# IMAT 2014 Q57 [Surface Area]

\frac{r}{2}, r, 2r, and 3r.

What is the sum of their surface areas?

A. 58 π r^2
B. 169 π r^2
C. 57 π r^2
D. 25 π r^2
E. 26 π r^2

The surface of a sphere is calculated with the formula: 4\pi r^2

1. sphere: P_1 = 4\pi (\frac{r}{2})^2 = 4 \pi (\frac{r^2}{4})=\pi r^2

2. sphere: P_2 = 4\pi r^2

3. sphere: P_3 = 4\pi (2r)^2 =4\pi 4 r^2 = 16\pi r^2

4. sphere: P_4 = 4\pi (3r)^2 =4\pi 9 r^2 = 36\pi r^2

Adding all the surfaces together, we will get the desired sum:

P = P_1 + P_2 + P_3 + P_4= \pi r^2 + 4\pi r^2 + 16\pi r^2 + 36\pi r^2 =
= \pi r^2 ( 1+4+16+36) = \pi r^2 57 \to P=57\pi r^2.

Which is the correct answer. (C)