Question

If the side of a cube is tripled, what is the ratio of the new total surface area to the original, and the ratio of the new volume to the original volume?

• A. 9 : 1 and 27 : 1
• B. 6 : 1 and 9 : 1
• C. 3 : 1 and 9 : 1
• D. 9 : 1 and 3 : 1

Answer 1

Answer:

A. 9 : 1 and 27 : 1

To find the ratios, let's assume the original side of the cube is $s$.

• Original side = $s$

• New side (tripled) = $3s$

1. Ratio of the New Total Surface Area to the Original

• Original Total Surface Area ($TSA_1$) = $6s^2$

• New Total Surface Area ($TSA_2$) = $6(3s)^2 = 6(9s^2) = 54s^2$

$\text{Ratio} = \frac{TSA_2}{TSA_1} = \frac{54s^2}{6s^2} = \frac{9}{1}$

• Answer (Surface Area): $9:1$

2. Ratio of the New Volume to the Original Volume

• Original Volume ($V_1$) = $s^3$

• New Volume ($V_2$) = $(3s)^3 = 27s^3$

$\text{Ratio} = \frac{V_2}{V_1} = \frac{27s^3}{s^3} = \frac{27}{1}$

• Answer (Volume): $27:1$