I wanted to talk about radiation levels in Japan
If two shapes at least have the same Shape, then we can say that they are similar. It is important to note that shape is determined by angles. Similar shapes can be created by enlarging or shrinking a figure. If two triangles have the same angles they will have the same shape; they will be similar. To enlarge something you change its dimensions or the lengths of its sides etc., but not its angles.
Now, if I take a shape and enlarge or shrink it (dilate it), how are its perimeter and area affected?
First, what do I mean by area?
The area of a shape is simply the number of square units it takes to cover the object; something like how many sticky notes will it take to cover it. So, let's think for a moment about these squares. We can start with a single square unit, or 1 unit2, and we can count the resulting units as we dilate the figure by uniformly increasing the width and length.
So, when you apply this to actual shapes you can see what happens as you dilate the each shape.
Here are the numbers:
| | Original | Multiply dimensions by 2 | Multiply dimensions by 3 | Multiply dimensions by 4 |
| Rectangle | P = 14 ft. A = 12 ft.2 | P = 14*2 = 28 ft. A = 12*22 = 48 ft.2 | P = 14*3 = 42 ft. A = 12*32 = 108 ft.2 | P = 14*4 = 28 ft. A = 12*42 = ft.2 |
| Circle | C = 2π in. A = π in.2 | C = 2π*2 = 4π in. A = π *22 = 4π in.2 | C = 2π*3 = 6π in. A = π *32 = 9π in.2 | C = 2π*4 = 8π in. A = π *42 = 16π in.2 |
| Triangle | P = 30 m A = 30 m2 | P = 30*2 = 60 m A = 30*22 = 120 m2 | P = 30*3 = 90 m A = 30*32 = 270 m2 | P = 30*4 = 120 m A = 30*42 = 480 m2 |
Albert Einstein claimed to have used this observation about dilation and area to prove the Pythagorean Theorem when he was 12 years old. (It should be noted that the proof he came up with was identical to one previously published, but it’s nice to use Einstein’s celebrity status)
Here is basically how Einstein's proof works:
The Pythagorean Theorem states that given any right triangle, the sum of the squares of the lengths of its legs are equal to the square of the length of the hypotenuse. So, in the diagram below we would want to show that a2 + b2 = c2
Now, we are going to use our model triangle to construct 3 other similar triangles. We can start by drawing a segment from point C perpendicular to the hypotenuse.
Notice that this gives us 2 similar triangles, a medium and small triangle, inside the original (large) triangle.
If two angles of one triangle are congruent to two angles of anther triangle, the remaining angles must be congruent also according to the Third Angle Theorem; thus, we can see the triangles are all similar.
In this situation Einstein made the following observation. Whatever the size of these three triangles, they are all just different size versions of each other. We can think of them all as different dilations of a single triangle whose hypotenuse, the side opposite the right angle, is one unit long.
Now, suppose that we let k represent the area of the triangle with the hypotenuse of 1. Based on our previous results we can observe the following:
Now, the area of the large (original) triangle equals c2k. We also see that the large triangle can be thought of as the small and medium triangle combined. So, a2k + b2k = c2k. Thus, we can see a2 + b2 = c2 by dividing by the common factor k.
