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 August 10th, 2009, 03:12 PM #1 Newbie   Joined: Aug 2009 Posts: 1 Thanks: 0 Geometric ratios How do I answer these questions? I would like the answer to so I can check the work but I need to know the steps please. 1. The volumes of two similar figures are 729 mm cubed and 2744 mm cubed. If the surface area of the larger figure is 393 mm squared what is the surface area of the smaller figure? 2. The surface area of two figures are 16 in squared and 25 inches squared. If the volume of he larger figure is 500 in cubed, what is the volume of the smaller figure? Answer must be in squared.
 August 10th, 2009, 03:44 PM #2 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 1. We can determine the surface area of such figure from using the formula of $(\frac{V_1}{V_2})^{\frac{1}{3}}=(\frac{A_1}{A_2})^ {\frac{1}{2}}$, where letting an such object having volume of $V_1$ and surface area of $A_1$, and the other object having volume of $V_2$ and surface area of $A_2$. Let $V_1=729\,mm^3$, $V_2=2744\,mm^3$ and $A_2=393\,mm^2$, because such an object has both volume of $V_2$ and surface area of $A_2$ and $V_2>V_1$. Solve for $A_1$ by plugging in the known values $\Rightarrow (\frac{729\,mm^3}{2744\,mm^3})^{\frac{1}{3}}=(\fra c{A_1}{393\,mm^2})^{\frac{1}{2}}\Rightarrow (\frac{9}{14})(sqrt{393}\,mm)=\sqrt{A_1}\Rightarro w\therefore A_1\approx162.41\,mm^2$.
 August 11th, 2009, 12:12 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,464 Thanks: 2038 2. Similarly, required volume = 500(16/25)^(3/2) in³, etc. (Note: cubed, not squared, since it's a volume.)

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# work done by the boy in the given figure is 250 J per second. the speed of the boy is 1.2m/s.what is the force applied by the boy is

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