Question

DIG DEEPER A rectangle has vertices 4(0, 0), B(0, 6), C(9, 6), and D(9.0). Explain how to dilate the rectangle to produce

an image whose area is twice the area of the original rectangle. Make a conjecture about how to dilate any polygon to
produce an image whose area is n times the area of the original polygon.

Make a conjecture about how to dilate any polygon to produce an image whose area is n times the area of the original
polygon.

Answer 1

From A to B is 6 units.

From B to C is 9 units.

The area of rectangle ABCD is 6*9 = 54 square units.

We want to double the area. Perhaps we can just double each of the side lengths. So instead of a 6 by 9 rectangle, we'd have a 12 by 18 rectangle.

New area = 18*12 = 216

But 216/54 = 4 i.e. 54*4 = 216 to show the new rectangle is 4 times as big, instead of twice as big. That's not what we want.

Instead of using the scale factor 2, let's use a scale factor of sqrt(2)

The new rectangle is now 6sqrt(2) units by 9sqrt(2) units.

Multiply those dimensions and you'll get 108 which is exactly twice as big compared to 54. This confirms the scale factor of sqrt(2) works if we want to double the area of the rectangle. It turns out it also works for any polygon and not just rectangles. This is because we can break up any polygon into smaller little grid squares.

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Conjecture:

In general, if we want a polygon image with n times the area of the original, then we use the scale factor sqrt(n)

This is because sqrt(n)*sqrt(n) = sqrt(n*n) = sqrt(n^2) = n is the area multiplier to go from the preimage to the image.