### Solution:

Let us find the area of part 1 first(i) Area of part 1

Let us assume that the triangle $\n \n \Delta ABF\n $ in different orientation as follows

We know that the all sides of square are equal so, let us assume that

$\n \n \u21d2BA=AF=a\n $

We are given that the length of diagonal as $\n \n 32cm\n $ so, we have

$\n \n \u21d2BF=32cm\n $

The Pythagoras theorem is given as $\n \n \n b\n 2\n \n =\n a\n 2\n \n +\n c\n 2\n \n \n $ .

By using the above theorem to triangle $\n \n \Delta ABF\n $ we get,

$\n \n \u21d2B\n A\n 2\n \n +A\n F\n 2\n \n =B\n F\n 2\n \n \n $

By substituting the required values we get

$\n \n \n \n \u21d2\n a\n 2\n \n +\n a\n 2\n \n =\n 32\n 2\n \n \n \n \n \n \n \u21d22\n a\n 2\n \n =1024\n \n \n \n \n \u21d2a=\n \n 512\n \n =16\n 2\n \n cm\n \n \n \n \n $

Let us assume that the area of part 1 as $\n \n \n A\n 1\n \n \n $ .

We know that the formula of area of triangle as

$\n \n \u21d2A=\n 1\n 2\n \n (base)(height)\n $

By using the above formula to $\n \n \Delta ABF\n $ we get,

$\n \n \u21d2\n A\n 1\n \n =\n 1\n 2\n \n (a)(a)=\n \n \n a\n 2\n \n \n 2\n \n \n $

By substituting the required values in above equation we get

$\n \n \n \n \u21d2\n A\n 1\n \n =\n \n \n \n (162\u2013\n 2\n \n )\n 2\n \n \n 2\n \n \n \n \n \n \n \u21d2\n A\n 1\n \n =\n \n 512\n 2\n \n =256c\n m\n 2\n \n \n \n \n \n \n $

Therefore the area of part 1 is $\n \n 256c\n m\n 2\n \n \n $ Now, let us solve the second part.

(i) Area of part 2

Let us assume that the area of part 2 as $\n \n \n A\n 2\n \n \n $

We know that the diagonal of a square divides the square into two triangles of same area that is

$\n \n \u21d2\n A\n 2\n \n =\n A\n 1\n \n =256c\n m\n 2\n \n \n $

Therefore the area of part 2 is $\n \n 256c\n m\n 2\n \n \n $ .

(iii) Area of part 3

Let us take the triangle $\n \n \Delta CDE\n $ and draw the height as shown below.

We are given that the dimensions of each triangle are $\n \n 8cm,6cm,6cm\n $ .

So, from the triangle $\n \n \Delta CDE\n $ we have,

$\n \n \n \n \u21d2CE=CD=6cm\n \n \n \n \n \u21d2DE=8cm\n \n \n \n \n $

We know that the altitude of isosceles triangle will be the median

So, the point ‘H’ is mid – point of ‘DE’

So, we can the length of ‘HE’ as

$\n \n \n \n \u21d2HE=\n \n DE\n 2\n \n \n \n \n \n \n \u21d2HE=\n 8\n 2\n \n cm=4cm\n \n \n \n \n $

The Pythagoras theorem is given as, $\n \n \n b\n 2\n \n =\n a\n 2\n \n +\n c\n 2\n \n \n $ .

By using the above theorem to triangle $\n \n \Delta CHE\n $ we get,

$\n \n \u21d2C\n H\n 2\n \n +H\n E\n 2\n \n =C\n E\n 2\n \n \n $

By substituting the required values we get

$\n \n \n \n \u21d2C\n H\n 2\n \n +\n 4\n 2\n \n =\n 8\n 2\n \n \n \n \n \n \n \u21d2C\n H\n 2\n \n =64\u221216\n \n \n \n \n \u21d2CH=\n \n 48\n \n =4\n 3\n \n cm\n \n \n \n \n $

We know that the formula of area of triangle as

$\n \n \u21d2A=\n 1\n 2\n \n (base)(height)\n $

By using the above formula to $\n \n \Delta CDE\n $ as $\n \n \n A\n 3\n \n \n $ ,

$\n \n \u21d2\n A\n 3\n \n =\n 1\n 2\n \n (DE)(CH)\n $

By substituting the required values in above equation we get

$\n \n \n \n \u21d2\n A\n 3\n \n =\n 1\n 2\n \n \n 8\n \n \n 4\n 3\n \n \n \n \n \n \n \n \u21d2\n A\n 3\n \n =16\n 3\n \n c\n m\n 2\n \n \n \n \n \n \n $

Therefore the area of third part is $\n \n 16\n 3\n \n c\n m\n 2\n \n \n $ .