#4**+10 **

I think that this is all that Heureka did. It is all that I would normally do.

if the ratio if a 1 dimensional measurement is $$a:b$$ (the radius is a 1 dimensional measurement)

then the ration of the 2 dimensional measurements (surface area) is just $$a^2:b^2$$

and the ratio of the 3 dimensional measurement (volume) is just $$a^3:b^3$$

so if ratio of radius is 1:2

then ratio of SA is 1:4

and ratio of Volume is 1:8

Melody
Oct 17, 2014

#1**+10 **

**(i) ratio among their surface areas is 4:1.**

** (ii) the ratio among their volumes is 8:1.**

heureka
Oct 17, 2014

#3**+10 **

**(i) ratio among their surface areas is 4:1.**

**$$r_a=2r_b$$**

**$$\begin{array}{rcl} Surface_A &=& 4\pi(2r_B)^2 \\ Surface_B &=& 4\pi(r_B)^2 \\ \hline \\ \frac{Surface_A}{Surface_B} &=& \frac{4\pi(2r_B)^2}{4\pi(r_B)^2 } \\\\ \frac{Surface_A}{Surface_B} &=& \frac{4\pi (2)^2 (r_B)^2}{4\pi(r_B)^2 }=2^2 = 4= \frac{4}{1} \\\\ \end{array}$$**

** (ii) the ratio among their volumes is 8:1.**

**$$\begin{array}{rcl} V_A &=& \frac{4}{3}\pi(2r_B)^3 \\\\ V_B &=& \frac{4}{3}\pi(r_B)^3 \\\\ \hline \\ \frac{V_A}{V_B} &=& \frac{\frac{4}{3}\pi(2r_B)^3}{\frac{4}{3}\pi(r_B)^3 } \\\\ \frac{V_A}{V_B} &=& \frac{\frac{4}{3}\pi(2)^3(r_B)^3}{\frac{4}{3}\pi(r_B)^3 } = 2^3 = 8 = \frac{8}{1} \\\\ \end{array}$$**

heureka
Oct 17, 2014

#4**+10 **

Best Answer

I think that this is all that Heureka did. It is all that I would normally do.

if the ratio if a 1 dimensional measurement is $$a:b$$ (the radius is a 1 dimensional measurement)

then the ration of the 2 dimensional measurements (surface area) is just $$a^2:b^2$$

and the ratio of the 3 dimensional measurement (volume) is just $$a^3:b^3$$

so if ratio of radius is 1:2

then ratio of SA is 1:4

and ratio of Volume is 1:8

Melody
Oct 17, 2014