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Irregular Hexagons $A$ and $B$ are geometrically similar. The shortest sides are $4$ inches and $3$ inches, respectively. If the area of hexagon $A$ is $48in^2$, what is the area of hexagon $B$?

I know the answer is $27 in^2$, but how do you get that?

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closed as off-topic by Saad, Leucippus, Xander Henderson, Chris Custer, Ahmad Bazzi Oct 1 '18 at 5:44

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Because if corresponding linear dimensions of $A$ and $B$ are in $4:3$ ratio, then their areas are in $4^2:3^2$ ratio.

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The areas of similar figures vary in proportion to the square of the proportion by which the lengths are stretched. For example, a square with one and one half times the side length has area $(3/2)^2$ as large.

Can you answer your question now?

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