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3 years ago

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The surface areas of two similar solids are 340 yd² and 1,158 yd². The volume of the larger solid is 1,712 yd³. What is the volu

me of the smaller solid?
272 yd³

170 yd³

928 yd³

754 yd³

1 answer:

ratelena [41]3 years ago

4 0

The ratio of surface area is equal to the ratio of the square of the corresponding dimensions. And ratio of volumes of two solids is equal to the cube of the ratio of the corresponding dimensions .

We start with the relation between ratio of surface area and ratio of corresponding sides. That is

$\frac{340}{1158}=(\frac{x}{y} )^2$

Here x and y are the corresponding sides .

$\frac{x}{y} =\sqrt{\frac{340}{1158} } =0.542$

Let the volume of the smaller one be v

$\frac{v}{1712}=0.542^3$

$v=1712*0.542^3=272$

So for the smaller solid, volume is 272 . And the correct option is the first option .

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Step-by-step explanation:

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Let $y=e^{rt}$ be it's solution.

We get,

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Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

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On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

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$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

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