Answer :
To solve the second part of the question given, we'll need to understand the relationship between similar shapes.
Similar shapes are shapes that have the same form, but not necessarily the same size. For shapes that are similar, the ratio of their corresponding linear dimensions are equal.
Since the two tins are similar and their corresponding linear dimensions are proportional, the ratio of their areas is equal to the square of the ratio of their corresponding linear dimensions. Moreover, because tins are three-dimensional objects and their volumes are directly proportional to the mass of fat they contain (assuming uniform density), the square of the ratio of their linear dimensions will be equal to the ratio of their masses.
Given that the masses of the fats in the smaller and larger tins are 500 g and 2000 g respectively, and that the base area of the smaller tin is 100 cm², we will find the area of the base of the larger tin.
First, calculate the ratio of the masses of the fats:
[tex]\[ \text{Ratio of masses} = \frac{\text{Mass of fat in larger tin}}{\text{Mass of fat in smaller tin}} \][/tex]
[tex]\[ \text{Ratio of masses} = \frac{2000}{500} \][/tex]
[tex]\[ \text{Ratio of masses} = 4 \][/tex]
Since the ratio of the areas is equal to the square of the ratio of their corresponding linear dimensions, and in this case, the ratio of the masses reflects the cubic ratio of their dimensions (because we are dealing with volume for the mass of fat), the ratio of the areas is actually equal to the ratio of the masses (since the question implicitly states that the only dimension that changes in the similar tins is the one related to the base area, not the height).
Thus, we have:
[tex]\[ \text{Area ratio} = \text{Ratio of masses} \][/tex]
Now we can calculate the area of the base of the larger tin by multiplying the ratio of the masses by the area of the base of the smaller tin:
[tex]\[ \text{Area of the base of the larger tin} = \text{Area ratio} \times \text{Area of the base of the smaller tin} \][/tex]
[tex]\[ \text{Area of the base of the larger tin} = 4 \times 100 \text{ cm}^2 \][/tex]
[tex]\[ \text{Area of the base of the larger tin} = 400 \text{ cm}^2 \][/tex]
So, the area of the base of the larger tin is 400 cm².
For the first part of the question, it appears that the question is incomplete. The capacity of the smaller pail would be found using information about the dimensions of the pail or another relevant method, but that information is not provided.
Similar shapes are shapes that have the same form, but not necessarily the same size. For shapes that are similar, the ratio of their corresponding linear dimensions are equal.
Since the two tins are similar and their corresponding linear dimensions are proportional, the ratio of their areas is equal to the square of the ratio of their corresponding linear dimensions. Moreover, because tins are three-dimensional objects and their volumes are directly proportional to the mass of fat they contain (assuming uniform density), the square of the ratio of their linear dimensions will be equal to the ratio of their masses.
Given that the masses of the fats in the smaller and larger tins are 500 g and 2000 g respectively, and that the base area of the smaller tin is 100 cm², we will find the area of the base of the larger tin.
First, calculate the ratio of the masses of the fats:
[tex]\[ \text{Ratio of masses} = \frac{\text{Mass of fat in larger tin}}{\text{Mass of fat in smaller tin}} \][/tex]
[tex]\[ \text{Ratio of masses} = \frac{2000}{500} \][/tex]
[tex]\[ \text{Ratio of masses} = 4 \][/tex]
Since the ratio of the areas is equal to the square of the ratio of their corresponding linear dimensions, and in this case, the ratio of the masses reflects the cubic ratio of their dimensions (because we are dealing with volume for the mass of fat), the ratio of the areas is actually equal to the ratio of the masses (since the question implicitly states that the only dimension that changes in the similar tins is the one related to the base area, not the height).
Thus, we have:
[tex]\[ \text{Area ratio} = \text{Ratio of masses} \][/tex]
Now we can calculate the area of the base of the larger tin by multiplying the ratio of the masses by the area of the base of the smaller tin:
[tex]\[ \text{Area of the base of the larger tin} = \text{Area ratio} \times \text{Area of the base of the smaller tin} \][/tex]
[tex]\[ \text{Area of the base of the larger tin} = 4 \times 100 \text{ cm}^2 \][/tex]
[tex]\[ \text{Area of the base of the larger tin} = 400 \text{ cm}^2 \][/tex]
So, the area of the base of the larger tin is 400 cm².
For the first part of the question, it appears that the question is incomplete. The capacity of the smaller pail would be found using information about the dimensions of the pail or another relevant method, but that information is not provided.
