To solve this problem, we need to use the properties of similar objects. When two objects are similar, the ratios of their corresponding linear dimensions are the same, and there are specific relationships between their volumes, surface areas, and masses. Here is the step-by-step solution:
1. Understand the relationship between masses:
Given:
- Mass of the larger block (M_large) = 275.6 g
- Mass of the smaller block (M_small) = 81.6 g
The ratio of their masses, which is also the ratio of their volumes since density is assumed to be constant, is calculated as follows:
[tex]\[
\text{Ratio of masses} = \frac{\text{M_large}}{\text{M_small}} = \frac{275.6}{81.6} \approx 3.377
\][/tex]
2. Determine the relationship between surface areas:
For similar objects, the surface area ratio (A_large / A_small) is the square of the ratio of their linear dimensions. Since the volumes ratio is the cube of the linear dimensions ratio, we use the following relationship for surface areas:
[tex]\[
\left(\frac{\text{A_large}}{\text{A_small}}\right) = \left(\frac{\text{ratio of volumes}}{}\right)^{\frac{2}{3}}
\][/tex]
Plugging in the ratio of the volumes calculated:
[tex]\[
\left(\frac{\text{A_large}}{\text{A_small}}\right) = (3.377)^{\frac{2}{3}}
\][/tex]
3. Calculate the surface area of the larger block:
Given:
- Surface area of the smaller block (A_small) = 340 cm²
We need to find the surface area of the larger block (A_large):
[tex]\[
\left(\frac{\text{A_large}}{\text{A_small}}\right) = 3.377^{\frac{2}{3}} \approx 2.251
\][/tex]
Thus:
[tex]\[
\text{A_large} = \text{A_small} \times 2.251 = 340 \times 2.251 \approx 765.37 \, \text{cm}²
\][/tex]
Therefore, the surface area of the larger block is approximately 765.37 cm².
