Math Problem Statement

Two solids are mathematically similar and have volumes 81 cm^3 and 24 cm^3. The surface area of the smaller solid is 44 cm^2. Calculate the surface area of the larger solid.

Solution

To solve this problem, we can use the properties of similar solids. When two solids are similar, the ratio of their volumes is the cube of the ratio of their corresponding linear dimensions, and the ratio of their surface areas is the square of the ratio of their corresponding linear dimensions.

Given:

  • Volume of the larger solid: V1=81cm3V_1 = 81 \, \text{cm}^3
  • Volume of the smaller solid: V2=24cm3V_2 = 24 \, \text{cm}^3
  • Surface area of the smaller solid: A2=44cm2A_2 = 44 \, \text{cm}^2

Step 1: Calculate the ratio of the volumes.

The ratio of the volumes of the two solids is: Volume ratio=V1V2=8124=278=(32)3\text{Volume ratio} = \frac{V_1}{V_2} = \frac{81}{24} = \frac{27}{8} = \left(\frac{3}{2}\right)^3

Step 2: Determine the ratio of the linear dimensions.

The linear dimension ratio is the cube root of the volume ratio: Linear dimension ratio=L1L2=2783=32\text{Linear dimension ratio} = \frac{L_1}{L_2} = \sqrt[3]{\frac{27}{8}} = \frac{3}{2}

Step 3: Calculate the surface area ratio.

The surface area ratio is the square of the linear dimension ratio: Surface area ratio=(L1L2)2=(32)2=94\text{Surface area ratio} = \left(\frac{L_1}{L_2}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}

Step 4: Calculate the surface area of the larger solid.

Let A1A_1 be the surface area of the larger solid. Using the surface area ratio: A1A2=94\frac{A_1}{A_2} = \frac{9}{4} A1=A2×94=44×94=44×2.25=99cm2A_1 = A_2 \times \frac{9}{4} = 44 \times \frac{9}{4} = 44 \times 2.25 = 99 \, \text{cm}^2

Thus, the surface area of the larger solid is 99 cm².

Would you like more details or have any questions?

Here are 5 related questions:

  1. If the volume of the smaller solid was 64 cm³, how would you calculate the surface area of the larger solid?
  2. What is the relationship between the surface area and volume ratios of similar solids?
  3. How would the surface area of the larger solid change if the linear dimension ratio was 43\frac{4}{3}?
  4. How does the concept of similarity apply to real-world problems involving geometric shapes?
  5. Can the surface area ratio ever be greater than the volume ratio? Why or why not?

Tip: When dealing with similar solids, always remember that the volume ratio is the cube of the linear ratio, while the surface area ratio is the square.

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Math Problem Analysis

Mathematical Concepts

Similarity of Solids
Volume and Surface Area Ratios

Formulas

Volume ratio formula
Surface area ratio formula for similar solids

Theorems

Properties of similar solids

Suitable Grade Level

High School