>>220037Consider your shape to be occupying x of the area of a square that completely encloses it, and having a perimeter y times the length of the perimeter of the square. If you stretch a square until it's twice the size, its area will be four times that of the original square, and the perimeter will be twice that of the original square. So then if you stretch your random shape the same way, it will obviously be:
- similar to the original, because you stretched it evenly on each axis
- still enclosed by the square, because it was enclosed before you stretched it, and you stretched it the same amount as the square
- have a perimeter that's x as much as the new square, because it's still similar and enclosed
- have an area that's y as much as the new square, because it's still similar and enclosed
So P(shape)=xP(square)
A(shape)=yA(square)=y( (P(square)/4)^2)
Where x and y depend on the shape.
Now let's remove the square. Because the perimeter of the shape is x times the perimeter of the square, we can replace any mention of the square with the shape's perimeter instead.
A(shape)=yA(square)
=y( (xP(square)/4x)^2)
=y/16x( (P(Shape))^2)
A(shape)=nP(Shape)^2
n=y/16x, but we don't care about the square, so we don't care about x or y. All we need to care about is that n is a constant, and it changes based on the shape of the shape, and not the size of the shape, so all similar shapes will have the same n.
Now we've worked out that A=nP^2, we can do our problem:
Shape 1: P=20, A=19.6
Triangle rule gives n=A/P^2
n=19.6/400=0.049
Shape2: P=34
A=nP^2=0.049*34*34
=56.644
