$A$ kite in the shape of a square with a diagonal $32\, cm$ and an isosceles triangle of base $8\, cm$ and sides $6\, cm$ each is to be made of three different shades as shown in the figure. How much paper of each shade has been used in it?

(N/A) Area of triangle $I$:
Diagonal $= 32\, cm$.
Since the diagonals of a square bisect each other at $90^\circ$,the height of triangle $I$ (which is half the diagonal) $= \frac{1}{2} \times 32\, cm = 16\, cm$.
The base of the triangle is the other diagonal of the square,which is also $32\, cm$.
Area of triangle $I = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 32\, cm \times 16\, cm = 256\, cm^2$.
Area of triangle $II$:
Since the diagonal of a square divides it into two congruent triangles,the area of triangle $II$ is equal to the area of triangle $I$.
Area of triangle $II = 256\, cm^2$.
Area of triangle $III$:
The triangle at the base has sides $a = 8\, cm, b = 6\, cm, c = 6\, cm$.
Semi-perimeter $s = \frac{a + b + c}{2} = \frac{8 + 6 + 6}{2} = 10\, cm$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{10(10-8)(10-6)(10-6)} = \sqrt{10 \times 2 \times 4 \times 4} = \sqrt{320} = 8\sqrt{5}\, cm^2$.
Taking $\sqrt{5} \approx 2.24$,Area $\approx 8 \times 2.24 = 17.92\, cm^2$.
Thus,the area of paper used for each shade is:
Shade $I = 256\, cm^2$,Shade $II = 256\, cm^2$,Shade $III = 17.92\, cm^2$.