Sinful area

Relevant wiki: Length and Area Problem Solving

Imagine the elephant as a sum of many many very small squares, so that the sum of their area is approximantely the sum of the tiny squares areas. If we magnify and look only at one square when we do the transformation of: Height-> Height , Width-> Width/2 then we see the area is preserved. So if we sum the area of the new squares we get the same area with the initial elephant.

Relevant wiki: Length and Area Problem Solving

the previous area and the new formed area is same.

just look carefully------the previous height and the new formed height is same....[for height]

and..the width of a new formed elephant is $\frac{1}{2}$ of the previous elephant.so, the width of the 2 elephant will be

$\frac{1}{2}+\frac{1}{2}=1$ or same of the previous width......[for width]

as, width and height both are same,the area must be same.

Relevant wiki: Area of a Rectangle

Elephant's height: X

Width: Y

XY = 4 ---------- Imagine the elephant as a rectangle figure.

1 half elephant = X ( $\frac{Y}{2}$ ) = $\frac{XY}{2}$ = $\frac{4}{2}$ = 2

There are two of the elephants, so 2*2 = 4

Yes, it is 4

Draw a rectangle around the elephant so that the elephant fits snugly inside i.e. the rectangle cannot be made any smaller without having it overlap with the elephant.

When we shrink the elephant widthwise by a factor of 2, the width of the rectangle will also get shrunk by a factor of 2. This type of transformation is called a distortion . Distortions preserve ratios of areas, so we can say that

$\dfrac{\text{Area of original elephant}}{\text{Area of rectangle}} = \dfrac{\text{Area of shrunken elephant}}{\text{Area of shrunken rectangle}}.$

The area of the shrunken rectangle will be one-half of the area of the original rectangle, so

$\dfrac{\text{Area of shrunken elephant}}{\text{Area of original elephant}} = \dfrac{1}{2}.$

Thus, the area of the shrunken elephant is exactly half of the original elephant. Since we have two of these shrunken elephants, the total area is equal to the area of the original elephant. We conclude that yes , the area will be exactly 4.

Addendum : This method of inscribing a figure inside of a rectangle can be used to show that the area of an ellipse with semi-major and semi-minor axes $a, b,$ respectively, is equal to $ab\pi.$

Since the width was halfed, on the second diagram e2 and e3 both are half of the first elephant(e1). a1 = h1×w1 (the first diagram ) a =h1 (w2+w3)--- w1 + w2 =[(1/2)w2 (1/2)w3] =(1)w1 & h1= h2 Therefore, a1 = a [h1×w1= h1×w1]= 4

If we half the width...area becomes half... Now two half widthed areas make the original area... (Do not consider the shape of elephant...it is the area that matters)

If we have a first area $a^2$ , than with half width: $a \times \frac{a}{2} = \frac{a^2}{2}$ . So we have half of the area. ${2} \times \frac{a^2}{2} = a^2$ . So, the correct answer is Yes, it is 4 .

Both figures had the same area as the original elephant, but when we halved each or their widths, their halves formed one whole width!

Divide the 2D figure...... Into lots of infinetisimally small squares..... (no. Of squares =n (say.... A very large no.)

Each squares having sides as 'h'(say) . Total area =n (h h)..........(i.e =4)

For the 2nd figure.... Total area of each part=n* (h*(h/2))

Therefore finally total area remains same. :)

The only hesitation I had was whether the tip of the trunk and the tip of the tail overlapped. Otherwise it's clear that this is an algebraic transformation, $\frac{a}{2}*2=1$

It doesn't matter what size or shape the shape is. Once it's area has been shrunken by a factor of two (half), it is $\frac{1}{2}$ of it's original size. However, if you join two of the same shrunken shapes together, it will be the size of the original shape.