cm, you get $0,37 h = 6$ and therefore $h \approx 16,2$ cm.

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