Given A1 = 12 cm^2 and A2 = 108 cm^2.

Welcome back Samson — missed you this week. Here's the solution to question 8. Step 1: Determine the ratio of the surface areas. Let $A_1$ be the surface area of the smaller bottle and $A_2$ be the surface area of the larger bottle. Given $A_1 = 12 \text{ cm}^2$ and $A_2 = 108 \text{ cm}^2$. The ratio of the surface areas is: $$ \frac{A_1}{A_2} = \frac{12 \text{ cm}^2}{108 \text{ cm}^2} = \frac{1}{9} $$ Step 2: Find the ratio of the corresponding linear dimensions (scale factor). For similar objects, the ratio of their surface areas is equal to the square of the ratio of their corresponding linear dimensions (scale factor, $k$). $$ \frac{A_1}{A_2} = k^2 $$ $$ \frac{1}{9} = k^2 $$ To find $k$, take the square root of both sides: $$ k = \sqrt{\frac{1}{9}} = \frac{1}{3} $$ So, the ratio of the linear dimensions of the smaller bottle to the larger bottle is $\frac{1}{3}$. Step 3: Calculate the volume of the smaller bottle. For similar objects, the ratio of their volumes is equal to the cube of the ratio of their corresponding linear dimensions. Let $V_1$ be the volume of the smaller bottle and $V_2$ be the volume of the larger bottle. Given $V_2 = 810 \text{ cm}^3$. $$ \frac{V_1}{V_2} = k^3 $$ $$ \frac{V_1}{810 \text{ cm}^3} = \left(\frac{1}{3}\right)^3 $$ $$ \frac{V_1}{810 \text{ cm}^3} = \frac{1}{27} $$ Now, solve for $V_1$: $$ V_1 = \frac{1}{27} \times 810 \text{ cm}^3 $$ $$ V_1 = \frac{810}{27} \text{ cm}^3 $$ $$ V_1 = 30 \text{ cm}^3 $$ The volume of the smaller bottle is $\boxed{\text{30 cm}^3}$. Send me the next one 📸