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If there was a lake that occupied $8$ squares, and the width/height of a square was $1km$, it would have an area of $8km^2$. What if the scale was $0.5km$ - the width/height of a square is only $0.5km$, would that mean the area of the lake will be $4km^2$ or $2km^2$?

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If you multiply linear dimensions by $k$, you multiply area by $k^2$. Here, if I understand the question correctly, you are multiplying linear dimensions by $0.5$. So area is multiplied by $(0.5)^2$, which is $0.25$. The resulting area is $2$.

In this case, we can argue more simply. Each of the $8$ little squares has area $(0.5)^2$, so the total area is $8(0.5)^2$.

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Hint: You double all lengths and you quadruple the area.

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Or half all the lengths and quarter the area. – mixedmath Aug 21 '12 at 3:23

A more complicated argument just to complement...

If you take any shape in $\mathbb{R}^n$ that has area, and you scale the coordinates by $\lambda_1,...,\lambda_n$ respectively, then the area will scale by $\lambda_1 \cdots \lambda_n$. This follows from the change of variables theorem since the Jacobian of $\phi(x) = (\lambda_1 x_1,...,\lambda_n x_n)$ is $J_\phi (x) = \lambda_1 \cdots \lambda_n$.

In your example the shape is in $\mathbb{R}^2$, and you have $\lambda_1 = \lambda_2 = \frac{1}{2}$. Hence the new area is $(\frac{1}{2})^2$ times the original area, ie, $2 \, \mathrm{km}^2$.

This works for any shape, of course.

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