Answer :
Let's analyze and compare the two rectangular solids as given in the question:
### Given Information:
- Two rectangular solids both have square bases with the same side length, so the side lengths of the bases are the same but the heights are different.
- The first rectangular solid has twice the height of the second rectangular solid.
### Statements:
1. The bases are congruent.
- Since both solids have square bases with the same side length, their bases are congruent.
2. The solids are similar.
- Two solids are similar if their corresponding dimensions are proportional. Because the height of the first solid is twice the height of the second solid, their dimensions are proportional, making the two solids similar.
3. The ratio of the volumes of the first solid to the second solid is 8:1.
- To find the volume ratio, let's compare the volumes:
- Let the side length of the base be [tex]\( x \)[/tex] and the height of the second solid be [tex]\( h \)[/tex].
- Volume of the second solid = [tex]\( x^2 \times h \)[/tex].
- Volume of the first solid (with twice the height) = [tex]\( x^2 \times 2h \)[/tex].
- Therefore, the ratio of the volumes is [tex]\(\frac{x^2 \times 2h}{x^2 \times h} = \frac{2}{1}\)[/tex], not 8:1.
4. The volume of the first solid is twice as much as the volume of the second solid.
- From the volume calculations above:
- Volume of the first solid = [tex]\( x^2 \times 2h \)[/tex].
- Volume of the second solid = [tex]\( x^2 \times h \)[/tex].
- So, the volume of the first solid is exactly twice that of the second solid.
5. If the dimensions of the second solid are [tex]\( x \)[/tex] by [tex]\( x \)[/tex] by [tex]\( h \)[/tex], the first solid has [tex]\( 4xh \)[/tex] more surface area than the second solid.
- Surface area of the second solid:
- Two bases: [tex]\( 2(x^2) \)[/tex]
- Four sides: [tex]\( 4(xh) \)[/tex]
- Total surface area = [tex]\( 2(x^2) + 4(xh) \)[/tex]
- Surface area of the first solid:
- Two bases: [tex]\( 2(x^2) \)[/tex]
- Four sides: [tex]\( 4(x \times 2h) = 8(xh) \)[/tex]
- Total surface area = [tex]\( 2(x^2) + 8(xh) \)[/tex]
- Difference in surface area between the first and the second solid = [tex]\( [2(x^2) + 8(xh)] - [2(x^2) + 4(xh)] = 4(xh) \)[/tex]
- Therefore, the difference in surface area is [tex]\( 4xh \)[/tex].
### Conclusion:
Given the explanations above, the following statements are true:
- The bases are congruent.
- The solids are similar.
- The volume of the first solid is twice as much as the volume of the second solid.
- If the dimensions of the second solid are [tex]\( x \)[/tex] by [tex]\( x \)[/tex] by [tex]\( h \)[/tex], the first solid has [tex]\( 4xh \)[/tex] more surface area than the second solid.
The only statement that is not true is:
- The ratio of the volumes of the first solid to the second solid is 8:1.
Therefore, the true statements are:
- The bases are congruent.
- The solids are similar.
- The volume of the first solid is twice as much as the volume of the second solid.
- If the dimensions of the second solid are [tex]\( x \)[/tex] by [tex]\( x \)[/tex] by [tex]\( h \)[/tex], the first solid has [tex]\( 4xh \)[/tex] more surface area than the second solid.
### Given Information:
- Two rectangular solids both have square bases with the same side length, so the side lengths of the bases are the same but the heights are different.
- The first rectangular solid has twice the height of the second rectangular solid.
### Statements:
1. The bases are congruent.
- Since both solids have square bases with the same side length, their bases are congruent.
2. The solids are similar.
- Two solids are similar if their corresponding dimensions are proportional. Because the height of the first solid is twice the height of the second solid, their dimensions are proportional, making the two solids similar.
3. The ratio of the volumes of the first solid to the second solid is 8:1.
- To find the volume ratio, let's compare the volumes:
- Let the side length of the base be [tex]\( x \)[/tex] and the height of the second solid be [tex]\( h \)[/tex].
- Volume of the second solid = [tex]\( x^2 \times h \)[/tex].
- Volume of the first solid (with twice the height) = [tex]\( x^2 \times 2h \)[/tex].
- Therefore, the ratio of the volumes is [tex]\(\frac{x^2 \times 2h}{x^2 \times h} = \frac{2}{1}\)[/tex], not 8:1.
4. The volume of the first solid is twice as much as the volume of the second solid.
- From the volume calculations above:
- Volume of the first solid = [tex]\( x^2 \times 2h \)[/tex].
- Volume of the second solid = [tex]\( x^2 \times h \)[/tex].
- So, the volume of the first solid is exactly twice that of the second solid.
5. If the dimensions of the second solid are [tex]\( x \)[/tex] by [tex]\( x \)[/tex] by [tex]\( h \)[/tex], the first solid has [tex]\( 4xh \)[/tex] more surface area than the second solid.
- Surface area of the second solid:
- Two bases: [tex]\( 2(x^2) \)[/tex]
- Four sides: [tex]\( 4(xh) \)[/tex]
- Total surface area = [tex]\( 2(x^2) + 4(xh) \)[/tex]
- Surface area of the first solid:
- Two bases: [tex]\( 2(x^2) \)[/tex]
- Four sides: [tex]\( 4(x \times 2h) = 8(xh) \)[/tex]
- Total surface area = [tex]\( 2(x^2) + 8(xh) \)[/tex]
- Difference in surface area between the first and the second solid = [tex]\( [2(x^2) + 8(xh)] - [2(x^2) + 4(xh)] = 4(xh) \)[/tex]
- Therefore, the difference in surface area is [tex]\( 4xh \)[/tex].
### Conclusion:
Given the explanations above, the following statements are true:
- The bases are congruent.
- The solids are similar.
- The volume of the first solid is twice as much as the volume of the second solid.
- If the dimensions of the second solid are [tex]\( x \)[/tex] by [tex]\( x \)[/tex] by [tex]\( h \)[/tex], the first solid has [tex]\( 4xh \)[/tex] more surface area than the second solid.
The only statement that is not true is:
- The ratio of the volumes of the first solid to the second solid is 8:1.
Therefore, the true statements are:
- The bases are congruent.
- The solids are similar.
- The volume of the first solid is twice as much as the volume of the second solid.
- If the dimensions of the second solid are [tex]\( x \)[/tex] by [tex]\( x \)[/tex] by [tex]\( h \)[/tex], the first solid has [tex]\( 4xh \)[/tex] more surface area than the second solid.
