In the given figure, ABCD is a square with diagonal 44 cm.

In the given figure, ABCD is a square with diagonal 44 cm.
∴ AB = BC = CD = DA.        ....(1)

In right angled ∆ABC,

$A C^{2}=A B^{2}+B C^{2}$         (Pythagoras Theorem)

$\Rightarrow 44^{2}=2 A B^{2}$

$\Rightarrow 1936=2 A B^{2}$

$\Rightarrow A B^{2}=\frac{1936}{2}$

$\Rightarrow A B^{2}=968$

$\Rightarrow A B=22 \sqrt{2} \mathrm{~cm}$           ...(2)

$\therefore$ Sides of square $=A B=B C=C D=D A=22 \sqrt{2} \mathrm{~cm}$

Area of square ABCD = (side)2

$=(22 \sqrt{2})^{2}$

$=968 \mathrm{~cm}^{2}$                ...(3)

Area of red portion $=\frac{968}{4}=242 \mathrm{~cm}^{2}$

Area of yellow portion $=\frac{968}{2}=484 \mathrm{~cm}^{2}$

Area of green portion $=\frac{968}{4}=242 \mathrm{~cm}^{2}$

Now, in ∆AEF,
The sides of the triangle are of length 20 cm, 20 cm and 14 cm.
∴ Semi-perimeter of the triangle is

$s=\frac{20+20+14}{2}=\frac{54}{2}=27 \mathrm{~cm}$

∴ By Heron's formula,

Area of $\Delta A E F=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{27(27-20)(27-20)(27-14)}$

$=\sqrt{27(7)(7)(13)}$

$=21 \sqrt{39}$

$=131.04 \mathrm{~cm}^{2} \quad \ldots(4)$

Total area of the green portion = 242 + 131.04 = 373.04 cm2

Hence, the paper required of each shade to make a kite is red paper 242 cm2, yellow paper 484 cm2 and green paper 373.04 cm2.