the surface areas of two similar figures are 36 in2 and 49 in2 . if the volume of the smaller figure is 648 in3, what is the volume of the larger figure?

Respuesta :

Area is a squared measure of the one-to-one measure of the figures. If the surface areas of the figures, from smaller to larger, are 36 and 49, by taking the square roots of those numbers, you get the perimeter measure of the figures, which is the one-to-one measure. The square root of 36 is 6, and the square root of 49 is 7. Now, in order to find the ratio of the volumes smaller to larger, we take that one-to-one measure and cube it, since volume is a cubed measure. We will set up the proportions to find the cubed measures first. My proportion will be set up with the smaller figure's info on top and the larger figure's info on bottom: [tex] \frac{6^3}{7^3}=\frac{216}{343} [/tex]. Those are the cubed ratios.  Now we will set up our equation to solve for the volume of the larger: [tex] \frac{216}{343}=\frac{648}{x} [/tex]. Now we can cross-multiply to solve for x, the volume of the larger figure. 216x = 222264. Dividing both sides by 216 gives us a volume for the larger figure of 1029 inches cubed. There you go!

Answer:

The volume of the larger figure will be 1029 cubic inches.

Step-by-step explanation:

Similar shapes have corresponding sides that are proportional.

Given is :

The surface areas of two similar figures are 36 square inches and 49 square inches. As area is a squared root. We will find the square roots.

The square roots will be :

[tex]\sqrt{36} =6[/tex]

[tex]\sqrt{49} =7[/tex]

The volume of the smaller figure is 648 cubic inches.

Lets say the volume of larger figure is L

The volume is a cubed unit.

So, equation forms:

[tex]\frac{6^{3} }{7^{3}}=\frac{648}{L}[/tex]

=> [tex]L=648\frac{7^{3} }{6^{3} }[/tex] =1029 cubic inches.

Therefore, the volume of the larger figure will be 1029 cubic inches.

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