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The Radius of sphere A is twice that of Sphere B. Find, (i) the ratio among their surface areas. (ii) the ratio among their volumes

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Thanks a lot.

Guest Oct 17, 2014

#4
+92217
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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
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#1
+19207
+10

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

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

heureka  Oct 17, 2014
#2
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but how plz explain with steps.

HamzaAbdullah  Oct 17, 2014
#3
+19207
+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
+92217
+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

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